\documentclass[handout]{beamer} \usepackage{beamerthemeshadow,psfrag,pst-all,eurosym,comment,soul,fancybox} \usepackage[english]{babel} % or whatever \usepackage[latin1]{inputenc} \usepackage{wasysym} % or whatever \usepackage{times,bm,ulem} \usepackage[T1]{fontenc} %\usepackage[usenames,dvipsnames]{xcolor} \newcommand{\announce}{\setbeamercolor{structure}{fg=green!60!black}} \newcommand{\sol}{\setbeamercolor{structure}{fg=purple!80!black}} \newcommand{\normal}{\setbeamercolor{structure}{fg=blue!60!green}} \newcommand{\seminar}{\setbeamercolor{structure}{fg=orange!60!red}} \newcommand{\vsp}{\vspace{0.3cm}} \newcommand{\vspl}{\vspace{0.15cm}} \newcommand{\xvec}{\boldsymbol{x}} \newcommand{\yvec}{\vec{y}} \newcommand{\Xvec}{\boldsymbol{X}} \newcommand{\Yvec}{\vec{Y}} \newcommand{\Evec}{\vec{E}} \newcommand{\tvec}{\bodsymbol{t}} \newcommand{\Tvec}{\bodsymbol{T}} \newcommand{\mutau}{\begin{pmatrix}\mu\\\tau\end{pmatrix}} \newcommand{\thetavec}{\vec{\theta}} \newcommand{\thetavechat}{\vec{\hat\theta}} \newcommand{\thetahat}{\hat\theta} \newcommand{\xvece}{\boldsymbol{x}=(x_1,x_2,\ldots,x_n)^T} \newcommand{\Xvece}{\boldsymbol{X}=(X_1,X_2,\ldots,X_n)^T} \newcommand{\thetavece}{\vec{\theta}=(\theta_1,\theta_2,\ldots,\theta_p)^T} \newcommand{\exxvec}{$x_1,x_2,\ldots,x_n$\xspace} \newcommand{\exxvecdot}{$x_1,x_2,\ldots,x_n$.\ } \newcommand{\exXvec}{$X_1,X_2,\ldots,X_n$\ } \newcommand{\comb}[2]{\,^{#1}\text{C}_{#2}} %\renewcommand{\jump}[1]{\textcolor{purple}{\boldsymbol{#1}}} \normal \title{ACC1012 / 1053:\\ Mathematics \& Statistics} \author{Dr. James Waldron, Dr. Kevin Wilson} %\institute{\textit{Newcastle University, Newcastle upon Tyne, U.K.}} \vspace{0.15cm} %\textcolor{blue}{\texttt{lee.fawcett@ncl.ac.uk}, \texttt{david.walshaw@ncl.ac.uk}}} \date{\textcolor{purple}{\textbf{Semester 2, 2017---2018}}} \usetheme[height=8mm]{Rochester} \setbeamercovered{transparent=9} \begin{document} \beamertemplatenavigationsymbolsempty \frame{\titlepage} %\begin{comment} \announce \frame{ \frametitle{Contact details} \begin{center} \begin{tabular}{l l} \textbf{Semester 1:} & Dr. James Waldron \\ & \\ \textbf{Office:} & Room 2.14 Herschel Building \\ & \\ \textbf{Phone:} & 0191 208 8019 \\ & \\ \textbf{Email:} & \texttt{james.waldron@ncl.ac.uk} \\ \end{tabular} \end{center} } \frame{ \frametitle{Contact details} \begin{center} \begin{tabular}{l l} \textbf{Semester 2:} & Dr. Kevin Wilson \\ & \\ \textbf{Office:} & Room 2.19 Herschel Building \\ & \\ \textbf{Phone:} & 0191 208 7311 \\ & \\ \textbf{Email:} & \texttt{kevin.wilson@ncl.ac.uk} \\ \end{tabular} \end{center} } \begin{frame} \frametitle{Timetable and Administrative arrangements} \begin{itemize} \item \textbf{Lectures} are on Fridays -- 9:30--10:30 in NUBS.1.03, Newcastle Business School \vspl %\textcolor{red}{Check your timetable every week!} \pause \pause \vspace{0.2cm} \item Each student should also attend one \textbf{workshop} \vspl \pause \begin{itemize} \item [--] Attendance is monitored in \underline{every} session! \pause \item [--] There are 8 workshops -- you have been allocated to \underline{one}: \pause \vspl {\scriptsize \begin{center} \begin{tabular}{|c|c c|l|} \hline Group &Day &Time &Place \\ \hline A &Mon &12--1 &Barbara Strang Teaching Centre 2.51 \\ B &Mon &1--2 &Barbara Strang Teaching Centre 2.51\\ C &Mon &4--5 & Barbara Strang Teaching Centre 2.41B\\ D &Mon &5--6 & Herschel Building TR1 (4th floor) \\ E &Tues &9--10 &Barbara Strang Teaching Centre 3.31\\ F &Tues &12--1 &Barbara Strang Teaching Centre 2.51 \\ G &Tues &2--3 &Barbara Strang Teaching Centre 2.51 \\ H &Thurs &4--5 & Barbara Strang Teaching Centre 3.31\\ \hline \end{tabular} \end{center} } \pause \vspl \item [--] These workshops are \underline{vital} for putting into practice what you've been taught in the lecture! \end{itemize} \end{itemize} \end{frame} \frame{ \frametitle{Timetable and Administrative arrangements} \begin{itemize} \item\textbf{Office hours:} Just pop by our office if you need to see us and if we're not busy, we'll be happy to help! \vsp \item \textbf{Revision classes} will be scheduled before the exam next summer \end{itemize} } \begin{frame} \frametitle{Assessment} Assessment is by: \vsp \begin{itemize} \item Professional Skills/Taxation component (\textbf{30\%}) \pause \vsp \item Mathematics \& Statistics component (\textbf{70\%}) \pause \vspl \begin{itemize} \item [--]Written exam at the end of the academic year \textcolor{red}{\bf (50\%)} \pause \vspl \item[--] Coursework: Eight small assignments across the year \textcolor{red}{\bf (20\%)}, consisting of computer work/written case study work \end{itemize} \end{itemize} \end{frame} \frame{ \frametitle{Coursework: Computer--based assessments (CBAs)} \begin{itemize} \item A bit like online tests, but not done under test conditions \pause \vspl \item Uses our \textit{Numbas} system, via Blackboard -- will demonstrate in a minute \pause \vspl \item You should all be familiar with \textit{Numbas} and Blackboard \pause \vspl \item There is a \textcolor{purple}{\bf practice} version for each CBA, where you can log-in and try the questions as many times as you like (the questions/numbers within questions will change slightly each time)\pause \vspl \item An \textcolor{purple}{\bf assessed} version of each CBA will be available for {\bf one week only}, during which the practice versions are unavailable -- you must log-in and complete this at some point before the deadline! \pause \vspl \item You will be reminded about upcoming deadlines in class \end{itemize} } \frame{ \frametitle{Coursework: Case studies} \begin{itemize} \item Require more thought than the CBA questions \pause \vspl \item Real--life accounting/finance scenarios and problems \pause \vspl \item More open--ended questions \pause \vspl \item Designed to make you \textcolor{purple}{\bf understand} the material, not just ``turn the handle''\pause \vspl \item You will work in small groups on these in the weekly workshops, under the supervision of me/assistants \pause \vspl \item You will hand in an individual report (about 5/6 sides of work including calculations) \pause \vspl \item Important that you attend all your workshops! \end{itemize} } \frame{ \frametitle{Deadlines: Semester 1} \begin{center} \begin{tabular}{|c|l|} \hline \textbf{Assessment} &\textbf{Deadline} \\ \hline\hline CBA &Practice mode opens Thurs 5th October 2017 \\ &Assessed mode opens Thurs 12th October 2017 \\ &\textcolor{red}{Deadline: 23:59 Wednesday 18th October} \\ \hline \pause Case Study &Start work in tutorials w/b Monday 9th October \\ &\textcolor{red}{Deadline: 4pm, Friday 3rd November} \\ \hline \pause Case Study &Start work in tutorials w/b Monday 20th November \\ &\textcolor{red}{Deadline: 4pm, Friday 12th January} \\ \hline \pause CBA &Practice mode opens Thurs 30th November 2017 \\ &Assessed mode opens Thurs 7th December 2017 \\ &\textcolor{red}{Deadline: 23:59 Wednesday 13th December} \\ \hline \end{tabular} \end{center} \textit{Don't worry -- you will be reminded about these in class!} } \frame{ \frametitle{Pre--requisites} \begin{itemize} \item A pass in GCSE Maths (grade C+) \pause \vspl \item Completion of the \textcolor{purple}{\bf summer revision booklet} -- if you've not looked at this by now, make sure you do before next week! \end{itemize} } \frame{ \frametitle{Other stuff} \begin{itemize} \item Notes (with gaps) will be handed out in lectures -- you should fill in the gaps during lectures \pause \vspace{0.15cm} \item A summarised version of the notes will be used in lectures as slides \pause \begin{itemize} \item [--]\color{blue!60!green}{\textbf{Listen and learn}} \item [--]\color{purple!80!black}{\textbf{Write down}} \item [--]\color{green!60!black}{\textbf{Announcements}} \end{itemize} \vspace{0.15cm} \pause \item These notes and slides will be posted on the \textbf{course website} after each topic is finished, along with any other course material -- such as problems sheets, supplementary handouts etc. \end{itemize} } \frame{ \frametitle{Other stuff} \begin{itemize} \item The course website can be found via the Blackboard page for ACC1012/ACC1053, or directly via: \begin{center} \large \textbf{\texttt{\textcolor{red}{www.mas.ncl.ac.uk/$\sim$nlf8/teaching/acc1012}}} \\ \end{center} \normalsize \vspace{0.2cm} \pause \item There will be practice questions in the lecture notes -- I will expect you to look at these before the workshops so you're ready for discussion...\pause \vspace{0.15cm} \item Please check your University email account regularly, as course announcements will often be made via email \vspace{0.15cm} \pause \item Course textbook: not compulsory that you buy it, but it could help! \end{itemize} } \frame{ \frametitle{Why Maths \& Stats?} \begin{itemize} \item Maths \& Stats has a huge role to play in the modern world of accounting and finance \pause \vspl \item Case in point: the recent global economic situation \pause \vspl \begin{itemize} \item [--] Mathematical/Statistical indicators are used to quantify the state of a country's economy and identify whether or not that economy is officially in recession \pause \vspl \item [--] Mathematical/statistical models are used to forecast periods of recession (and double--dip recession!) \end{itemize} \end{itemize} } \frame{ \frametitle{Why Maths \& Stats?} \begin{itemize} \item Mathematical tools (e.g. \textcolor{red}{Chaos Theory}) are used to map market trends \pause \vsp \item At a more basic level: \pause \vspl \begin{itemize} \item [--] Statistics and probability are used in everyday business and economics \pause \vspl \item [--] Employers love graduates with solid numeracy skills, as well as problem--solving skills and the ability to think logically \end{itemize} \end{itemize} } \frame{ \frametitle{Why Maths \& Stats?} \begin{itemize} \item In order to model real--life situations realistically, the maths involved can get quite sophisticated... \pause \vspl \item ...so we need to learn to walk before we can run:\pause \vspl \begin{itemize} \item [--]We will start with some very basic maths you would have seen before in your GCSE course, and build up quite gently... \pause \vspl \item [--]... it might take you a while to see the relevance of some of the stuff we do, but stick with it! \end{itemize} \end{itemize} } \normal \frame{ \frametitle{ACC1012 / ACC1053} \LARGE \begin{center} \textcolor{purple}{\bf \textit{Any questions...?}} \end{center} \normalsize } \end{document} \begin{comment} \seminar \frame{ \vspace{1cm} \huge \textbf{ACC1012/ACC1053} \vspace{1cm} \Large \textbf{Workshop 1} } \frame{ \frametitle{Workshop 1} \begin{itemize} \item Welcome to the first workshop! \pause \vspl \item These workshops will take place at this time every week -- you must attend the slot you have been allocated to! %\pause \vspl \item I will also see you in the lecture on Friday (\textcolor{blue}{\bf 1pm}, \textcolor{blue}{\bf Lecture Theatre 2}, on the \textcolor{blue}{\bf first floor} of \textcolor{blue}{\bf Claremont Tower}) \pause \vspl \item What you'll need: Calculator (see webpage for a list of permitted calculators), lecture notes, pens, ruler, paper,... \pause \vspl \item Usually, I will also ask you to work through some questions \textit{before} coming to seminars \end{itemize} } \frame{ \frametitle{How will the workshops `work'?} \begin{itemize} \item Not all taken by me \pause \vspl \item Attendance register \pause \vsp \item \textbf{First half:} \pause \vspl \begin{itemize} \item [--] Standard questions that were set the previous week \pause \vspl \item [--] More ``standard'' questions \pause \vspl \item [--] These questions aren't assessed -- model solutions will be made available during workshop/online at the end of the week \pause \vspl \item [--] Staff supervision \end{itemize} \end{itemize} } \frame{ \frametitle{How will the workshops `work'?} \begin{itemize} \item \textbf{Second half:} \pause \vspl \begin{itemize} \item [--] Work on case studies in small groups \pause \vspl \item [--] Group discussion \pause \vspl \item [--] Close staff supervision \pause \vspl \item [--] This work \textit{is} assessed -- through individual write--up and questions in the last workshop \end{itemize} \end{itemize} } \frame{ \frametitle{Case studies: More information} \begin{itemize} \item Groups of between 4 and 6: \pause allocate randomly/work with friends? \pause \vspl \item Will start the first one next week \pause \vspl \item Group discussion in the second half of each workshop until the work is due in \vspl \item Idea: to get you to \textcolor{blue}{\bf problem--solve}, interpret \textcolor{blue}{\bf real--life} scenarios, and to promote \textcolor{blue}{\bf deep learning} (as opposed to surface--learning) \pause \vspl \item Idea: to learn from each other! \pause \vspl \item More open--ended questions related to the scenario, rather than prescribed questions telling you exactly what to do \pause \vspl \item You might hate this at first, but give it a chance! It'll make you a much better person! \end{itemize} } \frame{ \frametitle{Case studies: Assessment} \begin{itemize} \item Individual written report after about 4/5 weeks' work in workshops \pause \vspl \item 2/3 pages, including write--up and maths \pause \vspl \item First one due in \textcolor{blue}{\bf 4pm, Friday 1st November} \pause \vspl \item In the last workshop before the written work is due in, you will each be asked a question by me/seminar assistant -- you will get a mark out of 3 for how well you answer this, and this will go towards your final mark for the case study \pause \vspl \item Each one (like \textit{all} of your homeworks for me) counts as 2.5\% of the module as a whole \pause \vspl \item You will be given an example case study, including write--up, to get you thinking about exactly what we're after \pause \vspl \item Feedback within two weeks \end{itemize} } \frame{ \frametitle{Other stuff} \begin{itemize} \item At some point ASAP, please make sure you can log into our \texttt{Moodle} system which hosts the Numbas CBAs (demo) \pause \vspl \item The first CBA will go live in \textcolor{blue}{\bf practice mode} next week... \pause \vspl \item ...then assessed mode the week after \pause \vspl \item And don't forget our first \textcolor{blue}{\bf lecture} on Friday at \textcolor{blue}{\bf 1pm}, \textcolor{blue}{\bf Lecture Theatre 2}, on the \textcolor{blue}{\bf first floor} of \textcolor{blue}{\bf Claremont Tower} \end{itemize} } \normal \frame{ \vspace{1cm} \huge \textbf{Chapter 1} \vspace{1cm} \Large \textbf{Linear and quadratic functions} } \frame{ \frametitle{What we'll cover...} \begin{itemize} \item What \textcolor{purple}{\bf are} linear/quadratic functions? \pause \vspl \item The role of the \textcolor{purple}{\bf gradient} and \textcolor{purple}{\bf intercept} \pause \vspl \item \textcolor{purple}{\bf Graphs} of linear/quadratic functions \pause \vspl \item Finding the \textcolor{purple}{\bf equation} of a line \pause \vspl \item \textcolor{purple}{\bf Solving} linear equations \pause \vspl \item \textcolor{purple}{\bf Simultaneous} linear equations \pause \vspl \item \textcolor{purple}{\bf Linear programming} \pause \vspl \item \textcolor{purple}{\bf Solving} quadratics using various methods \pause \vspl \item Chapters 1 and 2 from the course text would be useful here \end{itemize} } \frame{ \frametitle{1.1 Linear functions} \textcolor{purple}{\bf Linear functions} occur often in accounting and finance. \pause \vsp You should already be familiar with linear functions from your GCSE maths courses. \pause For example, \[ y=3x+4 \] is a linear function of $x$ which, when operated, gives us $y$.\pause \vsp We know this is a linear function because when we produce a graph of this function, we get a straight line. \pause \vsp For example, when $x=2$ we get \pause \[ y=3\times 2+4 \pause = 6+4 \pause = 10, \] bearing in mind the rules of ``\textsc{bidmas}'': \underline{\bf{B}}rackets, \underline{\bf{I}}ndices, \underline{\bf{D}}ivision, \underline{\bf{M}}ultiplication, \underline{\bf{A}}ddition, \underline{\bf{S}}ubtraction. } \frame{ \frametitle{1.1 Linear functions} Similarly, for other values of $x$: \pause \begin{center} \begin{tabular}{|c|c c c c c c c c|} \hline $x$ &0 &1 &\textbf{2} &3 &4 &5 &6 &7 \\ \hline $y$ &4 &7 &\textbf{10} &13 &16 &19 &22 &25 \\ \hline \end{tabular} \end{center} \pause } \frame{ \frametitle{1.1 Linear functions} Plotting $x$ against $y$ and joining up the points gives the graph shown below. \vspace{-0.5cm} %\begin{figure}[!h] \begin{center} \includegraphics[width=8cm,angle=-90]{graph1.ps} %\caption{Lee} \end{center} } \frame{ \frametitle{1.1 Linear functions} Simple linear functions have an \textcolor{purple}{\bf intercept} and a \textcolor{purple}{\bf gradient}. \pause \vspl \begin{itemize} \item Intercept: point at which the line would cut through the $y$--axis \pause -- in A\&F, often represents \textit{fixed costs} \pause \vspl \item Gradient: tells us how steep the line is, as well as the direction of the line \pause -- in A\&F, often represents \textit{ cost per unit made/sold} \end{itemize} } \frame{ \frametitle{1.1 Linear functions} Generally, the linear function \pause \[ y=mx + c \] \pause has intercept $c$ and gradient $m$. \vsp So, our old friend \[ y=3x +4 \] has gradient 3 and intercept 4 (we saw the intercept on the graph). \pause \vsp For every unit increase in $x$, $y$ increases by 3 units! } \frame{ \frametitle{1.1 Linear functions} Notice that the gradient is always the number that is ``stuck'' to the $x$; \pause that is, the gradient is the $x$ \textcolor{purple}{\bf coefficient}.\pause \vsp Notice also that the highest power in $x$ is a 1 -- because $x^{1}=x$. There are no terms in $x^{2}$ or $x^{3}$. Thus, we know we have a linear function. \pause \vsp The intercept is the ``other'' number, and is sometimes called the \textcolor{purple}{\bf constant}. \pause \vsp As we shall shortly see, if the $x$ coefficient is positive, then the graph has an uphill slope; if the $x$ coefficient is negative, then the graph has a downhill slope. \pause \vsp The bigger the coefficient, the steeper the slope, no matter which direction the slope goes in! \vsp We will come back to finding the equation of a line shortly. } \frame{ \frametitle{1.1.1 Solving linear equations} If we set our linear function equal to a number/constant, then we have a linear \textcolor{purple}{\bf equation}. \pause \vsp Solve each of the following linear equations for $x$ and $t$ respectively. \begin{itemize} \item [\textbf{1.}] $\displaystyle \frac{6}{x+4}=4$; \vsp \item [\textbf{2.}] $\displaystyle \frac{-4t+4}{5t-6} =-8$. \end{itemize} } \sol \frame{ \frametitle{Solution to (1)} \begin{eqnarray*} \frac{6}{x+4} &=& 4 \pause \\ & & \\ 6 &=&4(x+4) \pause \\ & & \\ 6 &=&4x + 16 \pause \\ & & \\ 6-16 &=& 4x \pause \\ & & \\ -10 &=&4x \pause \\ & & \\ x &=& -\frac{10}{4} = -2.5. \end{eqnarray*} } \frame{ \frametitle{Solution to (2)} \begin{eqnarray*} \frac{-4t+4}{5t-6} &=&-8 \pause \\ & & \\ -4t+4 &=&-8(5t-6) \pause \\ & & \\ -4t+4 &=&-40t+48 \pause \\ & & \\ -4t+40t &=&48-4 \pause \\ & & \\ 36t &=&44 \pause\\ & & \\ t &=& \frac{44}{36} = \frac{11}{9} = 1\frac{2}{9} = 1.222 \end{eqnarray*} } \frame{ \frametitle{1.1.2 Finding the equation of a line} Suppose we are told that the gradient of a line is 6. We are also told that the line passes through the point $(3,25)$. What is the equation of this line? \pause \vsp We are told that $m=6$, and so we have \[ y=6x+c. \] \pause So now we just need to find $c$! We are also told that the line passes through the point $(3,25)$, i.e. the point where $x=3$ and $y=25$. \pause \vsp Substituting these values into the above equation gives \pause \begin{eqnarray*} 25 &=& (6\times 3) + c \qquad \textrm{\sf i.e.} \\ \pause 25 &=& 18+c. \end{eqnarray*} \pause So obviously $c=7$, giving $y=6x+7$. } \frame{ \frametitle{1.1.2 Finding the equation of a line} Suppose we are told that a line passes through the points $(1,3)$ and $(5,35)$. What is the equation of the line? \pause \vsp Again, we need to find the gradient, $m$, and the intercept, $c$. \pause We know that \begin{eqnarray*} \textrm{\sf Gradient} &=& \frac{\textrm{\sf Change in $y$}}{\textrm{\sf Change in $x$}} \\ & & \\ \pause &=& \frac{35-3}{5-1} \\ & & \\ \pause &=& \frac{32}{4} = 8. \end{eqnarray*} } \frame{ \frametitle{1.1.2 Finding the equation of a line} So now we have \pause \[ y = 8x + c, \] and we just need to find $c$. \pause \vsp We know that the line passes through $(1,3)$, and so substituting $x=1$ and $y=3$ into the above equation gives \pause \begin{eqnarray*} 3 &=& (8 \times 1) + c \\ 3 &=& 8 + c \qquad \textrm{\sf i.e.} \\ 3-8 &=& c \qquad \textrm{\sf and so}\\ -5 &=& c.\pause \end{eqnarray*} So the equation of the line is $y=8x-5$. } \normal \frame{ \frametitle{Example 1.1} The United Nations believes the annual consumption of rice in India ($y$ kilograms per household) is a linear function of the unit cost ($x$ US \$). \vsp They also know that, when the unit cost of rice is \$12, the annual consumption of rice is 40.8kg per household; when this unit cost doubles, the corresponding consumption decreases to 21.6kg per household. \vsp \begin{itemize} \item [\textbf{(a)}] Using this information, obtain the linear function for demand in terms of cost. \item [\textbf{(b)}] Find the unit cost of rice if the demand is 25kg per household. \end{itemize} } \sol \frame{ \frametitle{Example 1.1: Solution to part (a)} %We are told that \begin{eqnarray*} x &:& \textsf{Cost (\$ US)}\qquad \textsf{and} \\ y &:& \textsf{Consumption (kg)} \end{eqnarray*} We are also told that \begin{itemize} \item when $x=12$, $y=40.8$ \pause $\longrightarrow (12,40.8)$, \item when $x=24$, $y=21.6$ \pause $\longrightarrow (24,21.6)$; thus \pause \vspl \end{itemize} \[ \textsf{Gradient} = \frac{40.8-21.6}{12-24} = -1.6. \pause \] Since we are asked to find the {\it linear} demand function, we know that $y=mx+c$ (where $m=-1.6$). Taking the point $(12,40.8)$: \pause \begin{eqnarray*} 40.8 &=&-1.6\times 12 + c \\ 40.8+1.6\times 12 &=&c\\ 60 &=&c. \end{eqnarray*} \pause So the linear demand function is $y=-1.6x+60$. } \frame{ \frametitle{Example 1.1: Solution to part (b)} If demand is 25kg per household, then $y=25$. Thus \pause \begin{eqnarray*} 25 &=&-1.6x + 60 \pause \\ 25-60 &=&-1.6x \pause \\ -35 &=&-1.6x \quad \textsf{so} \pause \\ x &=&21.875, \end{eqnarray*} or \$21.88. } \normal \frame{ \frametitle{1.1.3 Solving two linear equations simultaneously} Solve the following linear equations simultaneously; do \textit{not} use a graph. \begin{eqnarray*} 4x + 8y &=& 30 \\ 7x - 6y &=& 5. \end{eqnarray*} } \sol \frame{ \frametitle{Solution} Let's eliminate the $y$'s. \pause \vsp Multiplying the first equation by 3 and multiplying the second equation by 4 gives: \pause \begin{eqnarray*} 12x + 24y &=& 90 \\ 28x - 24y &=&20. \end{eqnarray*} \pause The signs of the $y$--coefficients are different, so we add: \pause \begin{eqnarray*} 12x + 24y &=& 90 \\ \underline{28x - 24y} &=&\underline{20} \\ \pause 40x + 0y &=& 110 \end{eqnarray*} \pause and so \[ x = \frac{110}{40} = \frac{11}{4} = 2.75. \] } \frame{ \frametitle{Solution} Substituting $x=2.75$ into the first equation gives:\pause \begin{eqnarray*} 4\times 2.75 + 8y&=& 30 \\ \pause 11 + 8y &=& 30 \\ \pause 8y &=&19 \pause \\ & & \\ y&=& \frac{19}{8} = 2.375. \end{eqnarray*} } \begin{comment} \frame{ \frametitle{Announcements} \begin{itemize} \item Numbas CBA1: \pause \vspl \begin{itemize} \item [--] Opens in \textcolor{purple}{\bf assessed mode} next week \pause \vspl \item [--] Practice mode will close at the end of this week \vspl \item [--] Deadline: \textbf{23:59 Sunday 25th October} \pause \end{itemize} \vsp \item Case studies -- same topics next week, example write--up available from course webpage \end{itemize} } \frame{ \frametitle{Recap} \begin{itemize} \item Linear functions: $y=mx+c$\pause \begin{itemize} \item [--] Gradient -- what is it? Interpetation? \textit{Practical} intepretation? \pause \item [--] Intercept -- what is it? Interpretation? \pause \item [--] How do we know it's linear? \pause \end{itemize} \vspl \item Finding the gradient: \pause \[ \frac{\textsf{Change in $y$}}{\textsf{Change in $x$}} \] \pause \item Finding the equation of the line \pause \begin{itemize} \item [--] Gradient--point \pause \item [--] Point--point \end{itemize} \end{itemize} } \frame{ \frametitle{Recap} Consider lines with the equations: \begin{eqnarray} y &=& 4x+17 \\ 3y&=&-12x+21 \\ 2x &=&10-y \end{eqnarray} Which is `steepest'? \pause \vsp \textbf{Gradient of (1):} $+4$ \pause \vspl \textbf{Gradient of (2):} \begin{eqnarray*} y &=&-4x+7 \quad \longrightarrow \textsf{Gradient} = -4 \end{eqnarray*} \pause \vspl \textbf{Gradient of (3):} \begin{eqnarray*} y &=&-2x+10 \quad \longrightarrow \textsf{Gradient} = -2 \end{eqnarray*} } \frame{ \frametitle{Recap} \begin{itemize} \item Solving linear equations \vsp \item Solving two linear equations simultaneously \end{itemize} } \normal \frame{ \frametitle{Example 1.2} A cookie company makes two types of biscuit: \textit{standard} and \textit{deluxe}. \pause Suppose the company makes $x$ batches of standard biscuits and $y$ batches of deluxe biscuits every day. \pause \vsp \begin{itemize} \item [\textbf{(a)}] Each batch of standard biscuits requires 5kg of flour; each batch of deluxe biscuits requires 8kg of flour. Every day, exactly 98kg of flour must be used. Write down a linear equation in $x$ and $y$ for the total amount of flour used each day (in kg). \vspl \item [\textbf{(b)}] Each batch of standard biscuits requires 2kg of butter; each batch of deluxe biscuits requires twice as much butter. Every day, exactly 44 kg of butter must be used. Write down a linear equation in $x$ and $y$ for the total amount of butter used each day (in kg). \vspl \item [\textbf{(c)}] How many batches of standard and deluxe biscuits can the company make each day? \end{itemize} } \sol \frame{ \frametitle{Example 1.2: Solution to (a) and (b)} It often helps if you draw up a table: \pause \begin{center} \begin{tabular}{|c| c c|} \hline &\textbf{Flour (kg)} &\textbf{Butter (kg)} \\ \hline Standard ($x$)& 5 &2 \\ Deluxe ($y$) &8 &4 \\ \hline \textbf{Total} &98 &44 \\ \hline \end{tabular} \end{center} \pause Then for flour (part (a)), we would have \pause \[ 5x+8y = 98. \pause \] For butter (part (b)), we have \pause \[ 2x+4y = 44. \] } \frame{ \frametitle{Example 1.2: Solution to (c)} If we keep the flour equation the same, but multiply the butter equation by 2, we get:\pause \begin{eqnarray*} 5x + 8y &=& 98 \\ 4x + 8y &=& 88 \end{eqnarray*} \pause Subtracting to eliminate the $y$'s gives: \pause \begin{eqnarray*} 1x + 0y &=& 10 \\ \pause x &=&10 \pause \end{eqnarray*} } \frame{ \frametitle{Example 1.2: Solution to (c)} Substituting $x=10$ into the flour equation gives \pause \begin{eqnarray*} 5\times 10 + 8y &=& 98 \pause \\ 50 + 8y &=& 98 \pause \\ 8y&=&48 \pause \\ y &=&6. \pause \end{eqnarray*} So, each day, the company must make 10 batches of standard biscuits and 6 batches of deluxe biscuits. } \normal \frame{ \frametitle{1.2 Quadratic functions} As you might imagine, not everything in real--life can be represented by a straight line! \pause \vsp At best, the linear functions we have looked at so far -- such as linear cost functions and linear functions for profit -- are simplifications. \pause \vsp With this in mind, in this section we will consider the role of \textcolor{purple}{\bf non--linear functions}. } \frame{ \frametitle{1.2 Quadratic functions} In particular, we will think about polynomial functions where the power in $x$ is greater than 1 (giving a non--linear graph). \pause \vsp The simplest case here is the \textcolor{purple}{\bf quadratic function}, where the highest power of $x$ in our polynomial is 2. \pause \vsp That is, at most we have an $x^{2}$ term. \pause For example, \[ y = 5x^{2}-2x + 6 \] is a quadratic, since the highest power of $x$ is 2. \pause The polynomial \[ y = 2x^{4}-3x^{2} + 9 \] is not a quadratic, since the highest power of x here is 4 (in fact, this is a \textcolor{purple}{\bf quartic}). } \frame{ \frametitle{1.2.1 Plotting quadratic functions} In the space below, plot the graph of the function \[ y=x^{2}-2x-15. \] } \sol \frame{ \frametitle{Solution} \begin{center} \begin{tabular}{|c|c c c c |c| c c c c c|} \hline $x$ &--4 &--3 &--2 &--1 &0 &1 &2 &3 &4 &5\\ \hline $y$ &\pause 9 &\pause 0 &\pause --7 &--12 &--15&--16&--15 &--12 &--7 &0 \\ \hline \end{tabular} \pause \vspace{0.6cm} \includegraphics[width=7cm,angle=-90]{graph10newb.ps} %\caption{Lee} \end{center} } \normal \frame{ \frametitle{Example 1.5} The estates manager of a zoo has 100 metres of fencing to construct a rectangular enclosure for some West African camels. \begin{itemize} \item [\textbf{(a)}] Formulate a non--linear function for the area of the enclosure in terms of its width, $x$ metres. \item [\textbf{(b)}] Produce a plot of your non--linear function in (a) (see overleaf). \item [\textbf{(c)}] Using your plot in (b), what length $x$ would give the optimal area for the camel enclosure? What area would this give? \end{itemize} } \sol \frame{ \frametitle{Example 1.5: Solution to part (a)} Consider the diagram: \pause \vspace{-2cm} \begin{center} \includegraphics[width=8cm,angle=-90]{box.ps} \end{center} \vspace{-2cm} The area is obviously $x\times y = xy$, but the question wants the solution in terms of $x$ only. } \frame{ \frametitle{Example 1.5: Solution to part (a)} Now \pause \[ \textsf{Perimeter} = x + x + y + y = 2x + 2y = 100. \] \pause Therefore we can write \pause \begin{eqnarray*} 2y &=& 100-2x \qquad \textsf{i.e.} \\ \pause y &=& 50-x. \end{eqnarray*} \pause So now \begin{eqnarray*} \textsf{Area} &=& x \times y \\ \pause &=&x(50-x) \\ \pause &=&50x-x^{2}. \end{eqnarray*} } \frame{ \frametitle{Example 1.5: Solution to part (b)} \begin{center} \includegraphics[width=8cm,angle=-90]{graph11sol.ps} %\caption{Lee} \end{center}} \frame{ \frametitle{Example 1.5: Solution to part (c)} From the graph, we can see that a length of $x=25m$ would give the largest area for the camel enclosure. \pause \vsp This would give an area of \pause \begin{eqnarray*} \textsf{Area} &=& 50\times 25 - 25^{2} \\ \pause &=& 625 \textsf{m}^{2}. \end{eqnarray*} } \normal \frame{ \frametitle{1.2.1 Plotting quadratic functions} As with linear functions, it would be useful to be able to plot or sketch a non--linear function without having to draw up a full table of results. \pause \vsp For a quadratic function, it might be useful to know where the curve cuts into the $x$--axis, as well as the $y$--intercept of the curve. \pause \vsp Consider the quadratic from the earlier example: $y=x^{2}-2x-15$. We know the $y$--intercept occurs when $x=0$, that is \[ y = 0^{2}-2\times 0 -15 = -15, \] and this can be seen in our plot of this function. \vsp \textcolor{white}{However, how can we tell, without drawing up a full table of results, where the curve will cut the $x$--axis? }} \frame{ \frametitle{1.2.1 Plotting quadratic functions} \begin{center} \includegraphics[width=8cm,angle=-90]{graph10newb.ps} %\caption{Lee} \end{center} } \frame{ \frametitle{1.2.1 Plotting quadratic functions} As with linear functions, it would be useful to be able to plot or sketch a non--linear function without having to draw up a full table of results. \vsp For a quadratic function, it might be useful to know where the curve cuts into the $x$--axis, as well as the $y$--intercept of the curve. \vsp Consider the quadratic from the earlier example: $y=x^{2}-2x-15$. We know the $y$--intercept occurs when $x=0$, that is \[ y = 0^{2}-2\times 0 -15 = -15, \] and this can be seen in our plot of this function. \vsp \textbf{However, how can we tell, without drawing up a full table of results, where the curve will cut the $x$--axis? }} \frame{ \frametitle{1.2.1 Plotting quadratic functions} The curve cuts the $x$--axis when $y=0$, that is when \[ x^{2}-2x-15 = 0, \] and so we would need to solve this equation for $x$. However, this is an example of a \textcolor{purple}{\bf quadratic equation}, and we haven't considered how to solve these yet... } \frame{ \frametitle{1.2.2 Quadratic equations} We can visualise the solution(s) for $x$ in the equation \pause \[ ax^{2}+bx+c = 0 \] \pause by looking at a plot of the quadratic and noting where the curve intersects/touches the line $y=0$, that is, the $x$--axis. \pause \vsp However, the accuracy of our solutions obtained in this way will depend on the accuracy of our graph. \pause \vsp We would also need to know what range of $x$--values to draw the graph over. \pause \vsp Two approaches for solving quadratics a bit more mathematically will now be considered. } \frame{ \frametitle{1.2.2 Quadratic equations} In both, we will assume the general form of a quadratic as \[ y = ax^{2}+bx+c; \] \pause the \textcolor{purple}{\bf discriminant}, $D$, is given by \pause \[ D = b^{2}-4ac, \] \pause and has important properties in terms of how we classify the solutions to our quadratic. } \frame{ \frametitle{1.2.2 Quadratic equations: Factorisation} If the discriminant is a perfect square -- that is, if the square root of $D$ is an integer -- then we can solve our quadratic by the method of \textcolor{purple}{\bf factorisation}.\pause \vsp Consider the following quadratic equations: \begin{itemize} \item [1.] $x^{2}-2x-15 = 0$ \item [2.] $x^{2}+5x+2=0$ \item [3.] $x^{2}-8x+12 = 0$ \item [4.] $x^{2}-5x-36=0$ \item [5.] $2x^{2}-7x+5=0$ \end{itemize} \vspace{0.5cm} \begin{itemize} \item [\textbf{(a)}] For each, find the discriminant $D$. \item [\textbf{(b)}] Which equations can be solved by factorisation? \item [\textbf{(c)}] Solve the equations identified in (b) by factorisation. \end{itemize} } \sol \frame{ \frametitle{Solution to (a)} \begin{itemize} \item [1.] $x^{2}-2x-15 = 0$ \pause \[ D = (-2)^{2} - (4\times 1 \times -15) = \pause 64 \]\pause \vsp \item [2.] $x^{2}+5x+2=0$ \pause \[ D = 5^{2} - (4\times 1\times 2) = \pause 17 \] \pause \vsp \item [3.] $x^{2}-8x+12 = 0$ \pause \[ D = (-8)^{2} - (4\times1\times12) = \pause 16 \] \end{itemize} } \frame{ \frametitle{Solution to (a)} \begin{itemize} \item [4.] $x^{2}-5x-36=0$ \pause \[ D = (-5)^{2} - (4\times 1 \times -36) = \pause 25+144 = 169 \] \pause \vsp \item [5.] $2x^{2}-7x+5=0$ \pause \[ D = (-7)^{2} -(4 \times 2 \times 5) =\pause 9 \] \end{itemize} } \frame{ \frametitle{Solution to (b)} Note that the discriminant for Equation (1) was $D=64$: \[ \sqrt{64} = 8, \] which is an integer value; thus $\boldsymbol{D}$ \textbf{is a perfect square}, and so we can solve this equation by factorisation. \pause \vsp In fact, we can solve all but equation (2) by factorisation -- equation 2 is the only one where the discriminant is {\it not} a perfect square. } \frame{ \frametitle{Solution to (c)} \textbf{Equation (1):} \vsp Sinced $D$ was a perfect square, we can factorise our quadratic; in fact, \[ x^{2}-2x-15 = (x-5)(x+3). \] \pause How did we do this? \pause \begin{itemize} \item The first term inside each pair of brackets must be $x$; \pause \vspl \item The other two numbers must add together to give the $x$--coefficient (e.g. {\color{red} $-5+3 =-2$})... \pause \vspl \item...and must multiply to give the constant (e.g. {\color{red}$-5 \times 3 = -15$}) \end{itemize} } \frame{ \frametitle{Solution to (c)} We can check our factorisation by expanding the brackets using the \textbf{F-O-I-L} rule, standing for ``\textbf{F}irst two'', ``\textbf{O}utside two'', ``\textbf{I}nside two'', ``\textbf{L}ast two'': \pause \begin{eqnarray*} \textbf{F} &:& x \times x = x^{2} \\ \pause \textbf{O} &:&x \times 3 = 3x \\\pause \textbf{I} &:&-5 \times x = -5x \\ \pause \textbf{L} &:&-5 \times 3 = -15, \pause \end{eqnarray*} giving \[ x^{2}+3x-5x-15 = \pause x^{2}-2x-15, \] which is exactly what we started with! } \frame{ \frametitle{Solution to (c)} So now we know that \[ x^{2}-2x-15 = (x-5)(x+3) = 0, \] we know that either $(x-5)=0$ or $(x+3)=0$. \pause So we have \begin{eqnarray*} x-5 = 0 &\textrm{\sf or} & x+3 = 0 \\ \pause x=5 &\textrm{\sf or} &x=-3, \end{eqnarray*} and these are the values we see on the graph on page 19. } \frame{ \frametitle{Solution to (c)} \textbf{Equation (3):} \begin{eqnarray*} x^{2}-8x+12 &=&0 \\ \pause (x-6)(x-2) &=&0,\pause \end{eqnarray*} so $x=6$ or $x=2$. } \frame{ \frametitle{Solution to (c)} \textbf{Equation (4): } \begin{eqnarray*} x^{2}-5x-36&=&0 \\ \pause (x-9)(x+4) &=&0, \pause \end{eqnarray*} so $x=9$ or $x=-4$. \pause \vsp \textbf{Equation (5):} \begin{eqnarray*} 2x^{2}-7x+5&=&0 \\ \pause (2x-5)(x-1)&=& 0, \pause \end{eqnarray*} so $(2x-5)=0$ or $(x-1)=0$. \pause From the first: \pause \[ 2x = 5 \pause \rightarrow x=2.5, \] and also $x=1$. } \normal \begin{comment} \frame{ \frametitle{Announcements: CBA1} \begin{itemize} \item Live right now in assessed mode \pause \vspl \item Have been some issues - check that I've replied to your emails \pause \vspl \item Deadline: \textcolor{red}{23:59, Sunday 25th October} \end{itemize} } \frame{ \frametitle{Announcements: Case studies} \begin{itemize} \item Decide on your $x$ and $y$ variables \pause \vspl \item See if you can find the co-ordinates of two points from which you can formulate a linear function. For example: \begin{center} \begin{tabular}{c|c} $x$ (Consumption / Selling price) & $y$ (Price / Units produced) \\ \hline $x_{1}$ &$y_{1}$ \\ $x_{2}$ &$y_{2}$ \\ \end{tabular} \end{center} \vspl \item Do this twice! We're looking for at least two linear functions in each case study... \pause \vspl \begin{itemize} \item [--]Alcohol: Thai government's research versus \textit{Singha}'s research \pause \vspl \item [--] iPhone: Supply versus demand \end{itemize} \pause \vspl \item Don't worry \textit{too much} about \textit{elasticity} or \textit{ceiling price} \pause \vspl \item Remember to look at example report; equation editor handout for MS WORD; graphical software in ``Videos'' \end{itemize} } \frame{ \frametitle{Non--linear functions: the story so far...} \begin{itemize} \item Any polynomial where the highest power of $x>1$ \pause \vspl \item Simplest non--linear function: \textcolor{purple}{\bf Quadratics}, where the highest power is 2 (i.e. a term in $x^{2}$) \pause \vspl \item General form: $ax^{2}+bx+c$, with \textcolor{purple}{\bf discriminant} \[ D=b^{2}-4ac \] \pause \vspl \item Should know how to \textit{plot} these \pause \vspl \item Should know how to \textit{solve} these when set equal to zero: \pause \vspl \begin{itemize} \item [--] by factorisation (when $D$ is a perfect square) \pause \vspl \item [--] by the \textcolor{purple}{\bf quadratic formula} \end{itemize} \end{itemize} } \normal \frame{ \frametitle{1.2.2 Quadratic equations: formula} \noindent If we cannot factorise our quadratic, then we can use the \textcolor{purple}{\bf quadratic formula} in order to find the solution(s) to our quadratic equation. \pause \vsp For example, consider equation (2) in the last question: \[ x^{2}+5x+2 = 0. \] The discriminant \[ D = 5^{2} - 4\times 1 \times 2 = 17, \] which is not a perfect square. } \frame{ \frametitle{1.2.2 Quadratic equations: formula} All is not lost -- we can use the quadratic formula! \pause \vsp \begin{itemize} \item Developed by \textcolor{purple}{\bf Indian mathematicians} around 628AD \pause \vspl \item The formula as we know it today was not published until 1637 by a \textcolor{purple}{\bf French mathematician}\pause \vspl \item In fact, the formula can \textit{always} be used -- whether the discriminant is a perfect square or not! \end{itemize} } \frame{ \frametitle{1.2.2 Quadratic equations: formula} The solution(s) to the equation \[ ax^{2}+bx+c=0, \] with discriminant $D=b^{2}-4ac$, are given by the formula \pause \[ x=\frac{-b\pm \sqrt{D}}{2a}. \] } \frame{ \frametitle{1.2.2 Quadratic equations: formula} Consider the quadratic we plotted on page 19: \[ y=x^{2}-2x-15. \] \pause The graph showed that when $y=0$ the solutions were $x=5$ and $x=-3$.\pause \vsp Note that $a=1$, $b=-2$ and $c=-15$. Thus the discriminant is \[ D = (-2)^{2} - 4 \times 1 \times -15 = 64. \] \pause So by the formula, we have \begin{eqnarray*} x &=& \frac{-(-2)\pm \sqrt{64}}{2 \times 1} \\ \pause & & \\ &=& \frac{2 \pm 8}{2}. \end{eqnarray*}} \frame{ \frametitle{1.2.2 Quadratic equations: formula} So the solutions are given by \[ x = \frac{2+8}{2} = \frac{10}{2} = 5, \] \pause or \[ x=\frac{2-8}{2} = \frac{-6}{3} = -3, \] \pause exactly the solutions we obtained graphically (and by factorisation)! } \sol \frame{ \frametitle{1.2.2 Quadratic equations: formula} Solve the quadratic equation \[ x^{2}+5x+2 = 0. \] \pause \vsp \textbf{Solution} \pause \vsp The discriminant is \pause \[ D = 5^{2} - (4\times 1 \times 2) = 17. \] \pause Thus, we have \pause \[ x =\frac{-5\pm\sqrt{17}}{2}, \] } \frame{ \frametitle{1.2.2 Quadratic equations: formula} giving \pause \begin{eqnarray*} x &=&\frac{-5+\sqrt{17}}{2} \\ & & \\ &=& -0.44 \pause \end{eqnarray*} \textbf{or} \pause \begin{eqnarray*} x &=&\frac{-5-\sqrt{17}}{2} \\ & & \\ &=& -4.56 \pause \end{eqnarray*} } \normal \frame{ \frametitle{Example 1.6} A company models its annual profits using the function \[ P(x) = x^{2}+20x-300, \] where $P$ represents profits and $x$ is the number of units sold. In 2011 their profits were \pounds 167,700. How many units of their product did they sell? } \sol \frame{ \frametitle{Example 1.6: Solution} We set the equation equal to 167700 and solve for $x$: \pause \begin{eqnarray*} x^{2}+20x-300 &=& 167700 \pause \\ x^{2}+20x-300-167700 &=& 0 \pause \\ x^{2}+20x-168000 &=&0 \pause \end{eqnarray*} Now the discriminant is \pause \[ D = 20^{2} - (4 \times 1 \times -168000) = 672400, \]\pause and so \begin{eqnarray*} x &=& \frac{-20 \pm \sqrt{672400}}{2} \pause \\ \pause & & \\ &=& \frac{-20 \pm 820}{2}. \end{eqnarray*} } \frame{ \frametitle{Example 1.6: Solution} So \[ x =\frac{-20+820}{2} = 400 \pause \] or \[ x = \frac{-20-820}{2} = -420. \] Obviously we can't sell a negative number of units, and so $x=400$.\pause \vsp Note the discriminant was a perfect square and so we could have solved this quadratic via factorisation. } \normal \frame{ \frametitle{Example 1.7} The manager of a vineyard thinks there is a relationship between the average temperature during the grape--growing season ($t^{\textrm{o}}\textrm{C}$) and the quality of the grapes produced and hence the selling price of resulting bottles of wine ($p$ pounds). } \frame{ \frametitle{Example 1.7} Over a ten year period the following relationship is revealed: \begin{center} \psfrag{5}[][]{\sf \tiny 5} \psfrag{6}[][]{\sf \tiny 6} \psfrag{7}[][]{\sf \tiny 7} \psfrag{8}[][]{\sf \tiny 8} \psfrag{9}[][]{\sf \tiny 9} \psfrag{10}[][]{\sf \tiny 10} \psfrag{16}[][]{\sf \tiny 16} \psfrag{18}[][]{\sf \tiny 18} \psfrag{20}[][]{\sf \tiny 20} \psfrag{22}[][]{\sf \tiny 22} \psfrag{24}[][]{\sf \tiny 24} \psfrag{t}[][]{\sf \tiny Temperature ($t^{\textrm{o}}\textrm{C}$)} \psfrag{p}[][]{\sf \tiny Price (\pounds $p$)} \includegraphics[width=9cm,height=6cm]{Rplot1.ps} \end{center} } \frame{ \frametitle{Example 1.7} \begin{itemize} \item [\textbf{(a)}] Which model would best suit the relationship shown? \begin{itemize} \item [*] $p = 0.2t+4.4$ \item [*] $p = 0.2t^{2}-8t+70$ \item [*]$p = -0.2t^{2}+8t-70$ \end{itemize} \vsp \item [\textbf{(b)}] Suppose the manager needs to sell this variety of wine in 2013 for at least \pounds 9 per bottle. What is the critical range of temperatures required to meet this price? \end{itemize} } \sol \frame{ \frametitle{Example 1.7: Solution to part (a)} Clearly, it cannot be the first equation, as this is a linear equation and we have a non--linear relationship. \pause \vsp Of the two remaining quadratics, it must be the last one to give the ``upside--down'' curve we see in the graph (negative coefficient for $t^{2}$). } \frame{ \frametitle{Example 1.7: Solution to part (b)} We set $p=9$ and solve for $t$: \pause \begin{eqnarray*} -0.2t^{2}+8t-70 &=& 9 \pause \\ -0.2t^{2}+8t -70-9 &=&0 \pause \\ -0.2t^{2}+8t - 79 &=&0. \end{eqnarray*} Now $a=-0.2$, $b=8$ and $c=-79$, giving \pause \[ D = 8^{2} - (4\times -0.2 \times -79) = 0.8. \pause \] } \frame{ \frametitle{Example 1.7: Solution to part (b)} So \[ x = \frac{-8 \pm \sqrt{0.8}}{-0.4}, \] \pause giving \[ x = \frac{-8+\sqrt{0.8}}{-0.4} = 17.76 \]\pause and \[ x = \frac{-8-\sqrt{0.8}}{-0.4} = 22.24. \] \pause So any temperatures between $17.8^{\textrm{o}}\textrm{C}$ and $22.2^{\textrm{o}}\textrm{C}$ should give bottles of wine that sell for at least \pounds 9. } %\frame{ %\frametitle{Example 1.7: Solution to part (b)} %\begin{center} %\psfrag{16}[][]{\sf \tiny 16} %\psfrag{18}[][]{\sf \tiny 18} %\psfrag{20}[][]{\sf \tiny 20} %\psfrag{22}[][]{\sf \tiny 22} %\psfrag{24}[][]{\sf \tiny 24} %\psfrag{5}[][]{\sf \tiny 5} %\psfrag{6}[][]{\sf \tiny 6} %\psfrag{7}[][]{\sf \tiny 7} %\psfrag{8}[][]{\sf \tiny 8} %\psfrag{9}[][]{\sf \tiny 9} %\psfrag{10}[][]{\sf \tiny 10} %\psfrag{p}[][]{$p$} %\psfrag{t}[][]{$t$} %\includegraphics[width=11cm,height=7cm]{wine1.ps} %\end{center} %} %\frame{ %\frametitle{Example 1.7: Solution to part (b)} %\begin{center} %\psfrag{16}[][]{\sf \tiny 16} %\psfrag{18}[][]{\sf \tiny 18} %\psfrag{20}[][]{\sf \tiny 20} %\psfrag{22}[][]{\sf \tiny 22} %\psfrag{24}[][]{\sf \tiny 24} %\psfrag{5}[][]{\sf \tiny 5} %\psfrag{6}[][]{\sf \tiny 6} %\psfrag{7}[][]{\sf \tiny 7} %\psfrag{8}[][]{\sf \tiny 8} %\psfrag{9}[][]{\sf \tiny 9} %\psfrag{10}[][]{\sf \tiny 10} %\psfrag{p}[][]{$p$} %\psfrag{t}[][]{$t$} %\includegraphics[width=11cm,height=7cm]{wine2.ps} %\end{center} %} \frame{ \frametitle{Example 1.7: Solution to part (b)} \begin{center} \psfrag{16}[][]{\sf \tiny 16} \psfrag{18}[][]{\sf \tiny 18} \psfrag{20}[][]{\sf \tiny 20} \psfrag{22}[][]{\sf \tiny 22} \psfrag{24}[][]{\sf \tiny 24} \psfrag{5}[][]{\sf \tiny 5} \psfrag{6}[][]{\sf \tiny 6} \psfrag{7}[][]{\sf \tiny 7} \psfrag{8}[][]{\sf \tiny 8} \psfrag{9}[][]{\sf \tiny 9} \psfrag{10}[][]{\sf \tiny 10} \psfrag{p}[][]{$p$} \psfrag{t}[][]{$t$} \includegraphics[width=11cm,height=7cm]{wine3.ps} \end{center} } \normal \frame{ \frametitle{Summary} \begin{itemize} \item If $D$ is a perfect square, then we can solve the quadratic by \textcolor{purple}{\bf factorisation}; otherwise, we can use the \textcolor{purple}{\bf formula} (in fact, the formulae will always work!)\pause \vspl \item If $D>0$, our quadratic will have \textcolor{purple}{\bf two real solutions} -- that is, the curve will cut the $x$--axis in two places\pause \vspl \item If $D=0$, then our quadratic will have \textcolor{purple}{\bf repeated solutions} -- that is, our graph only just touches the $x$--axis \pause \vspl \item If $D<0$, then we cannot operate the quadratic formula as we cannot take the square root of a negative -- at least, not in the real number system. Actually, such a quadratic has \textcolor{purple}{\bf complex roots} making use of the complex number system, but this goes beyond the scope of ACC1012/53. \end{itemize} } \frame{ \frametitle{1.3 Higher--order polynomials} In more complicated accounting and finance problems, linear and/or quadratic functions might still not be adequate enough to model our situation realistically. Higher--order polynomials can be used, such as \textcolor{purple}{\bf cubics}: \pause \[ y = ax^{3}+bx^{2}+cx +d, \] or even \textcolor{purple}{\bf quartics}: \pause \[ y=ax^{4}+bx^{3}+cx^{2}+dx + e; \]\pause We will return to such functions in Chapter 3 of this course. } \frame{ \begin{center} Now back to page 8... \end{center} } \normal \frame{ \frametitle{1.1.4 Linear programming} \noindent \textcolor{purple}{\bf Dynamic programming} techniques were developed during the Second World War by a group of American mathematicians.\pause \vsp They sought to produce mathematical models of situations in which all the \textcolor{purple}{\bf requirements}, \textcolor{purple}{\bf constraints} and \textcolor{purple}{\bf objectives} were expressed as algebraic equations. \pause \vsp They then developed methods for obtaining the \textcolor{purple}{\bf optimal solution} -- the maximum or minimum value of a required function. } \frame{ \frametitle{1.1.4 Linear programming} In this section, we will attempt to formulate the requirements, constraints and objectives of real--life accounting and finance problems as linear equations. \pause \vsp This branch of dynamic programming is referred to as \textcolor{purple}{\bf linear programming}, for obvious reasons. \pause \vsp Thus, all algebraic expressions will be of the form \begin{eqnarray*} (\textrm{\sf a number})x + (\textrm{\sf a number})y; \end{eqnarray*} \pause for example, \begin{eqnarray*} \textrm{\sf Profit} &=& 4x - 3y \end{eqnarray*} is a linear equation for profit in terms of $x$ and $y$, where $x$ might represent our expenditure on advertising and $y$ might be our costs. } \frame{ \frametitle{1.1.4 Linear programming} \noindent Linear programming methods are some of the most widely used methods employed to solve management and economic problems. \vsp They have been applied in a variety of contexts, some of which will be discussed in this chapter, with enormous savings in money and resources. \vsp The first step is to \textcolor{purple}{\bf formulate} a problem as a linear programming problem; \pause the second step is to \textcolor{purple}{\bf solve} the problem. } \frame{ \frametitle{1.1.4 Linear programming} As well as drawing upon our recent work on linear equations, in this section we also need to think about linear \textcolor{purple}{\bf inequalities}; \pause for example, \begin{eqnarray*} 16x +18y &\leq& 25 \end{eqnarray*} is a linear inequality in $x$ and $y$. The role of inequalities will become apparent as we work through some examples. } \frame{ \frametitle{1.1.4 Formulating linear programming problems} \noindent The first step in formulating a linear programming problem is to determine which quantities you need to know to solve the problem. These are called the \textcolor{purple}{\bf decision variables}. \pause \vsp \noindent The second step is to decide what the \textcolor{purple}{\bf constraints} are in the problem. For example, there may be a limit on resources. \pause \vsp \noindent The third step is to determine the obective to be achieved. This is the quantity to be maximised or minimised, that is, \textit{optimised}. The function of the decision variables that is to be optimised is called the \textcolor{purple}{\bf objective function}. } \frame{ \frametitle{Example 1.3: The chair manufacturer} \noindent A manufacturer makes two kinds of chairs -- \textcolor{purple}{\textbf{A}} and \textcolor{purple}{\textbf{B}}. Each type of chair has to be processed in two departments -- \textrm{\textbf{I}} and \textrm{\textbf{II}}. \pause \vspace{0.3cm} \noindent Chair \textcolor{purple}{\textbf{A}} spends 3 hours in department \textrm{\textbf{I}} and 2 hours in department \textrm{\textbf{II}}. Chair \textcolor{purple}{\textbf{B}} spends 3 hours in department \textrm{\textbf{I}} and 4 hours in department \textrm{\textbf{II}}. } \frame{ \frametitle{Example 1.3: The chair manufacturer} \noindent The time available in department \textrm{\textbf{I}} in any given month is 120 hours, and the time available in department \textrm{\textbf{II}} in the same month is 150 hours. \pause \vspace{0.3cm} \noindent Chair \textcolor{purple}{\textbf{A}} has a selling price of \pounds 10 and chair \textcolor{purple}{\textbf{B}} of \pounds 12. } \frame{ \frametitle{Example 1.3: The chair manufacturer} \noindent The manufacturer wishes to maximise his income. \pause \vspace{0.5cm} \textbf{How many of each type of chair should be made?} } \frame{ \frametitle{Example 1.3: The chair manufacturer} \noindent You'll notice that there's a lot of information given in the question -- this is typical of a linear programming problem. Sometimes it's easier to summarise the information given in a table: \pause \vspace{0.3cm} \begin{center} \small \begin{tabular}{|c| c| c| c|} \hline Chair &Time in dept. \textrm{I} &Time in dept. \textrm{II} &Selling price \\ \hline \textcolor{purple}{\textbf{A}} &3 &2 &10 \\ \textcolor{purple}{\textbf{B}} &3 &4 &12 \\ \hline Time limits &120 &150 & \\ \hline \end{tabular} \normalsize \end{center} } \frame{ \frametitle{Example 1.3: The chair manufacturer} \noindent To formulate this linear programming problem, we consider the following three steps: \pause \begin{itemize} \item [\textbf{1.}] What are the \textcolor{purple}{\textbf{decision variables}}? (i.e. which quantities do you need to know in order to solve the problem?) \pause \item [\textbf{2.}] What are the \textcolor{purple}{\textbf{constraints}}? \pause \item [\textbf{3.}] What is the \textcolor{purple}{\textbf{objective}}? \end{itemize} } \frame{ \frametitle{Step 1: Decision variables} \noindent Read through the question and identify the things you'd like to know. You can usually do this by going straight to the last sentence of the question: \pause \begin{center} \textbf{``How many of each chair should be made...''} \end{center} \pause Thus, we'd like to know \begin{itemize} \item the number of type \textcolor{purple}{\textbf{A}} chairs to make, and \item the number of type \textcolor{purple}{\textbf{B}} chairs to make. \end{itemize} } \frame{ \frametitle{Step 1: Decision variables} \noindent These are our decision variables, and are usually denoted with lower case letters. Thus, our decision variables are \pause \begin{eqnarray*} x&=& \textrm{\sf number of type \textcolor{purple}{\textbf{A}} chairs made} \qquad \textrm{\sf and} \\ \pause y&=& \textrm{\sf number of type \textcolor{purple}{\textbf{B}} chairs made.} \end{eqnarray*} } \frame{ \frametitle{Step 2: Constraints} \noindent This is probably the hardest bit! Consider what could happen in each department. \pause \vspace{0.3cm} \noindent For example, if we focus on what could happen in department \textrm{\textbf{I}}: \pause \begin{center} \begin{tabular}{l l} Since: &the production of 1 type \textcolor{purple}{\textbf{A}} chair uses 3 hours,\\ \pause then: &the production of $x$ type \textcolor{purple}{\textbf{A}} chairs takes $\textrm{\sf 3}x$ hours.\\ & \\ \pause Similarly: &the production of 1 type \textcolor{purple}{\textbf{B}} chair uses 3 hours,\\ \pause so: &the production of $y$ type \textcolor{purple}{\textbf{B}} chairs takes $\textrm{\sf 3}y$ hours.\\ \end{tabular} \end{center} } \frame{ \frametitle{Step 2: Constraints} \noindent The total time used in department \textrm{\textbf{I}} is therefore \pause \begin{eqnarray*} (\textrm{\sf 3}x + \textrm{\sf 3}y) \textrm{ \sf hours}. \end{eqnarray*} \pause Since only 120 hours are available in department \textrm{\textbf{I}}, one constraint is \pause \begin{eqnarray*} (\textrm{\sf 3}x+\textrm{\sf 3}y) \textrm{ \sf hours} &\leq& \textrm{\sf 120} \textrm{ \sf hours}, \qquad \textrm{ \sf or just} \\ \pause (\textrm{\sf 3}x+\textrm{\sf 3}y) &\leq& \textrm{\sf 120}. \end{eqnarray*} } \frame{ \frametitle{Step 2: Constraints} \noindent Considering department \textrm{\textbf{II}} in a similar way, we get: \begin{center} \begin{tabular}{l l} Since: &the production of 1 type \textcolor{purple}{\textbf{A}} chair uses 2 hours,\\ \pause then: &the production of $x$ type \textcolor{purple}{\textbf{A}} chairs takes 2$x$ hours. \\ \pause & \\ Similarly: &the production of 1 type \textcolor{purple}{\textbf{B}} chair uses 4 hours,\\ \pause so: &the production of $y$ type \textcolor{purple}{\textbf{B}} chairs takes 4$y$ hours.\\ \end{tabular} \end{center} } \frame{ \frametitle{Step 2: Constraints} \noindent The total time used is therefore \pause \begin{eqnarray*} (\textrm{ \sf 2}x+\textrm{\sf 4}y) \textrm{ \sf hours}. \end{eqnarray*} \pause Since only 150 hours are available in department \textrm{\textbf{II}}, a second constraint is \begin{eqnarray*} (\textrm{\sf 2}x+\textrm{ \sf 4}y) \textrm{ \sf hours} &\leq& \textrm{ \sf 150} \textrm{ \sf hours}, \qquad \textrm{\sf or just} \\ \pause (\textrm{\sf 2}x + \textrm{\sf 4}y) &\leq& \textrm{\sf 150}. \end{eqnarray*} } \frame{ \frametitle{Step 2: Constraints} \noindent We're still not done! We can't make a negative number of chairs, so we also have: \pause \begin{eqnarray*} x &\geq& \textrm{\sf 0} \qquad \textrm{\sf and} \\ y &\geq& \textrm{\sf 0}. \end{eqnarray*} \pause \noindent These are called the \textbf{non--negativity constraints}. } \frame{ \frametitle{Step 3: Objective function} \noindent Our objective here is to \textbf{maximise income}. \pause \vspace{0.3cm} \noindent If we make $x$ type \textcolor{purple}{\textbf{A}} chairs, then we get \pounds$\textrm{\sf 10} \times x = \textsf{\pounds} \textrm{\sf 10}x$, since each type \textcolor{purple}{\textbf{A}} chair sells for \pounds$\textrm{\sf 10}$. \pause \vspace{0.3cm} \noindent Similarly, if we make $y$ type \textcolor{purple}{\textbf{B}} chairs, then we get \pounds $\textrm{\sf 12} \times y = \textsf{\pounds} \textrm{\sf 12}y$, since each type \textcolor{purple}{\textbf{B}} chair sells for \pounds$\textrm{\sf 12}$. } \frame{ \frametitle{Step 3: Objective function} \noindent The total income is then \pause \begin{eqnarray*} \textsf{\pounds} Z &=& \textsf{\pounds} (\textrm{\sf 10}x + \textrm{\sf 12} y). \end{eqnarray*} \pause \noindent The aim is to maximise income, so we'd like to maximise \pause \begin{eqnarray*} Z &=& \textrm{\sf 10}x + \textrm{\sf 12}y, \end{eqnarray*} \pause where $Z$ is the objective function. } \frame{ \frametitle{Example 1.3: The chair manufacturer} \noindent \textbf{Thus, to summarise, we have the following linear programming problem:} \pause \vspace{0.3cm} \noindent Maximise $Z=\textrm{\sf 10}x+\textrm{\sf 12}y$ subject to the constraints \pause \begin{eqnarray*} \textrm{\sf 3}x+\textrm{\sf 3}y &\leq& \textrm{\sf 120}, \\ \pause \textrm{\sf 2}x+\textrm{\sf 4}y&\leq& \textrm{\sf 150}, \\ \pause x &\geq& \textrm{\sf 0} \qquad \textrm{\sf and} \\ \pause y &\geq& \textrm{\sf 0}. \end{eqnarray*} } \frame{ \frametitle{Example 1.4: Replica sports shirts} \textit{Sportizus} Clothing Company produce replica football shirts and replica rugby shirts for sale on the high street. Each shirt produced goes through a \textcolor{purple}{\bf sewing process} and a \textcolor{purple}{\bf transfer process}. \pause \vsp \noindent Each football shirt requires \textbf{8} minutes of \textcolor{purple}{\bf sewing time} and \textbf{9} minutes for the \textcolor{purple}{\bf transfer process}, whereas rugby shirts each require \textbf{5} minutes of \textcolor{purple}{\bf sewing time} and \textbf{15} minutes for the \textcolor{purple}{\bf transfer process}. \pause \vsp In any given day, the \textcolor{purple}{\bf total time} available for the sewing process and transfer process is \textbf{10} hours and \textbf{15} hours respectively. \pause } \frame{ \frametitle{Example 1.4: Replica sports shirts} \noindent To meet current demand, \textit{Sportizus} must produce at least \textbf{30} football shirts and \textbf{10} rugby shirts each day. \pause \vsp The company sells football shirts and rugby shirts at a \textcolor{purple}{\bf profit} of \pounds 22 and \pounds 16 respectively. \pause \vsp \noindent\textit{ How many of each type of shirt should \textit{Sportizus} produce in order to maximise profits?} } \sol \frame{ \frametitle{Example 1.4: Replica sports shirts} Let's start off with a table which summarises the question: \pause \begin{center} \begin{tabular}{|c| c| c|c|} \hline &Sewing (mins) &Transfer (minutes) &Profit ($P$)\\ \hline & & & \\ Football& \textbf{8} &\textbf{9} & \textbf{22} \\ & & & \\ \hline & & & \\ Rugby &\textbf{5} &\textbf{15} & \textbf{16}\\ & & & \\ \hline & & &\\ Total time &\textbf{600} &\textbf{900} & \\ & & & \\ \hline \end{tabular} \end{center} } \frame{ \frametitle{Step 1: Decision variables} The decision variables are the number of football and rugby shirts to make. Let \begin{eqnarray*} x &=& \textrm{\sf \textcolor{blue}{number of football shirts to make}} \qquad \textrm{\sf and} \\ & & \\ y &=& \textrm{\sf \textcolor{blue}{number of rugby shirts to make.}} \end{eqnarray*} } \frame{ \frametitle{Step 2: Constraints} The constraints are: \begin{eqnarray*} \textrm{\sf \textbf{sewing}}: 8x+5y &\leq& 600 \qquad \textrm{\sf and} \\ \pause & & \\ \textrm{\sf \textbf{transfer}}: 9x+15y&\leq& 900 \pause \end{eqnarray*} We also have the \textcolor{purple}{\bf non--negativity contrainits}. However, we are also told that we must make at least 30 football shirts and 10 rugby shirts, giving \pause \begin{eqnarray*} x &\geq& 30 \qquad \textrm{\sf and} \\ & & \\ y &\geq& 10 \end{eqnarray*} } \frame{ \frametitle{Step 3: Objective function} The aim is to \textcolor{purple}{\bf maximise profit} -- call this $P$. \pause \vsp We know that we make \pounds 22 for each football shirt that we make and sell. Since we make (and sell) $x$ football shirts, this will give \pause \[ \textrm{\sf \pounds} 22x \qquad \textrm{\sf profit.} \]\pause We also know that we make \pounds 16 for each rugby shirt we make and sell. Since we make (and sell) $y$ rugby shirts, this will give \pause \[ \textrm{\sf \pounds} 16y \qquad \textrm{\sf profit.} \] \pause Thus total profit is \pause \[ \textrm{\sf \pounds} P = 22x+16y, \] } \frame{ \frametitle{Summary} \textbf{Thus, to summarise:} \pause \vsp Maximise $P=22x+16y$ subject to the constraints \pause \begin{eqnarray*} 8x+5y &\leq& 600, \\ \pause & & \\ 9x+15y &\leq& 900, \pause \\ & & \\ x &\geq& 30 \qquad \textrm{\sf and} \\ & & \\ y &\geq& 10. \end{eqnarray*} } \begin{comment} \frame{ \frametitle{Announcements} \begin{itemize} \item Chapter 1 learning outcomes -- will give out today -- prize question! \pause \vspl \item CBA1: Well done! \pause \begin{center} \begin{tabular}{c|c c|c c|c|} \hline &$n$ &$n^{*}$ &Mean &Median &Max \\ \hline 2013 &210 &2 &94 &100 &100 \\ \hline \pause 2014 &210 &4 &92 &100 &100 \\ \hline \pause 2015 &243 &11 &94 &100 &100 \\ \hline \end{tabular} \end{center} \vspl \item Case studies: come to a close next week... \pause \vspl \begin{itemize} \item[--] We will be asking you all an individual question in next week's workshop which counts towards your final mark \pause \item[--] Individual written report (4 pages max) due in by \textbf{4pm next Friday}: \pause \vspl \item [--] Must include a completed cover sheet \pause \vspl \item [--] Must be posted into the Stage 1 homework submission letterbox in the General Office on the 3rd floor of the \textbf{Herschel Building} \end{itemize} \end{itemize} } \frame{ \frametitle{Announcements} \textbf{Alcohol case study} \begin{center} {\it ``I heard that there were mistakes in the alcohol case study and that you were going to scrap it and give us all a new one'' } \pause \vsp \textbf{ \LARGE NO!} \pause \end{center} \vsp \begin{itemize} \item Some confusion over ``elasticity'' \pause \vspl \item Different interpretations of this (Mathematics/Economics) \pause \vspl \item It's up to you what you do here -- you will not be penalised! \end{itemize} } \normal \frame{ \frametitle{Solving linear programming problems} \noindent In the previous section we discussed how to \textcolor{purple}{\bf formulate} a linear programming problem. \vsp We now consider how to \textcolor{purple}{\bf solve} such problems. \vsp Linear programming problems that involve only two decision variables, $x$ and $y$, may be solved \textcolor{purple}{\bf graphically} -- that is, by drawing lines associated with the constraints, identifying a region of possible solutions and then locating a point in this region that has a particular property. \vsp We will now re--visit the two examples from the previous section. } \frame{ \frametitle{Example 1.3 revisited (the chair manufacturer)} \noindent Recall that the problem posed in \textcolor{purple}{\bf example 1.3} could be summarised in the following way: \vsp Maximise $Z=10x + 12y$ subject to the following constraints: \begin{eqnarray*} 3x+3y &\leq& 120,\\ 2x+4y &\leq& 150, \\ x &\geq& 0 \qquad \textrm{\sf and} \\ y &\geq& 0. \end{eqnarray*} } \frame{ \frametitle{Example 1.3 revisited (the chair manufacturer)} \noindent To find the \textcolor{purple}{\bf feasible region} for this problem (i.e. the region on a graph which satisfies all of our inequalities), we proceed by indicating, on a diagram, the region for which all of the inequalities hold. } \frame{ \frametitle{Example 1.3 revisited (the chair manufacturer)} \noindent The first inequality is $3x+3y \leq 120$. \pause To show this on a graph, we first need to plot the line $3x+3y\boldsymbol{=}120$. \begin{itemize} \item When $x=0$, we have \begin{eqnarray*} 3 \times 0 + 3y &=& 120 \qquad \textrm{\sf i.e.} \\ 3y &=& 120 \qquad \textrm{\sf i.e.} \\ y &=& 40. \end{eqnarray*} \item When $y=0$, we have \begin{eqnarray*} 3x + 3 \times 0 &=& 120 \qquad \textrm{\sf i.e.} \\ 3x &=& 120 \qquad \textrm{\sf i.e.} \\ x &=&40. \end{eqnarray*} \end{itemize} The points $(0,40)$ and $(40,0)$ have been plotted on the graph on page 16, and shading used to illustrate the inequality. } \frame{ \frametitle{Example 1.3 revisited (the chair manufacturer)} \begin{center} \includegraphics[angle=-90,width=12cm]{graph7.ps} \end{center} } \frame{ \frametitle{Example 1.3 revisited (the chair manufacturer)} Now consider the second inequality $2x + 4y \leq 150$. \pause Again, to show this on a diagram, we first need to plot the line $2x+4y\boldsymbol{=}150$.\pause \begin{itemize} \item When $x=0$, we have \pause \begin{eqnarray*} 2 \times 0 + 4y &=& 150 \qquad \textrm{\sf i.e.} \\ 4y &=& 150 \qquad \textrm{\sf i.e.} \\ y &=& 37.5. \end{eqnarray*} \item When $y=0$, we have \begin{eqnarray*} 2x + 4 \times 0 &=& 150 \qquad \textrm{\sf i.e.} \\ 2x &=& 150 \qquad \textrm{\sf i.e.} \\ x &=& 75. \end{eqnarray*} \end{itemize} \noindent On the same graph, you should plot the points $(0,37.5)$ and $(75,0)$, and shade accordingly. } \sol \frame{ \frametitle{Example 1.3 revisited (the chair manufacturer)} \begin{center} \includegraphics[angle=-90,width=12cm]{graph7a.ps} \end{center} } \frame{ \frametitle{Example 1.3 revisited (the chair manufacturer)} \begin{center} \includegraphics[angle=-90,width=12cm]{graph7b.ps} \end{center} } \frame{ \frametitle{Example 1.3 revisited (the chair manufacturer)} \begin{center} \includegraphics[angle=-90,width=12cm]{graph7c.ps} \end{center} } \frame{ \frametitle{Example 1.3 revisited (the chair manufacturer)} \begin{center} \includegraphics[angle=-90,width=12cm]{graph7d.ps} \end{center} } \normal \frame{ \frametitle{Example 1.3 revisited (the chair manufacturer)} We also need to respect the non--negativity constraints, and so we shade out the negative portion of the graph.} \sol \frame{ \frametitle{Example 1.3 revisited (the chair manufacturer)} \begin{center} \includegraphics[angle=-90,width=12cm]{graph7e.ps} \end{center} } \frame{ \frametitle{Example 1.3 revisited (the chair manufacturer)} \begin{center} \includegraphics[angle=-90,width=12cm]{graph7f.ps} \end{center} } \normal \frame{ \frametitle{Example 1.3 revisited (the chair manufacturer)} The unshaded region in your graph shows the feasible region associated with our set of inequalities. \vsp \textcolor{white}{What we must do now is find the point in that region which meets our objective -- i.e. the point in that region which \textcolor{white}{\bf maximises income}. } \vsp \textcolor{white}{One way of doing this is to also plot the \textcolor{white}{\bf objective function}. } } \sol \frame{ \frametitle{Example 1.3 revisited (the chair manufacturer)} \begin{center} \includegraphics[angle=-90,width=12cm]{graph7g.ps} \end{center} } \normal \frame{ \frametitle{Example 1.3 revisited (the chair manufacturer)} The unshaded region in your graph shows the feasible region associated with our set of inequalities. \vsp What we must do now is find the point in that region which meets our objective -- i.e. the point in that region which \textcolor{purple}{\bf maximises income}. \pause \vsp One way of doing this is to also plot the \textcolor{purple}{\bf objective function}. } \frame{ \frametitle{Example 1.3 revisited (the chair manufacturer)} Now our objective function is \pause \begin{eqnarray*} Z&=& 10x + 12y, \end{eqnarray*} where $Z$ is our income. \pause \vsp When $Z$ takes different values we get a family of \textcolor{purple}{\bf parallel straight lines}. \pause \vsp We need to choose a starting value for $Z$ in order to be able to plot the objective function. } \frame{ \frametitle{Example 1.3 revisited (the chair manufacturer)} It's often a good idea to try a value which is a multiple of both the coefficients of $x$ and $y$. \pause \vsp The coefficient of $x$ is 10 and the coefficient of $y$ is 12, so we could try a starting value of $Z=120$. \pause Thus, the objective function is now \begin{eqnarray*} 10x + 12y &=& 120. \end{eqnarray*} \pause We can plot this line in the same way as before -- i.e. consider what happens when $x$ and $y$ are zero.} \frame{ \frametitle{Example 1.3 revisited (the chair manufacturer)} \begin{itemize} \item When $x=0$, we have \pause \begin{eqnarray*} 10 \times 0 + 12y &=&120 \qquad \textrm{\sf i.e.} \\ 12y &=& 120 \qquad \textrm{\sf i.e.} \\ y &=&10. \pause \end{eqnarray*} \item When $y=0$, we have \begin{eqnarray*} \pause 10x + 12 \times 0 &=& 120 \qquad \textrm{\sf i.e.} \\ \pause 10x &=& 120 \qquad \textrm{\sf i.e.} \\ \pause x &=& 12. \end{eqnarray*} \end{itemize} \pause This line should also plotted on your graph. \begin{itemize} \item Notice that this line does not give the \textcolor{purple}{\bf optimal income}\pause \item The origin represents zero income, and we want to move as far away from this as possible. \end{itemize} } \sol \frame{ \frametitle{Example 1.3 revisited (the chair manufacturer)} \begin{center} \includegraphics[angle=-90,width=12cm]{graph7h.ps} \end{center} } \normal \frame{ \frametitle{Example 1.3 revisited (the chair manufacturer)} You should see from your graph that the point within the feasible region that is furthest away from the origin, \textcolor{purple}{\bf but parallel to the objective function}, is $\boldsymbol{x=5}$ and $\boldsymbol{y=35}$. \vsp \textcolor{white}{Thus, to maximise income, we should make \textcolor{white}{\bf 5 type A chairs} and \textcolor{white}{\bf 35 type B chairs}.} \vsp \textcolor{white}{This will give an income of} \begin{eqnarray*} && \\ && \\ && \\ && \end{eqnarray*} \textcolor{white}{i.e. \pounds 470.} } \sol \frame{ \frametitle{Example 1.3 revisited (the chair manufacturer)} \begin{center} \includegraphics[angle=-90,width=12cm]{graph7i.ps} \end{center} } \normal \frame{ \frametitle{Example 1.3 revisited (the chair manufacturer)} You should see from your graph that the point within the feasible region that is furthest away from the origin, \textcolor{purple}{\bf but parallel to the objective function}, is $\boldsymbol{x=5}$ and $\boldsymbol{y=35}$. \vsp Thus, to maximise income, we should make \textcolor{purple}{\bf 5 type A chairs} and \textcolor{purple}{\bf 35 type B chairs}. \pause \vsp This will give an income of \pause \begin{eqnarray*} Z &=& 10x+12y \qquad \textrm{\sf i.e.}\\ \pause Z &=& 10 \times 5 + 12 \times 35 \qquad \textrm{\sf i.e.} \\ \pause Z &=& 50 + 420 \\ \pause &=& 470, \pause \end{eqnarray*} i.e. \pounds 470. } \frame{ \frametitle{Example 1.4 revisited (\textit{Sportizus})} Recall the example about \textit{Sportizus}, a company who produce replica football and rugby shirts. \pause We summarised this problem as follows. \vsp Maximise $P=22x+16y$ subject to the following constraints: \begin{eqnarray*} 8x+5y &\leq& 600, \\ 9x+15y &\leq& 900, \\ x&\geq& 30 \qquad \textrm{\sf and} \\ y&\geq& 10, \end{eqnarray*} where $x$ and $y$ are the number of football and rugby shirts to make, respectively. \pause \vsp Let's now solve this problem graphically.} \sol \frame{ \frametitle{Example 1.4 revisited (\textit{Sportizus})} \noindent The first inequality is $8x+5y\leq 600$: \pause \begin{itemize} \item When $x=0$, we have \pause \begin{eqnarray*} 5y &=& 600 \qquad \textrm{\sf i.e.} \\ \pause y &=&120. \pause \end{eqnarray*} \item When $y=0$, we have \pause \begin{eqnarray*} 8x &=&600 \qquad \textrm{\sf i.e.}\\ \pause x&=& 75.\pause \end{eqnarray*} \end{itemize} So plot $(0,120)$, $(75,0)$. } \frame{ \frametitle{Example 1.4 revisited (\textit{Sportizus})} \begin{center} \includegraphics[angle=-90,width=12cm]{graph8a.ps} \end{center} } \frame{ \frametitle{Example 1.4 revisited (\textit{Sportizus})} \begin{center} \includegraphics[angle=-90,width=12cm]{graph8b.ps} \end{center} } \frame{ \frametitle{Example 1.4 revisited (\textit{Sportizus})} \begin{center} \includegraphics[angle=-90,width=12cm]{graph8c.ps} \end{center} } \frame{ \frametitle{Example 1.4 revisited (\textit{Sportizus})} The second inequality is $9x+15y\leq 900$: \pause \begin{itemize} \item When $x=0$, we have \pause \begin{eqnarray*} 15y &=&900 \qquad \textrm{\sf i.e.} \\ \pause y &=&60. \pause \end{eqnarray*} \item When $y=0$, we have \pause \begin{eqnarray*} 9x &=&900 \qquad \textrm{\sf i.e.} \\ \pause x &=&100. \end{eqnarray*} \end{itemize} So plot $(0,60)$ and $(100,0)$. } \frame{ \frametitle{Example 1.4 revisited (\textit{Sportizus})} \begin{center} \includegraphics[angle=-90,width=12cm]{graph8d.ps} \end{center} } \frame{ \frametitle{Example 1.4 revisited (\textit{Sportizus})} \begin{center} \includegraphics[angle=-90,width=12cm]{graph8e.ps} \end{center} } \frame{ \frametitle{Example 1.4 revisited (\textit{Sportizus})} \begin{center} \includegraphics[angle=-90,width=12cm]{graph8f.ps} \end{center} } \frame{ \frametitle{Example 1.4 revisited (\textit{Sportizus})} We also require $x\geq 30$ and $y\geq 10$. \pause \vsp Thus, on the graph, you should also shade out all $x$ values less than 30 and all $y$ values less than 10, to leave $x\geq 30$ and $y\geq 10$ \textit{unshaded}. } \frame{ \frametitle{Example 1.4 revisited (\textit{Sportizus})} \begin{center} \includegraphics[angle=-90,width=12cm]{graph8g.ps} \end{center} } \frame{ \frametitle{Example 1.4 revisited (\textit{Sportizus})} \begin{center} \includegraphics[angle=-90,width=12cm]{graph8h.ps} \end{center} } \frame{ \frametitle{Example 1.4 revisited (\textit{Sportizus})} \begin{center} \includegraphics[angle=-90,width=12cm]{graph8i.ps} \end{center} } \frame{ \frametitle{Example 1.4 revisited (\textit{Sportizus})} \begin{center} \includegraphics[angle=-90,width=12cm]{graph8j.ps} \end{center} } \frame{ \frametitle{Example 1.4 revisited (\textit{Sportizus})} \begin{center} \includegraphics[angle=-90,width=12cm]{graph8k.ps} \end{center} } \frame{ \frametitle{Example 1.4 revisited (\textit{Sportizus})} Finally, we need to plot the \textcolor{purple}{\bf objective function}. \pause Here, we have \[ P=22x+16y; \] \pause as with the previous example, we need a starting value for $P$ in order to plot this line. \pause \vsp You should notice that starting with $P=22\times 16 =352$ will give $(0,22)$ and $(16,0)$, and so we also plot this line on the graph. } \frame{ \frametitle{Example 1.4 revisited (\textit{Sportizus})} \begin{center} \includegraphics[angle=-90,width=12cm]{graph8l.ps} \end{center} } \frame{ \frametitle{Example 1.4 revisited (\textit{Sportizus})} \begin{center} \includegraphics[angle=-90,width=12cm]{graph8m.ps} \end{center} } \frame{ \frametitle{Example 1.4 revisited (\textit{Sportizus})} To maximise profit, $\boldsymbol{x=60}$ and $\boldsymbol{y=24}$, i.e. we should make \textcolor{purple}{\bf 60 football shirts} and \textcolor{purple}{\bf 24 rugby shirts} each day. \pause \vsp This will give a maximum daily profit of \pause \begin{eqnarray*} P &=&22x + 16y\\ \pause &=&22 \times 60 + 16 \times 24 \\ \pause &=&\textrm{\sf \pounds 1704}. \end{eqnarray*}} \normal \frame{ \frametitle{Algebraic solutions for linear programming problems} \noindent Notice that in both the examples we have looked at so far, the graphical representation indicates that the solution is at the intersection of two lines on the graph. \pause \vsp Thus, we could find also the solution \textcolor{purple}{\bf algebraically}, as the solution to a pair of \textcolor{purple}{\bf simultaneous equations}. } \frame{ \frametitle{Algebraic solutions for linear programming problems} For example, for \textit{Sportizus}, the solution was at the intersection of the two lines \begin{eqnarray*} 8x+5y &=& 600 \\ 9x+15y &=&900. \end{eqnarray*} \pause Multyplying the first equation by 3 will allow us to eliminate $y$ on subtraction: \pause \begin{eqnarray*} 24x + 15y &=& 1800 \\ \underline{9x + 15y} &=&\underline{900} \\ 15x + 0y &=& 900, \pause \end{eqnarray*} and so $x=900/15 = 60$. \pause Substituting this into the first equation gives:\pause \begin{eqnarray*} 480 + 5y &=& 600 \\ \pause 5y &=& 120 \\ \pause y &=&24. \end{eqnarray*} } \normal \frame{ \vspace{1cm} \huge \textbf{Chapter 2} \vspace{1cm} \Large \textbf{Optimisation using calculus} } \begin{comment} \frame{ \frametitle{Announcements} \begin{itemize} \item Case study work: report due in by \textbf{4pm today} \pause \vspl \begin{itemize} \item [--] 4 pages \pause \vspl \item [--] Cover sheet \pause \vspl \item [--] Maths \& Stats Stage 1 homework submission letterbox, 3rd floor Herschel \pause \vspl \item [--] No electronic submission \end{itemize} \vsp \item Workshops next week will focus on content from the lectures -- no case study work for two weeks \pause \item Lecture attendance register starts today: make sure you sign in! \end{itemize} } \normal \frame{ \frametitle{What we'll cover...} \begin{itemize} \item The gradient of a \textcolor{purple}{\bf curve}? Ideas of \textcolor{purple}{\bf rates of change}\pause \vspl \item The method of \textcolor{purple}{\bf differentiation}: first principles \pause \vspl \item Differentiation: \textcolor{purple}{\bf general results} for polynomials \pause \vspl \item Finding the derivative of non--linear functions from \textcolor{purple}{\bf real--life} scenarios \pause \vspl \item \textcolor{purple}{\bf Optimisation} using the differential calculus, and the role of the second derivative \pause \vspl \item Chapters 3 and 4 in the \textcolor{purple}{\bf course text} \end{itemize} } \normal \frame{ \frametitle{2. Optimisation using calculus} \noindent An important topic in many disciplines is the study of how quickly quantities \textcolor{purple}{\bf change over time}.\pause \vsp For example, in order to estimate the future demand for a commodity, we need information about \textcolor{purple}{\bf rates of change}.\pause \vsp As we shall see in this chapter, we can also use such information to solve optimisation problems: \pause for example, we might be interested in the number of items that need to be produced in order to \textcolor{purple}{\bf maximise profit} or \textcolor{purple}{\bf minimise loss}. } \frame{ \frametitle{2. Optimisation using calculus} \noindent We have already considered rates of change in Chapter 1. \pause \vsp In the equation of the straight line \[ y=mx + c, \pause \] the gradient $m$ represents the rate of change of $y$ with respect to $x$, and we thought about how to calculate $m$ using the formula \pause \[ \textrm{\sf Gradient}= \frac{\textrm{\sf Change in $y$}}{\textrm{\sf Change in $x$}}. \] } \frame{ \frametitle{Example} %\begin{figure}[!h] \begin{center} \includegraphics[width=8cm,angle=-90]{graph1.ps} %\caption{Lee} \end{center}} \frame{ \frametitle{2. Optimisation using calculus} In Chapter 1 we also considered some \textcolor{purple}{\bf non--linear functions}, focussing primarily on quadratic functions. \pause \vsp We will begin Chapter 2 with an examination of how to find the \textcolor{purple}{\bf gradient of a non--linear function}, before thinking about how this can be used to aid the process of \textcolor{purple}{\bf optimisation}. } \frame{ \frametitle{2.1 Differentiation} The concept used to describe the rate of change of a function (linear or non--linear) is the \textcolor{purple}{\bf derivative}, which is \underline{the} central concept in mathematical analysis. \vsp In this section we will define the derivative of a function, as well as present some of the important rules for calculating it. \vsp \textcolor{purple}{\bf Isaac Newton} (1642--1727) and \textcolor{purple}{\bf Gottfried Leibniz} (1646--1716) discovered most of these general rules independently of each other. \vsp This initiated the \textcolor{purple}{\bf differential calculus}, which has been the foundation for the development of modern science. \vsp It has also been of central importance to the theoretical development of modern economics. } \frame{ \frametitle{2.1.1 The gradient of a curve} A curve does not have a constant gradient -- its direction is \textcolor{purple}{\bf continuously changing}, and so its gradient will continuously change too. \pause \vsp So we look at the gradient of the curve at a \textcolor{purple}{\bf particular point on the curve}, rather than calculate the gradient of an entire line as with a linear function. } \frame{ \frametitle{2.1.1 The gradient of a curve} Look at the graph of $y=x^{2}$ shown below. \pause \vsp Let's suppose we are interested in the gradient of this curve at the point $x=3$. \pause \vsp Informally, we can say a couple of things about the gradient at this point:\pause \vspl \begin{itemize} \item The gradient will be \textcolor{purple}{\bf positive} -- as you look at the graph from left to right at this point, the curve goes uphill \pause \vspl \item The graph is \textcolor{purple}{\bf symmetric} about the point $x=0$: the gradient at $x=-3$ will be the same as that at $x=3$ (just negative)\pause \vspl \item As we move from $x=3$ to $x=0$, the gradient gets \textcolor{purple}{\bf less and less steep}, and ``flattens out'' completely at $x=0$ \end{itemize} } \frame{ \frametitle{2.1.1 The gradient of a curve} \begin{center} \includegraphics[width=8cm,angle=-90]{diff1.ps} %\caption{Lee} \end{center} } \frame{ \frametitle{2.1.1 The gradient of a curve} We know how to find the gradient of a straight line from Chapter 1, so we can \textcolor{purple}{\bf approximate} the gradient of the curve at $x=3$ by drawing a chord on the curve. \pause \vsp For example, draw a chord on the curve between the points $x=3$ and $x=5$, and find the gradient of this chord \textit{[the co-ordinates will be (3,9) and (5,25)]}: {\color{white} \[ \textrm{\sf Gradient} = \frac{25-9}{5-3} = \frac{16}{2} = 8. \]} } \sol \frame{ \frametitle{2.1.1 The gradient of a curve} \begin{center} \includegraphics[width=8cm,angle=-90]{diff2.ps} %\caption{Lee} \end{center} } \normal \frame{ \frametitle{2.1.1 The gradient of a curve} We know how to find the gradient of a straight line from Chapter 1, so we can \textcolor{purple}{\bf approximate} the gradient of the curve at $x=3$ by drawing a chord on the curve. \vsp For example, draw a chord on the curve between the points $x=3$ and $x=5$, and find the gradient of this chord \textit{[the co-ordinates will be (3,9) and (5,25)]}: \pause \[ \textrm{\sf Gradient} = \frac{25-9}{5-3} =\pause \frac{16}{2} = 8. \] } \frame{ \frametitle{2.1.1 The gradient of a curve} Now decrease the length of the chord and bring it \textcolor{purple}{\bf closer to the curve} -- for example, draw the chord between the points $x=3$ and $x=4$ -- and find the gradient \textit{[The co-ordinates will be (3,9) and (4,16)]}: {\color{white} \[ \textrm{\sf Gradient} = \frac{16-9}{4-3} = \frac{7}{1} = 7. \]} } \sol \frame{ \frametitle{2.1.1 The gradient of a curve} \begin{center} \includegraphics[width=8cm,angle=-90]{diff3.ps} %\caption{Lee} \end{center} } \normal \frame{ \frametitle{2.1.1 The gradient of a curve} Now decrease the length of the chord and bring it \textcolor{purple}{\bf closer to the curve} -- for example, draw the chord between the points $x=3$ and $x=4$ -- and find the gradient \textit{[The co-ordinates will be (3,9) and (4,16)]}: \pause \[ \textrm{\sf Gradient} = \frac{16-9}{4-3} = \pause \frac{7}{1} = 7. \] } \normal \frame{ \frametitle{2.1.1 The gradient of a curve} Now decrease the length of the chord again and bring it \textcolor{purple}{\bf even closer to the curve} -- draw the chord between the points $x=3$ and $x=3.5$ -- and find the gradient \textit{[The co-ordinates will be (3,9) and (3.5,12.25)]}: {\color{white} \[ \textrm{\sf Gradient} = \frac{12.25-9}{3.5-3} = \pause \frac{3.25}{0.5} = 6.5. \]}} \sol \frame{ \frametitle{2.1.1 The gradient of a curve} \begin{center} \includegraphics[width=8cm,angle=-90]{diff4.ps} %\caption{Lee} \end{center} } \frame{ \frametitle{2.1.1 The gradient of a curve} Now decrease the length of the chord again and bring it \textcolor{purple}{\bf even closer to the curve} -- draw the chord between the points $x=3$ and $x=3.5$ -- and find the gradient \textit{[The co-ordinates will be (3,9) and (3.5,12.25)]}: \pause \[ \textrm{\sf Gradient} = \frac{12.25-9}{3.5-3} = \pause \frac{3.25}{0.5} = 6.5. \] } \sol \frame{ \frametitle{2.1.1 The gradient of a curve} \begin{center} \begin{tabular}{|c|c|c|} \hline Point of interest &Point 2 &Gradient \\ \hline $(3,9)$ &$(5,25)$ &$\displaystyle \frac{25-9}{5-3} = \textbf{8}$ \\ & & \\ $(3,9)$ &$(4,16)$ &$\displaystyle \frac{16-9}{4-3} = \textbf{7}$ \\ & & \\ $(3,9)$ &$(3.5,12.25)$ &$\displaystyle \frac{12.25-9}{3.5-3}=\textbf{6.5}$ \\ \pause & & \\ $(3,9)$ &$(3.25,10.5625)$ &$\displaystyle \frac{10.5625-9}{3.25-3} = \textbf{6.25}$\\ \pause & & \\ $(3,9)$ &$(3.1,9.61)$ &$\displaystyle \frac{9.61-9}{3.1-3} = \textbf{6.1}$ \\ \pause & & \\ $(3,9)$ &$(3.01,9.0601)$ &$\displaystyle \frac{9.0601-9}{3.01-3} = \textbf{6.01}$ \\ & & \\ \hline \end{tabular} \end{center}} \normal \frame{ \frametitle{2.1.1 The gradient of a curve} So it looks like at $x=3$ the gradient converges to 6. \pause \vsp We can \textcolor{purple}{\bf generalise} this approach \textcolor{purple}{\bf algebraically}.\pause \vsp Suppose the point of interest has co--ordinates $(x,x^{2})$ and point 2 has an $x$ co--ordinate close to that of the point of interest, say $x+\delta$. Then the $y$ co--ordinate of point 2 is \[ y = (x+\delta)^{2}. \]} \frame{ \frametitle{2.1.1 The gradient of a curve} Thus, the gradient is given by \pause \begin{eqnarray*} \textrm{\sf Gradient} &=& \frac{\textrm{\sf Change in $y$}}{\textrm{\sf Change in $x$}} \\ \pause & & \\ &=& \frac{(x+\delta)^{2} - x^{2}}{(x+\delta)-x} \\ & & \\ \pause &=& \frac{(x+\delta)(x+\delta)-x^{2}}{x + \delta - x} \\ & & \\ \pause &=& \frac{x^{2}+2\delta x + \delta^{2} - x^{2}}{\delta} \\ & &\\ \pause &=& \frac{2\delta x + \delta^{2}}{\delta} = 2x+\delta. \end{eqnarray*}} \frame{ \frametitle{2.1.1 The gradient of a curve} Now if we let $\delta$ get smaller and smaller, i.e. ``tend to zero'', we are left with just \pause \[ \textrm{\sf Gradient} = 2x. \]\pause So the gradient, or \textcolor{purple}{\bf derivative}, of the curve $y=x^{2}$ is $2x$. } \frame{ \frametitle{2.1.1 The gradient of a curve} We usually denote this quantity as $\displaystyle \frac{dy}{dx}$, pronounced ``dee y by dee x'', and so when \pause \begin{eqnarray*} y &=& x^{2} \\ & & \\ \pause \frac{dy}{dx} &=& 2x. \end{eqnarray*} } \frame{ \frametitle{2.1.1 The gradient of a curve} We can now use this derivative to find the gradient of the curve at any point $x$. \pause \vsp For example, when $x=3$, the gradient is \[ \frac{dy}{dx} = 2x = 2\times 3 = 6, \] \pause exactly the answer we found from first principles in the table on page 41! } \sol \frame{ \frametitle{2.1.1 The gradient of a curve} What is the gradient of the curve $y=x^{2}$ at the points $x=4$, $x=1$, $x=0$ and $x=-4$? Do your answers make sense in relation to the graph of this function? \vspace{1cm} \textbf{Solution} \begin{eqnarray*} x=4 &:& \frac{dy}{dx} = 2\times 4 = 8 \pause \\ & & \\ x=1&:& \frac{dy}{dx} = 2\times 1 = 2 \pause \\ & & \\ x=0&:& \frac{dy}{dx} = 2\times 0 =0 \pause \\ & & \\ x=-1 &:& \frac{dy}{dx} = 2\times -1 = -2. \end{eqnarray*} } \frame{ \frametitle{2.1.1 The gradient of a curve} Looking at the graph, these values make perfect sense: as we approach zero, we can see the curve becomes less and less steep, until at the origin the gradient is zero. \pause \vsp Also, the curve is symmetric about $x=0$ and so the gradient at $x=\pm 1$ will be the same, just negated. } \normal \frame{ \frametitle{2.1.2 Some general results for polynomials} In the previous section we found that, when \begin{eqnarray*} y &=&x^{2}, \\ & & \\ \frac{dy}{dx} &=& 2x. \end{eqnarray*} \pause Notice that the power of $x$ has been ``brought down'', in front of the $x$, and the power itself has reduced by one ($2x$ is actually $2x^{1}$). } \frame{ \frametitle{2.1.2 Some general results for polynomials} In general, when \begin{eqnarray*} y &=& x^{n}, \\ & & \\ \pause \frac{dy}{dx} &=&nx^{n-1}. \end{eqnarray*} \pause More generally still, if \begin{eqnarray*} y &=& kx^{n}, \qquad \textrm{\sf then} \\ & & \\ \pause \frac{dy}{dx} &=&nkx^{n-1}. \end{eqnarray*}} \frame{ \frametitle{2.1.2 Examples} Find $\displaystyle \frac{dy}{dx}$ for each of the following: \vspl \begin{itemize} \item [\textbf{1.}] $y=x^{3}$ \item [\textbf{2.}] $y=x^{5}$ \item [\textbf{3.}] $y=2x^{3}$ \item [\textbf{4.}] $y=4x^{3}-7x^{2}+2x$ \item [\textbf{5.}] $y=5x^{2}+2$ \item [\textbf{6.}] $y=\sqrt{x}+1$ \item [\textbf{7.}] $\displaystyle y=\frac{1}{x}$ \end{itemize} } \sol \frame{ \frametitle{Solution to examples} \begin{itemize} \item [\textbf{1.}] We have $y=x^{3}$. Bringing the power down, and then reducing the power by one, gives \[ \frac{dy}{dx} = 3x^{2}. \] \pause \vsp \item [\textbf{2.}] We have $y=x^{5}$. Bringing the power down, and then reducing the power, gives \pause \[ \frac{dy}{dx} = 5x^{4}. \] \end{itemize} } \frame{ \frametitle{Solution to examples} \begin{itemize} \item [\textbf{3.}] We have $y=2x^{3}$. Bringing the power down, and then reducing the power by one, gives \[ \frac{dy}{dx} = \textbf{3}\times 2x^{2} = 6x^{2}. \]\pause \vsp \item [\textbf{4.}] We have $y=4x^{3}-7x^{2}+2x^{1}$. Bringing the powers of $x$ down, and then reducing the powers, gives \pause \begin{eqnarray*} \frac{dy}{dx} &=& \textbf{3}\times 4x^{2} \pause - \textbf{2}\times 7x^{1}\pause + \textbf{1}\times 2x^{0} \\ &=&12x^{2} \pause - 14x \pause + 2. \end{eqnarray*} \end{itemize} } \frame{ \frametitle{Solution to examples} \begin{itemize} \item [\textbf{5.}] We have $y=5x^{2}+2$, \pause i.e. $y=5x^{2}+2x^{0}$. Thus \pause \begin{eqnarray*} \frac{dy}{dx} &=& \textbf{2}\times 5x^{1} + \textbf{0}\times 2x^{-1} \pause \\ &=& 10x, \end{eqnarray*} \pause \textcolor{purple}{\bf so constants just differentiate to zero!} \end{itemize} } \frame{ \frametitle{Solution to examples} \begin{itemize} \item [\textbf{6.}] We have $y=\sqrt{x}+1 \pause = x^{0.5}+1$. Thus \pause \begin{eqnarray*} \frac{dy}{dx} &=& 0.5x^{-0.5} \pause \\ &=& \frac{1}{2} x^{-\frac{1}{2}} \pause \\ & & \\ &=& \frac{1}{2}\times \frac{1}{x^{\frac{1}{2}}} \pause \\ & & \\ &=&\frac{1}{2} \times \frac{1}{\sqrt{x}} \pause \\ & & \\ &=& \frac{1}{2\sqrt{x}}. \end{eqnarray*} \end{itemize} } \frame{ \frametitle{Solution to examples} \begin{itemize} \item [\textbf{7.}] We have $\displaystyle y=\frac{1}{x} \pause = x^{-1}$. Thus \pause \begin{eqnarray*} \frac{dy}{dx} &=& -1x^{-2} \pause \\ &=& -\frac{1}{x^{2}}. \end{eqnarray*} \end{itemize} } \normal \frame{ \frametitle{Example 2.1} The managing director of a company is interested in modelling the relationship between her company's annual profits (\pounds $P$ million) and their annual advertising budget (\pounds $A$ thousand). \vsp The graph below shows how the managing director believes $P$ changes with $A$. } \frame{ \frametitle{Example 2.1} \begin{center} \psfrag{0}[]{\sf \tiny 0} \psfrag{10}[]{\sf \tiny 10} \psfrag{20}[]{\sf \tiny 20} \psfrag{30}[]{\sf \tiny 30} \psfrag{40}[]{\sf \tiny 40} \psfrag{-400}[]{\sf \tiny -400} \psfrag{-300}[]{\sf \tiny -300} \psfrag{-200}[]{\sf \tiny -200} \psfrag{-100}[]{\sf \tiny -100} \psfrag{A}[]{\sf \tiny \pounds $A$ thousand} \psfrag{P}[]{\sf \tiny \pounds $P$ million} \includegraphics[width=12cm,height=7cm]{example3.1.ps} \end{center} } \frame{ \frametitle{Example 2.1} \begin{itemize} \item [\textbf{(a)}] You work as part of a team of business analysts for this company. A colleague proposes the following three profit functions: \begin{itemize} \item [*]$P = 80-4.3A$ \item [*]$P= 2.1+13.5A+0.5A^{2}$ \item [*]$P=2.1+6.3A-0.13A^{2}+0.0008A^{3}-0.0002A^{4}$ \end{itemize} \vsp Which of these do you think would be most appropriate in this situation? Why? \end{itemize} } \sol \frame{ \frametitle{Solution to Example 2.1(a)} We clearly do not have a linear function, and so this rules out the first. \pause \vsp Our graph shows that the function will not be symmetric, and so we cannot have a quadratic. \pause \vsp Thus the quartic must be the most suitable function. } \normal \frame{ \frametitle{Example 2.1} \begin{itemize} \item [\textbf{(b)}] For the profit function you have chosen in part (a), find $\displaystyle \frac{dP}{dA}$. \end{itemize} } \sol \frame{ \frametitle{Solution to Example 2.1(b)} We have $P=2.1+6.3A^{1}-0.13A^{2}+0.0008A^{3}-0.0002A^{4}$. \vsp Thus \pause \begin{eqnarray*} \frac{dP}{dA} &=& \textbf{1}\times 6.3A^{0}\pause -\textbf{2}\times 0.13A^{1}\pause + \textbf{3}\times 0.0008A^{2} \pause - \textbf{4}\times 0.0002A^{3} \pause \\ & & \\ &=&6.3 - 0.26A + 0.0024A^{2} - 0.0008A^{3}. \end{eqnarray*} } \normal \frame{ \frametitle{Example 2.1} \begin{itemize} \item [\textbf{(c)}] Use your answer to part (b) to find the gradient of your profit function when the company spends (i) \pounds 10,000; (ii) \pounds 15,250; (iii) \pounds 20,000 and (iv) 40,000 on advertising each year. Do these values match up with what you see in the graph? \end{itemize} } \sol \frame{ \frametitle{Solution to Example 2.1(c)} {\scriptsize \begin{eqnarray*} A=10 &:& \pause \frac{dP}{dA} = 6.3-0.26\times 10 + 0.0024\times 10^2 -0.0008\times 10^3 = \textbf{3.14} \\ \pause & & \\ A=15.25 &:& \pause \frac{dP}{dA} = ... = \textbf{0.056} \pause \\ & & \\ A=20 &:& \pause \frac{dP}{dA} = ... = \boldsymbol{-4.34} \pause \\ & & \\ A=40 &:& \pause \frac{dP}{dA} = ... = \boldsymbol{-51.46} \end{eqnarray*}} } \normal \frame{ \frametitle{Example 2.1} \begin{itemize} \item [\textbf{(d)}] What will be the value of $\frac{dP}{dA}$ when the company optimises their advertising budget (i.e. maximises profit)? \end{itemize}} \sol \frame{ \frametitle{Solution to Example 2.1(d)} From the graph, we can see that the company will maximise their profit at the peak of the curve. \pause \vsp This is known as a \textcolor{purple}{\bf turning point} in the graph. \pause \vspl \begin{itemize} \item \textbf{Before} this point, the gradient ($\frac{dP}{dA}$) is positive \pause \vspl \item \textbf{After} this point, the gradient ($\frac{dP}{dA}$) is negative \pause \vspl \item \textbf{At} this point $\frac{dP}{dA} = 0$ \end{itemize} } \begin{comment} \frame{ \frametitle{Announcements} \begin{itemize} \item \textbf{Case study 1} \pause \vspl \begin{itemize} \item [--] Currently being marked \pause \vspl \item [--] Marks so far look generally good! \end{itemize} \pause \vsp \item \textbf{Prize question winners!} \pause \vspl \begin{itemize} \item [--] Abigail Farnworth \pause \vspl \item [--] Mingrui Li \pause \vspl \item [--] Wine, text book, University-approved calculator \end{itemize} \end{itemize} } \normal \normal \frame{ \frametitle{Example 2.2} A small firm employs five machine operators. \vspl For a particular contract, the firm believes it will produce $Q=\frac{1}{2}\sqrt M$ units of a commodity, where $M$ machine operators are used. \vspl The cost, per operator, is \euro 40 and the price obtained per unit is \euro 160. \vsp \begin{itemize} \item [\textbf{(a)}] Formulate a non--linear function for the firm's profit, $\pi$, in terms of $M$ only. \end{itemize} } \sol \frame{ \frametitle{Solution to Example 2.2(a)} \begin{eqnarray*} \textrm{\sf Profit} &=& \textrm{\sf Total income} - \textrm{\sf Total costs} \\ & & \\ \pi &=& 160Q -40 M \pause \\ & & \\ &=& 160\left(\frac{1}{2}\sqrt{M}\right) - 40M \pause \\ & & \\ &=&80\sqrt{M} - 40M. \end{eqnarray*} } \normal \frame{ \frametitle{Example 2.2} \begin{itemize} \item [\textbf{(b)}] Find $\frac{d \pi}{dM}$, and solve $\frac{d\pi}{dM} = 0$ for $M$. Comment. \end{itemize} } \sol \frame{ \frametitle{Solution to Example 2.2(b)} We have $\pi = 80\sqrt{M} - 40M = 80M^{0.5}-40M$. Thus \pause \begin{eqnarray*} \frac{d\pi}{dM} &=& 0.5\times 80M^{-0.5}-40 \pause \\ &=& 40 \times \frac{1}{M^{0.5}} - 40 \pause \\ &=& \frac{40}{\sqrt{M}} - 40. \end{eqnarray*} \pause When $\frac{d\pi}{dM} = 0$ we get \pause \begin{eqnarray*} \frac{40}{\sqrt{M}} &=& 40 \pause \\ & & \\ 40 &=&40\sqrt{M} \pause \\ 1 &=& \sqrt{M}, \end{eqnarray*} and so $M=1$. \pause This means that the turning point in our graph will occur when we use 1 machine operator.} \normal \frame{ \frametitle{Example 2.2} \begin{itemize} \item [\textbf{(c)}] Shown below is a plot of $\pi$ against $M$. Can you see how this corresponds to your answer to part (b)? \begin{center} \psfrag{0}[]{\sf \tiny 0} \psfrag{1}[]{\sf \tiny 1} \psfrag{2}[]{\sf \tiny 2} \psfrag{3}[]{\sf \tiny 3} \psfrag{4}[]{\sf \tiny 4} \psfrag{5}[]{\sf \tiny 5} \psfrag{-20}[]{\sf \tiny -20} \psfrag{-10}[]{\sf \tiny -10} \psfrag{10}[]{\sf \tiny 10} \psfrag{20}[]{\sf \tiny 20} \psfrag{30}[]{\sf \tiny 30} \psfrag{40}[]{\sf \tiny 40} \psfrag{L}[]{\sf \tiny $M$} \psfrag{P}[]{\sf \tiny $\pi$} \includegraphics[width=11cm,height=6cm]{example3.2.ps} \end{center} \end{itemize} } \sol \frame{ \frametitle{Solution to Example 2.2(c)} It is clear from the graph that the turning point in this function occurs at $M=1$ -- our calculations show that this is the point at which the gradient is zero. } \normal \frame{ \frametitle{2.2 Optimisation using differentiation} \noindent Suppose a company believes there is a non--linear relationship between it's monthly advertising budget (\pounds $x$ thousand) and their monthly profit (\pounds $y$ thousand). \pause \vsp In particular, their analyst believes that the following cubic function explains how $y$ varies with $x$: \[ y = -7.36+9.42x-1.17x^{2}+0.04x^{3}. \] } \frame{ \frametitle{2.2 Optimisation using differentiation} The company's monthly advertising budget cannot exceed \pounds 17,000; their current contractual arrangements for commercial TV and radio advertising means they must always spend at least \pounds 4,000 on advertising every month. \pause \vsp A graph of this cubic, between the values of $x=4$ and $x=17$, is shown overleaf. } \frame{ \frametitle{2.2 Optimisation using differentiation} \begin{center} \psfrag{0}[]{\sf \tiny 0} \psfrag{2}[]{\sf \tiny 2} \psfrag{4}[]{\sf \tiny 4} \psfrag{6}[]{\sf \tiny 6} \psfrag{8}[]{\sf \tiny 8} \psfrag{10}[]{\sf \tiny 10} \psfrag{12}[]{\sf \tiny 12} \psfrag{14}[]{\sf \tiny 14} \psfrag{16}[]{\sf \tiny 16} \psfrag{x}[]{\sf \tiny $x$} \psfrag{y}[]{\sf \tiny $y=-7.36+9.42x-1.17x^{2}+0.04x^{3}$} \includegraphics[width=11cm,height=6cm]{cubic.ps} \end{center} } \frame{ \frametitle{2.2 Optimisation using differentiation} Some things to note from the graph: \pause \begin{itemize} \item There are two \textcolor{purple}{\bf turning points} on the graph -- one at about $x=6$ and the other at about $x=14$; \pause \vspl \item One is known as a \textcolor{purple}{\bf local maximum}, the other is a \textcolor{purple}{\bf local minimum}; \pause \vspl \item In this example, one of these points corresponds to maximum profit, the other corresponds to minimum profit; \pause \vspl \item At both turning points, the gradient of the curve -- and hence $\displaystyle \frac{dy}{dx}$ -- is \textcolor{purple}{\bf zero}. \end{itemize} } \frame{ \frametitle{2.2 Optimisation using differentiation} To find the exact $x$ co--ordinate of each turning point, we can equate the derivative to zero and solve for $x$. \pause \vsp For example, we know that \pause \begin{eqnarray*} \frac{dy}{dx} &=& 9.42-2\times1.17x + 3\times 0.04x^{2}\\ \pause &=& 9.42 -2.34x+0.12x^{2}. \end{eqnarray*} \pause At each of the turning points shown in the graph, we know that the gradient is zero. \pause Thus, we know that \[ 9.42 - 2.34x +0.12x^{2} = 0, \] and we know how to solve such a quadratic equation from the material in Chapter 1! } \frame{ \frametitle{2.2 Optimisation using differentiation} Using the quadratic formula, we know that \[ x = \frac{-b \pm \sqrt{D}}{2a}, \] \pause where $D$ is the discriminant and is equal to $b^{2}-4ac$; $a$ and $b$ are the coefficients of the $x^{2}$ and $x$ terms, respectively, and $c$ is the constant. \pause \vsp In this example, $a=0.12$, $b=-2.34$ and $c=9.42$, giving \pause \[ D = (-2.34)^{2} - (4\times 0.12 \times 9.42) = 0.954, \] } \frame{ \frametitle{2.2 Optimisation using differentiation} and so \[ x = \frac{2.34\pm\sqrt{0.954}}{0.24}, \pause \] giving \pause \begin{eqnarray*} x &=& \frac{2.34+\sqrt{0.954}}{0.24} = 13.81971, \qquad \textrm{\sf or} \\ \pause & & \\ x &=& \frac{2.34-\sqrt{0.954}}{0.24} = 5.680295. \end{eqnarray*} } \frame{ \frametitle{2.2 Optimisation using differentiation} These are the \textcolor{purple}{\bf precise} values of $x$ at which the curve has zero gradient -- i.e. the advertising expenditure that will minimise/maximise profit -- and this can be seen in the graph above. \pause \vsp So, spending \pounds 5,680 per month on advertising seems to be the optimal strategy, which will give monthly profits of \pause \[ y = -7.36+9.42(5.680)-1.17(5.680^{2})+0.04(5.680^{3}) = \textsf{\pounds} 15, 729. \] } \frame{ \frametitle{2.2 Optimisation using differentiation} Spending \pounds 13,820 per month on advertising seems to give the worst outcome, with monthly profits of just \pause \[ y = -7.36+9.42(13.820)-1.17(13.820^{2})+0.04(13.820^{3}) = \textsf{\pounds} 4,944. \] } \frame{ \frametitle{2.2 Optimisation using differentiation} It is obvious from our graph which of the turning points gives the maximum profit and which gives the minimum profit. \pause \vsp However, if we didn't have a graph of the function, the \textcolor{purple}{\bf second derivative test} can help us here. \pause \vsp The second derivative of a function, sometimes written as \[ \frac{d^{2}y}{dx^{2}}, \] (``dee 2 y by dee x squared'') is just the derivative of the derivative. \pause } \frame{ \frametitle{2.2 Optimisation using differentiation} If $x$ represents the $x$ co--ordinate of our turning point, i.e. $\displaystyle \frac{dy}{dx}(x) = 0$, then: \pause \begin{itemize} \item If $\displaystyle \frac{d^{2}y}{dx^{2}}(x)<0$, our turning point is a \textcolor{purple}{\bf maximum}; \pause \vspl \item If $\displaystyle \frac{d^{2}y}{dx^{2}}(x)>0$, our turning point is a \textcolor{purple}{\bf minimum}. \end{itemize} } \frame{ \frametitle{2.2 Optimisation using differentiation} In this example, \[ \frac{dy}{dx}=9.42 -2.34x+0.12x^{2}. \] \pause Thus, \[ \frac{d^{2}y}{dx^{2}} = -2.34+0.24x. \] \pause At the first turning point, $x=5.680295$, giving \pause \[ \frac{d^{2}y}{dx^{2}} = -2.34+0.24\times 5.680295 = -0.9767292, \] \pause which is negative, and so here we have a \textcolor{purple}{\bf maximum turning point}. } \frame{ \frametitle{2.2 Optimisation using differentiation} At the second turning point $x=13.81971$, giving \pause \[ \frac{d^{2}y}{dx^{2}} = -2.34+0.24\times 13.81971 = 0.9767304, \] \pause which is positive, and so here we have a \textcolor{purple}{\bf minimum turning point}. } \frame{ \frametitle{2.2 Optimisation using differentiation} \textbf{To summarise:} \pause \begin{itemize} \item There are two turning points: one at $x=5.680295$ and one at $x=13.81971$ \pause \vspl \item At $x=5.680295$, the second derivative is negative and so we have a local maximum \pause \vspl \item At $x=13.81971$ the second derivative is positive and so we have a local minimum \pause \vspl \item The local maximum/minimum correspond to maximum/minimum profit, and so spending \pounds 5,680 pounds on advertising is the optimal strategy, giving profits of just under \pounds 16,000. \pause \vspl \item The worst strategy is to spend \pounds 13,820 on advertising, which will give a profit of just under \pounds 5,000. \end{itemize} } \frame{ \frametitle{Example 2.3} \textit{Choctastic!} are a chocolate manufacturer. \vsp One of their gift boxes has 27 chocolates in it, requiring a volume of 1350 cubic centimetres. \vsp The company would like to work out the optimal dimensions of the required box to minimise the amount of packaging used (and hence minimise production costs). } \frame{ \frametitle{Example 2.3} The diagram below shows a ``net'' of the box that will be used; $x$ and $y$ are both in centimetres. \begin{center} \includegraphics[height=8cm,angle=-90]{choc.ps} \end{center} } \sol \frame{ \frametitle{Example 2.3: Solution to part (a)} \begin{itemize} \item [\textbf{(a)}] Write down an expression for the volume of the box, in terms of $x$ and $y$. \pause \vspace{0.75cm} \textbf{Solution} \vsp The volume is given by \pause \begin{eqnarray*} \textsf{length} \times \textsf{width} \times \textsf{height} &=& 1350 \pause \\ & &\\ y \times x \times x &=& 1350 \pause \\ x^{2}y &=& 1350. \end{eqnarray*} \end{itemize} } \frame{ \frametitle{Example 2.3: Solution to part (b) (1/2)} \begin{itemize} \item [\textbf{(b)}] Formulate an expression for the surface area $S$ of the box, in terms of $x$ only. \pause \vspace{0.75cm} \textbf{Solution} \pause \vsp Looking at the net, there are four rectangles with area $xy$ and two squares with area $x^{2}$. So the total surface area is \pause \[ S = 4xy+2x^{2}. \] \end{itemize} } \frame{ \frametitle{Example 2.3: Solution to part (b) (2/2)} Now from (a), we know that \pause \[ y = \frac{1350}{x^{2}}. \] \pause Substituting this into the equation for $S$ gives \pause \begin{eqnarray*} S &=& 4x\left(\frac{1350}{x^{2}}\right)+2x^{2} \pause \\ & & \\ &=& \frac{4x \times 1350}{x^{2}} + 2x^{2} \pause \\ & & \\ &=& \frac{5400}{x} + 2x^{2} \pause = 5400x^{-1}+2x^{2}. \end{eqnarray*} } \normal \frame{ \frametitle{Example 2.3} \begin{itemize} \item [\textbf{(c)}] Find $\displaystyle \frac{dS}{dx}$, and hence show that the optimal strategy is to use a cuboid box for the chocolates. What are the dimensions of this cuboid? \end{itemize} } \sol \frame{ \frametitle{Example 2.3: Solution to part (c) (1/3)} We have \[ S = 5400x^{-1}+2x^{2}. \]\pause Thus, \[ \frac{dS}{dx} = \pause -5400x^{-2} \pause +4x. \]\pause At turning points the gradient is zero. Thus \pause \begin{eqnarray*} -\frac{5400}{x^{2}}+4x &=& 0 \pause \\ & & \\ 4x &=&\frac{5400}{x^{2}} \end{eqnarray*} } \frame{ \frametitle{Example 2.3: Solution to part (c) (2/3)} So \pause \begin{eqnarray*} 4x^{3} &=&5400 \pause \\ x^{3} &=& 1350 \pause \\ x &=&^{3}\sqrt{1350} \pause = 11.052 \textsf{ cm}. \end{eqnarray*} } \frame{ \frametitle{Example 2.3: Solution to part (c) (3/3)} When $x=11.052$, \pause \[ y = \frac{1350}{x^{2}} = \frac{1350}{11.052^{2}} = 11.052 \textsf{ cm}; \] with reference to the net, we see this will give a cuboid shape for the box! } \normal \frame{ \frametitle{Example 2.3} \begin{itemize} \item [\textbf{(d)}] Use the second derivative test to show that you have minimised the amount of packaging used. \end{itemize} } \sol \frame{ \frametitle{Example 2.3: Solution to part (d) (1/1)} We know that \[ \frac{dS}{dx} = -5400x^{-2}+4x. \]\pause Thus \pause \[ \frac{d^{2}S}{dx^{2}} = 10800x^{-3}+4. \] \pause At our turning point, $x=11.052$, giving \pause \[ \frac{d^{2}S}{dx^{2}} = \frac{10800}{11.052^{3}}+4 = 12, \] \pause which is positive -- this confirms that our turning point is a \textcolor{purple}{\bf minimum}. } \normal \frame{ \frametitle{2.3 Partial differentiation} As you will see in some of your other courses, most relationships in Economics, Accounting and Finance involve \textcolor{purple}{\bf more than two variables}. \vsp For example, the demand for a good depends not only on its own price but also on the price of substitutable or complementary goods, incomes of consumers, advertising expenditure and so on. \vsp Likewise, the output from a production process depends on a variety of inputs, including land, capital and labour. \vsp To analyse general economic behaviour we must extend the concept of differentiation to several variables. } \sol \frame{ \frametitle{Examples} \begin{itemize} \item [\textbf{1.}] If $f(x,y) = xy+2y$, evaluate \begin{itemize} \item [(a)] $f(3,4)$; \item [(b)] $f(4,3)$. \end{itemize} \end{itemize} \pause \begin{eqnarray*} f(3,4) &=& 3\times 4 + 2\times 4 = 20 \\ \pause f(4,3) &=& 4 \times 3 + 2\times 3 =18 \end{eqnarray*} } \frame{ \frametitle{Examples} \begin{itemize} \item [\textbf{2.}] If $g(x_1,x_2,x_3)=x_{1}^2+x_{2}-3x_{3}$, evaluate \begin{itemize} \item [(a)] $g(5,6,10)$; \item [(b)] $g(0,0,0)$. \end{itemize} \end{itemize} \pause \begin{eqnarray*} g(5,6,10) &=& 5^2+6-(3\times 10) = 1 \\ \pause g(0,0,0) &=& 0. \end{eqnarray*} } \normal \frame{ \frametitle{2.3 Partial differentiation} Given a function of two variables, for example \[ z=f(x,y), \] we can determine two first--order derivatives. \pause The \textit{partial derivative of $f$ with respect to $x$} is written as \[ \frac{\partial z}{\partial x} \quad \textrm{\sf or} \quad \frac{\partial f}{\partial x} \quad \textrm{\sf or} \quad f_{x}, \] and is found by differentiating $f$ with respect to $x$, with $y$ held constant. \pause \vsp Similarly, the \textit{partial derivative of $f$ with respect to $y$} is written as \[ \frac{\partial z}{\partial y} \quad \textrm\sf {or} \quad \frac{\partial f}{\partial y} \quad \textrm{\sf or} \quad f_{y}, \] and is found by differentiating $f$ with respect to $y$, with $x$ held constant.} \sol \frame{ \frametitle{Examples} Find the first--order partial derivatives of the functions \begin{itemize} \item [(a)] $f(x,y)=x^2+y^3$; \pause \vsp \textbf{Solution} \pause \begin{eqnarray*} \frac{\partial f}{\partial x} &=& 2x;\\ \pause && \\ \frac{\partial f}{\partial y} &=& 3y^{2}. \end{eqnarray*} \end{itemize} } \frame{ \frametitle{Examples} Find the first--order partial derivatives of the functions \begin{itemize} \item [(b)] $f(x,y)=x^2y$; \vsp \textbf{Solution} \pause \begin{eqnarray*} \frac{\partial f}{\partial x} &=& 2xy; \\ \pause \frac{\partial f}{\partial y} &=& x^{2}. \end{eqnarray*} \end{itemize} } \frame{ \frametitle{Examples} Find the first--order partial derivatives of the functions \begin{itemize} \item [(c)] $f(x,y)=x^2y^3-10x$. \vsp \textbf{Solution} \pause \begin{eqnarray*} \frac{\partial f}{\partial x} &=& 2xy^{3}-10; \pause\\ \frac{\partial f}{\partial y} &=& 3x^{2}y^{2}. \end{eqnarray*} \end{itemize} } \normal \frame{ \frametitle{2.3 Partial differentiation} In general, when we differentiate a function of two variables, the thing we end up with is itself a function of two variables. \pause \vsp This suggests the possibility of differentiating a second time.\pause \vsp In fact, there are four \textit{second--order partial derivatives}. } \frame{ \frametitle{2.3 Partial differentiation} We write: \[ \frac{\partial^{2}z}{\partial x^{2}} \quad \textrm{\sf or} \quad \frac{\partial^{2}f}{\partial x^{2}} \quad \textrm{\sf or} \quad f_{xx} \] for the function obtained by differentiating twice with respect to $x$,\pause \[ \frac{\partial^{2}z}{\partial y^{2}} \quad \textrm{\sf or} \quad \frac{\partial^{2}f}{\partial y^{2}} \quad \textrm{\sf or} \quad f_{yy} \] for the function obtained by differentiating twice with respect to $y$; } \frame{ \frametitle{2.3 Partial differentiation} \[ \frac{\partial^{2}z}{\partial y \partial x} \quad \textrm{\sf or} \quad \frac{\partial^{2}f}{\partial y \partial x} \quad \textrm{\sf or} \quad f_{yx} \] for the function obtained by differentiating first with respect to $x$ and then with respect to $y$, and \pause \[ \frac{\partial^{2}z}{\partial x \partial y} \quad \textrm{\sf or} \quad \frac{\partial^{2}f}{\partial x \partial y} \quad \textrm{\sf or} \quad f_{xy} \] for the function obtained by differentiating first with respect to $y$ and then with respect to $x$. } \frame{ \frametitle{Examples} Find expressions for the second--order partial derivatives $f_{xx}$, $f_{yy}$, $f_{yx}$ and $f_{xy}$ for the functions \begin{itemize} \item [(a)] $f(x,y)=x^2+y^3$; \item [(b)] $f(x,y)=x^2y$. \end{itemize} } \sol \frame{ \frametitle{Examples} For part (a), we know that \[ f_{x} = 2x \quad \textsf{and} \quad f_{y} = 3y^{2}. \] \pause Thus \begin{eqnarray*} f_{xx} &=& 2; \\ \pause f_{yy} &=& 6y; \\ \pause f_{xy} &=& 0; \\ \pause f_{yx} &=& 0. \end{eqnarray*} } \frame{ \frametitle{Examples} For part (b), we know that \[ f_{x} = 2xy \quad \textsf{and} \quad f_{y} = x^{2}. \] \pause Thus \begin{eqnarray*} f_{xx} &=& 2y; \\ \pause f_{yy} &=& 0; \\ \pause f_{xy} &=&2x; \\ \pause f_{yx} &=&2x. \end{eqnarray*} } \begin{comment} \frame{ \frametitle{Plan for today} \begin{itemize} \item \textbf{Attendance list:} don't forget to sign in \pause (you can't email us at a later date and tell us you forgot to sign in. Well, you \textit{can} do this, but nothing will be done about it!) \pause \vspl \item \textbf{Chapter 2:} Question 5, 6 and 7 on page 59 of the lecture notes (we might run through some of this stuff at the board in around 15/20 minutes) \pause \vspl \item \textbf{Chapter 3:} lecture notes will be given out so we can make a prompt start in the lecture this Friday \end{itemize} } \frame{ \frametitle{Plan for today} \begin{itemize} \item \textbf{Case study 1:} \begin{itemize} \item [--] Some general feedback \pause \vspl \item [--] Work is now available to collect from the ACC1012/1053 homework collection box, on the 3rd floor of the Herschel Building \pause \end{itemize} \vspl \item \textbf{Case study 2:} please form groups; we will then give out the case study material and you must sign up using the group membership forms \end{itemize} } \frame{ \frametitle{Case study 1: General feedback} \begin{center} \begin{tabular}{|c c|c c|c c|c|}\hline $n$ &$n^{*}$ &Mean &Median &LQ &UQ &SD \\ \hline 243 &5 &20.8 &21 &20 &22 &2.45 \\ \hline \end{tabular} \end{center} \pause \begin{center} \includegraphics[width=7cm,height=6.65cm]{cs1marks.ps} \end{center} \pause } \frame{ \frametitle{Case study 1: iPhone feedback} \begin{itemize} \item Generally very well done \vspl \item Nearly all students were able to form linear functions for supply and demand \vspl \item Most students used simultaneous equations to find the optimum price \vspl \item Most students forgot to give a practical interpretation of the gradient for supply and demand equations and thus lost 2 marks \vspl \item Only a handful of students commented on the suitability of the linear model \end{itemize} } \frame{ \frametitle{Case study 1: alcohol feedback} \begin{itemize} \item Once again, generally well done! \vspl \item Students forgot to give practical interpretation of the gradient \vspl \item Students struggled with the concept of a ceiling price and most did not calculate it explicitly \vspl \item Most students did not comment on the suitability of the linear model \vspl \item Otherwise well done, plots produced were fine \vspl \item Some extra research was impressive \end{itemize} } \normal \frame{ \vspace{1cm} \huge \textbf{Chapter 3} \vspace{1cm} \Large \textbf{Collecting, presenting and }\\ \textbf{summarsing data} } \frame{ \frametitle{3. Data: collection, presentation and summaries} \textcolor{purple}{\textbf{Data}} are the key to many important management decisions. \vsp \pause \begin{itemize} \item Is a new product selling well? \pause \vspl \item Do potential customers like the new advertising campaign? \end{itemize} \vsp \noindent These and many other questions can be answered with \textcolor{purple}{\textbf{data}}. } \frame{ \frametitle{What we'll cover...} \begin{itemize} \item The different \textcolor{purple}{\bf kinds of data} that exist \pause \vsp \item How to \textcolor{purple}{\bf get} data \pause \vsp \item How to \textcolor{purple}{\bf look at} the data we've got \pause \vsp \item How to \textcolor{purple}{\bf summarise} our data numerically \end{itemize} } \frame{ \frametitle{Example: Sizing clothes} \begin{center} \includegraphics[width=4.5cm]{clothing-sizes.ps}\qquad \includegraphics[width=3.5cm]{dress-size.ps} \end{center} \begin{itemize} \item Most clothing now comes in essentially standard sizes $\ldots$ \item but where do these standards come from? \item By sampling from the population as a whole, standards can be set around the most common sizes. \item We cannot say that an individual is \alert{\textit{exactly}} a standard size; \item However, we \alert{\textit{can}} say that they will \alert{\textit{probably}} fall within a range either side of a standard size. \end{itemize} } \frame{ \frametitle{Example: Car maintenance} \begin{center} \includegraphics[width=4cm]{newcar.ps}\qquad \includegraphics[width=4cm]{car-maintenance.ps} \end{center} \begin{itemize} \item Just imagine you're buying a new car $\ldots$ \pause \item ...it might be useful to know how much it will cost to run over the next few years. \pause \item Obviously this cannot be predicted exactly -- \begin{itemize} \item [--]each car will be slightly different; \item [--]each user will be slightly different. \pause \end{itemize} \item Collecting data from people who have bought similar cars will give us some idea! \end{itemize} } \frame{ \frametitle{Example: Profit versus advertising} \begin{center} \psfrag{10}[][]{\sf \tiny 10} \psfrag{15}[][]{\sf \tiny 15} \psfrag{20}[][]{\sf \tiny 20} \psfrag{25}[][]{\sf \tiny 25} \psfrag{30}[][]{\sf \tiny 30} \psfrag{35}[][]{\sf \tiny 35} \psfrag{50}[][]{\sf \tiny 50} \psfrag{55}[][]{\sf \tiny 55} \psfrag{60}[][]{\sf \tiny 60} \psfrag{65}[][]{\sf \tiny 65} \psfrag{P}[][]{\sf \tiny $P$ (Profit)} \psfrag{A}[][]{\sf \tiny $A$ (Advertising)} \psfrag{E}[][]{\sf \tiny $P=50.3+0.67A$} \includegraphics[width=9cm,height=6cm]{determine.ps} \end{center} } \frame{ \frametitle{Example: Profit versus advertising} \begin{center} \psfrag{10}[][]{\sf \tiny 10} \psfrag{15}[][]{\sf \tiny 15} \psfrag{20}[][]{\sf \tiny 20} \psfrag{25}[][]{\sf \tiny 25} \psfrag{30}[][]{\sf \tiny 30} \psfrag{35}[][]{\sf \tiny 35} \psfrag{50}[][]{\sf \tiny 50} \psfrag{55}[][]{\sf \tiny 55} \psfrag{60}[][]{\sf \tiny 60} \psfrag{65}[][]{\sf \tiny 65} \psfrag{P}[][]{\sf \tiny $P$ (Profit)} \psfrag{A}[][]{\sf \tiny $A$ (Advertising)} \psfrag{E}[][]{\sf \tiny $P=50.3+0.67A$} \includegraphics[width=9cm,height=6cm]{determine2.ps} \end{center} } \frame{ \frametitle{3.1 Important definitions} The quantities measured in a study are called \textcolor{purple}{\textbf{random variables}}. \pause \vspace{0.3cm} A particular outcome is called an \textcolor{purple}{\textbf{observation}}. \vspace{0.3cm} \pause A collection of such observations is the \textcolor{purple}{\textbf{data}}. \vspace{0.3cm} \pause The collection of all possible outcomes is the \textcolor{purple}{\textbf{population}}. } \frame{ \frametitle{3.1 Important definitions} Suppose we are interested in the height of students on A\&F courses at Newcastle $\ldots$ \begin{center} \includegraphics[width=1.5cm]{measuring-height.ps} \end{center} \begin{itemize} \pause \item Our \textcolor{purple}{\textbf{random variable}} is ``the height of students on A\&F courses at Newcastle''. \pause \item If Joe Bloggs is an A\&F student, and we measured his height, then that value would be a single \textcolor{purple}{\textbf{observation}}. \pause \item If we measured the height of every first year A\&F student, we would have a collection of such observations which would be our \textcolor{purple}{\textbf{data}}. \pause \item This would be a \textcolor{purple}{\textbf{sample}} from the \textcolor{purple}{\textbf{population}} which consists of all students registered on A\&F degrees. \end{itemize} } \frame{ \frametitle{3.1 Important definitions} \noindent Ideally, to get a true idea of what is going on, we'd like to observe the whole population (take a \textcolor{purple}{\textbf{census}}). However, this can be difficult: %\pause \begin{itemize} \item If the population is huge, then this would take ages! \item \textit{And} it would be very costly! \item In reality, we usually observe a subset of the population $\ldots$ but how do we choose who to observe? \end{itemize} \begin{center} \includegraphics[width=3.5cm]{sample.ps} \end{center} } \frame{ \frametitle{3.1 Important definitions {\small \textcolor{orange}{(Section 2.1, summer revision booklet)}}} \noindent There are two types of variables: \textcolor{purple}{\textbf{qualitative}} and \textcolor{purple}{\textbf{quantitative}}. %\pause \vspace{0.2cm} \begin{itemize} \item \textit{Qualitative} variables have non--numerical outcomes, and are usually \textcolor{purple}{\textbf{categorical}} \pause \begin{itemize} \item [--]Sex of a person (categories: male or female); \item [--]Colour of a car (categories: \textcolor{red}{red}, black, \textcolor{gray}{silver}, \textcolor{blue}{blue}, ... -- could code this to \textcolor{red}{1}, 2, \textcolor{gray}{3}, \textcolor{blue}{4}, ...); \item [--]Mode of transport. \end{itemize} \vspace{0.2cm} \pause \item \textit{Quantitative} variables have numerical outcomes with a natural ordering \pause \begin{itemize} \item [--]people's height \item [--]number of defective components in a batch \end{itemize} \end{itemize} } \frame{ \frametitle{3.1 Important definitions {\small \textcolor{orange}{(Section 2.1, summer revision booklet)}}} We are most interested in \textcolor{purple}{\textbf{quantitative}} variables. \vspace{0.5cm} \pause These variables can be subdivided into two types: \pause \begin{center} \textcolor{purple}{\textbf{discrete}} and \textcolor{purple}{\textbf{continuous}}. \end{center} \pause \vspace{0.2cm} \textbf{Discrete random variables} \begin{itemize} \item can only take a sequence of distinct values (usually integers); \item are usually \textcolor{purple}{\bf{countable}} -- e.g.\ the number of people attending a tutorial group; \item can be \textcolor{purple}{\textbf{ordinal}} -- where the outcomes are ordered. \end{itemize} } \frame{ \frametitle{3.1 Important definitions {\small \textcolor{orange}{(Section 2.1, summer revision booklet)}}} \medskip \textbf{Continuous random variables} \begin{itemize} \item can take any value over some continuous scale -- e.g.\ height or weight. \item can be \textcolor{purple}{\textbf{measured}} to a very high degree of accuracy (provided we have the equipment to do so) $\ldots$ \item $\ldots$ however, we can never say \textit{precisely} how much someone weighs, for example, \item might be measured to the nearest whole number -- and so could ``look'' discrete -- be careful! \end{itemize} \begin{center} \includegraphics[width=3cm]{tape-measure.ps} \includegraphics[width=3cm]{scales.ps} \includegraphics[width=3cm]{stop-watch.ps} \end{center} } \sol \frame{ \frametitle{3.1 Important definitions} \vspace{-2cm} \begin{center} \includegraphics[width=8.5cm,angle=-90]{datatree.ps} \end{center} } \normal \frame{ \frametitle{3.2 Sampling} We can rarely observe the whole population. Instead we observe some sub--set of this (called the \textcolor{purple}{\textbf{sample}}). %\pause \vspace{0.3cm} The difficulty is in obtaining a \textcolor{purple}{\textbf{representative}} sample. %\pause \vspace{0.3cm} For example, if you were to ask people leaving a gym if they took exercise this would produce a \textcolor{purple}{\textbf{biased}} sample and would not be representative of the population as a whole. %\pause \vspace{0.3cm} \textbf{The importance of obtaining a representative sample cannot be stressed too highly}. } \frame{ \frametitle{3.2 Sampling} \noindent There are three general forms of sampling techniques: \vsp \pause \begin{itemize} \item [\textbf{1.}]\textcolor{purple}{\textbf{\textit{Random sampling}}} -- where the members of the sample are chosen by some random (i.e.\ unpredictable) mechanism. \pause \vspl \item [\textbf{2.}] \textcolor{purple}{\textbf{\textit{Quasi--random sampling}}} -- where the mechanism for choosing the sample is only partly random. \pause \vspl \item [\textbf{3.}]\textcolor{purple}{\textbf{\textit{Non--random sampling}}} -- where the sample is specifically selected rather than randomly selected. \end{itemize} } \frame{ \frametitle{3.2.1 Simple random sampling...} This method is the \textcolor{purple}{\textbf{simplest to understand}}. \pause \vspace{0.3cm} \noindent If we had a population of 200 students we could put all their names into a hat and draw out 20 names as our sample. \begin{center} \includegraphics[width=1.75cm]{hat.ps} \end{center} \vspace{0.3cm}\pause We cannot predict with certainty which 20 names will be drawn, and so the sample is unpredictable, and therefore completely \textcolor{purple}{\textbf{random}}. \pause \vspace{0.3cm} Each name has an \textcolor{purple}{\textbf{equally likely chance}} of being drawn. \vspace{0.3cm} Furthermore each possible \textcolor{purple}{\textbf{sample of 20}} has an equal chance of being selected. } \frame{ \frametitle{3.2.1 Simple random sampling...} \vspace{0.3cm} %This defines a \textcolor{purple}{\textbf{simple random sample}}. \vspace{0.3cm} %\pause In reality the drawing of the names would be done by a computer and the population and samples would be considerably larger. } \frame{ \frametitle{... and some disadvantages} We often don't have a \textcolor{purple}{\textbf{complete list}} of the population. \pause \vspace{0.3cm} For example, if you were surveying the market for some new software, the population would be everybody with a compatible computer! \pause \vspace{0.3cm} Not all elements of the population are \textcolor{purple}{\textbf{equally accessible}}. \pause \vspace{0.3cm} Purely by chance, you could pick an \textcolor{purple}{\textbf{unrepresentative sample}}! } \frame{ \frametitle{3.2.2 Stratified sampling...} This is a form of random sample where clearly defined groups, or \textcolor{purple}{\textbf{strata}}, exist within the population. \pause \vspace{0.3cm} If we know the overall proportion of the population that falls into each of these groups, we can take a simple random sample from each of the groups and then adjust the results according to the known proportions. \pause \vspace{0.3cm} For example, if we assume that the population is 55\% female and 45\% male and we wanted a sample of 1000, we would \pause \begin{itemize} \item decide to have 550 females and 450 males in our sample, and then \pause \item pick the members of our sample from their respective groups randomly. \end{itemize} } \frame{ \frametitle{... and some disadvantages} We need clear information on the \textcolor{purple}{\textbf{size}} and \textcolor{purple}{\textbf{composition}} of each stratum which can be difficult to obtain. \vspace{0.3cm} \noindent We still need a \textcolor{purple}{\textbf{list of the entire population}} to sample from. } \frame{ \frametitle{3.2.3 Systematic sampling...} This is a form of \textcolor{purple}{\textbf{quasi--random sampling}} which can be used when the population is clearly structured. \pause \vspace{0.3cm} For example, if you were interested in obtaining a 10\% sample from a batch of components being manufactured, you would \begin{itemize} \item select the first component at random, and then \pause \item pick every 10th item to come off the production line after that \end{itemize} \begin{center} \includegraphics[width=3cm]{conveyorbelt.ps} \end{center} \vspace{0.3cm} This scheme is \textcolor{purple}{\textbf{very easy to implement}}! } \frame{ \frametitle{... and some disadvantages} This method is \textcolor{purple}{\textbf{not entirely random}}! \pause \vspace{0.3cm} If there is a pattern in the process, it is very easy to obtain a \textcolor{purple}{\textbf{biased sample}}. } \begin{frame} \frametitle{Other types of sampling} Read through Sections 3.2.4 to 3.2.8 of the notes before your tutorial session next week. \medskip \begin{itemize} \item Multi-stage sampling \medskip \item Cluster sampling \medskip \item Judgemental sampling \medskip \item Accessibility sampling \medskip \item Quota sampling \end{itemize} \begin{center} \includegraphics[width=3cm]{bookworm.ps} \end{center} \end{frame} \frame{ \frametitle{3.2.9 Sample size} \begin{center} \includegraphics[width=5cm]{sample_size_tshirt.ps} \end{center} \pause \medskip \begin{itemize} \item Larger samples will generally give more precise information about the population. \item Expense and time tend to limit the size of the sample. \item E.g.\ national opinion polls often rely on samples in the region of just 1000. \end{itemize} } \frame{ \frametitle{3.3 Frequency tables} \begin{itemize} \item After collecting data, the first stage of any analysis is to present them in a simple, easily understood way. \pause \vsp \item \textcolor{purple}{\textbf{Tables}} are perhaps the simplest means of presenting data. \pause \vsp \item Tables can be informative, but some tables can be difficult to interpret, especially if they contain vast amounts of data. \end{itemize} } \frame{ \frametitle{3.3 Frequency tables} \begin{itemize} \item \textcolor{purple}{\textbf{Frequency tables}} are amongst the most commonly-used tables and are perhaps the easiest to understand. \pause \vsp \item They can be used with continuous, discrete, categorical and ordinal data. \pause \vsp \item Frequency tables have uses in some of the techniques we will see in the next lecture. \end{itemize} } \frame{ \frametitle{3.3.1 Frequency tables for categorical data} The following table presents the modes of transport used daily by 30 students to get to and from University. \begin{center} \begin{tabular}{|c|c||c|c||c|c|} \hline \textcolor{magenta}{\textbf{Student}} & \textcolor{magenta}{\textbf{Mode}} & \textcolor{magenta}{\textbf{Student}} & \textcolor{magenta}{\textbf{Mode}} & \textcolor{magenta}{\textbf{Student}} & \textcolor{magenta}{\textbf{Mode}}\\ \hline 1 & Car & 11 & Walk & 21 & Walk\\ 2 & Walk & 12 & Walk & 22 & Metro\\ 3 & Car & 13 & Metro & 23 & Car\\ 4 & Walk & 14 & Bus & 24 & Car\\ 5 & Bus & 15 & Train & 25 & Car\\ 6 & Metro & 16 & Bike & 26 & Bus\\ 7 & Car & 17 & Bus & 27 & Car\\ 8 & Bike & 18 & Bike & 28 & Walk\\ 9 & Walk & 19 & Bike & 29 & Car\\ 10 & Car & 20 & Metro & 30 & Car\\ \hline \end{tabular} \end{center} } \frame{ \frametitle{3.3.1 Frequency tables for categorical data} \noindent The table obviously contains much information. However, it is difficult to see which method of transport is the most widely used. \vspace{0.3cm} \pause \textbf{Idea:} count the number of students using each mode of transport! \pause \begin{center} \begin{tabular}{|c | c|} \hline \textcolor{magenta}{\textbf{Mode}} & \textcolor{magenta}{\textbf{Frequency}}\\ \hline Car & 10\\ Walk & 7\\ Bike & 4\\ Bus & 4\\ Metro & 4\\ Train & 1\\ \hline \textbf{Total} & 30\\ \hline \end{tabular} \end{center} This gives us a much clearer picture of the methods of transport used. } \frame{ \frametitle{3.3.1 Frequency tables for categorical data} \noindent Also of interest might be the \textcolor{purple}{\textbf{\textit{relative}}} frequency of each of the modes of transport. \vspace{0.3cm} This is simply the frequency expressed as a proportion of the total number of students surveyed. If this is given as a percentage, as here, this is known as the \textcolor{purple}{\textbf{percentage relative frequency}}. \begin{center} \begin{tabular}{|c | c | c|} \hline \textcolor{blue}{\textbf{Mode}} & \textcolor{blue}{\textbf{Frequency}} & \textcolor{blue}{\textbf{Relative Frequency (\%)}}\\ \hline Car & 10 & 33.3\\ Walk & 7 & 23.4\\ Bike & 4 & 13.3\\ Bus & 4 & 13.3\\ Metro & 4 & 13.3\\ Train & 1 & 3.4\\ \hline \textbf{Total} & 30 & 100\\ \hline \end{tabular} \end{center} } \frame{ \frametitle{3.3.2 Frequency tables for count data} \begin{center} \includegraphics[width=4cm]{showroom.ps} \end{center} The following table shows the raw data for car sales at a new car showroom over a two week period in July. \begin{center} \begin{tabular}{|c|c|c|c|} \hline \textcolor{blue}{\textbf{Date}} & \textcolor{blue}{\textbf{Cars Sold}} & \textcolor{blue}{\textbf{Date}} & \textcolor{blue}{\textbf{Cars Sold}}\\ \hline 1st July & ~9 & 8th July & 10\\ 2nd July & ~8 & 9th July & ~5\\ 3rd July & ~6 & 10th July & ~8\\ 4th July & ~7 & 11th July & ~4\\ 5th July & ~7 & 12th July & ~6\\ 6th July & 10 & 13th July & ~8\\ 7th July & 11 & 14th July & ~9\\ \hline \end{tabular} \end{center} } \frame{ \frametitle{Example 3.1} Present these data in a relative frequency table by number of days on which different numbers of cars were sold. } \sol \frame{ \frametitle{Solution to Example 3.1} \begin{center} \begin{tabular}{|c|c|c|c|} \hline \textcolor{blue}{\textbf{Cars Sold}} & \textcolor{blue}{\textbf{Tally}} & \textcolor{blue}{\textbf{Frequency}} & \textcolor{blue}{\textbf{Relative Frequency \%}}\\ \hline 1 & &\textcolor{white}{0} &\textcolor{white}{0} \\ 2 & &\textcolor{white}{0} &\textcolor{white}{0} \\ 3 & &\textcolor{white}{0} &\textcolor{white}{0} \\ 4 &\textcolor{white}{l} &\textcolor{white}{1} &\textcolor{white}{7.14} \\ 5 &\textcolor{white}{l} &\textcolor{white}{1} &\textcolor{white}{7.14} \\ 6 &\textcolor{white}{ll} &\textcolor{white}{2} &\textcolor{white}{14.29} \\ 7 &\textcolor{white}{ll} &\textcolor{white}{2} &\textcolor{white}{14.29} \\ 8 &\textcolor{white}{lll} &\textcolor{white}{3} &\textcolor{white}{21.43} \\ 9 &\textcolor{white}{ll} &\textcolor{white}{2} &\textcolor{white}{14.29} \\ 10 &\textcolor{white}{ll} &\textcolor{white}{2} &\textcolor{white}{14.29}\\ 11 &\textcolor{white}{l} &\textcolor{white}{1} &\textcolor{white}{7.14} \\ \hline \textbf{Totals} &\textcolor{white}{14} &\textcolor{white}{14} &\textcolor{white}{100} \\ \hline \end{tabular} \end{center} } \frame{ \frametitle{Solution to Example 3.1} \begin{center} \begin{tabular}{|c|c|c|c|} \hline \textcolor{blue}{\textbf{Cars Sold}} & \textcolor{blue}{\textbf{Tally}} & \textcolor{blue}{\textbf{Frequency}} & \textcolor{blue}{\textbf{Relative Frequency \%}}\\ \hline 1 & &\textcolor{white}{0} &\textcolor{white}{0} \\ 2 & &\textcolor{white}{0} &\textcolor{white}{0} \\ 3 & &\textcolor{white}{0} &\textcolor{white}{0} \\ 4 &l &\textcolor{white}{1} &\textcolor{white}{7.14} \\ 5 &l &\textcolor{white}{1} &\textcolor{white}{7.14} \\ 6 &ll &\textcolor{white}{2} &\textcolor{white}{14.29} \\ 7 &ll &\textcolor{white}{2} &\textcolor{white}{14.29} \\ 8 &lll &\textcolor{white}{3} &\textcolor{white}{21.43} \\ 9 &ll &\textcolor{white}{2} &\textcolor{white}{14.29} \\ 10 &ll &\textcolor{white}{2} &\textcolor{white}{14.29}\\ 11 &l &\textcolor{white}{1} &\textcolor{white}{7.14} \\ \hline \textbf{Totals} &14 &\textcolor{white}{14} &\textcolor{white}{100} \\ \hline \end{tabular} \end{center} } \frame{ \frametitle{Solution to Example 3.1} \begin{center} \begin{tabular}{|c|c|c|c|} \hline \textcolor{blue}{\textbf{Cars Sold}} & \textcolor{blue}{\textbf{Tally}} & \textcolor{blue}{\textbf{Frequency}} & \textcolor{blue}{\textbf{Relative Frequency \%}}\\ \hline 1 & &0 &\textcolor{white}{0} \\ 2 & &0 &\textcolor{white}{0} \\ 3 & &0 &\textcolor{white}{0} \\ 4 &l &1 &\textcolor{white}{7.14} \\ 5 &l &1 &\textcolor{white}{7.14} \\ 6 &ll &2 &\textcolor{white}{14.29} \\ 7 &ll &2 &\textcolor{white}{14.29} \\ 8 &lll &3 &\textcolor{white}{21.43} \\ 9 &ll &2 &\textcolor{white}{14.29} \\ 10 &ll &2 &\textcolor{white}{14.29}\\ 11 &l &1 &\textcolor{white}{7.14} \\ \hline \textbf{Totals} &14 &14 &\textcolor{white}{100} \\ \hline \end{tabular} \end{center} } \frame{ \frametitle{Solution to Example 3.1} \begin{center} \begin{tabular}{|c|c|c|c|} \hline \textcolor{blue}{\textbf{Cars Sold}} & \textcolor{blue}{\textbf{Tally}} & \textcolor{blue}{\textbf{Frequency}} & \textcolor{blue}{\textbf{Relative Frequency \%}}\\ \hline 1 & &0 &0 \\ 2 & &0 &0 \\ 3 & &0 &0 \\ 4 &l &1 &7.14 \\ 5 &l &1 &7.14 \\ 6 &ll &2 &14.29 \\ 7 &ll &2 &14.29 \\ 8 &lll &3 &21.43 \\ 9 &ll &2 &14.29\\ 10 &ll &2 &14.29\\ 11 &l &1 &7.14 \\ \hline \textbf{Totals} &14 &14 &100 \\ \hline \end{tabular} \end{center} } \normal \frame{ \frametitle{3.3.3 Frequency tables for continuous data} With discrete data it is easy to count the quantities in the defined categories. With \textcolor{purple}{\textbf{continuous data}} this is not possible. \pause \vsp For example, the following data set represents the service time in seconds for callers to a credit card call centre. \begin{center} \begin{tabular}{|c c c c c|} \hline 214.8 & 220.6 & 216.7 & 195.1 & 211.4\\ 195.8 & 201.1 & 185.8 & 183.4 & 178.8\\ 196.3 & 199.7 & 206.7 & 203.8 & 203.1\\ 200.8 & 201.3 & 205.6 & 181.6 & 201.7\\ 180.2 & 193.3 & 188.2 & 199.9 & 204.7\\ 198.3 & 193.1 & 204.0 & 197.2 & 193.5\\ 205.5 & 217.5 & 208.8 & 197.7 & 212.3\\ 209.9 & 197.6 & 204.9 & 203.1 & 192.9\\ 208.9 & 202.0 & 195.0 & 192.7 & 219.8\\ 208.8 & 200.7 & 191.9 & 188.8 & 206.8\\ \hline \end{tabular} \end{center} } \frame{ \frametitle{3.3.3 Frequency tables for continuous data} This is what we do... \begin{enumerate} \item[\textbf{1.}] Divide the range of the variable into smaller ranges called \textcolor{purple}{\textbf{class intervals}}. \pause \vspl \item[\textbf{2.}] There should be \textcolor{purple}{\textbf{no gaps}} between these intervals. \pause \vspl \item[\textbf{3.}] The class interval width should be a convenient number (e.g. 5, 10, 100, depending on the data). \pause \vspl \item[\textbf{4.}] You should aim for no more than about \textcolor{purple}{\textbf{ten to fifteen classes}}. \end{enumerate} } \frame{ \frametitle{Example 3.2} Create a frequency table for the call centre data. \vsp Also, find the relative frequencies in each class interval. } \sol %\setbeamercolor{structure}{fg=red!80!black} \frame{ \frametitle{Solution to Example 3.2} For the call centre data, we might get: \begin{center} \begin{tabular}{|c|c|c|c|} \hline \textcolor{blue}{\textbf{Class Interval}} & \textcolor{blue}{\textbf{Tally}} & \textcolor{blue}{\textbf{Frequency}} & \textcolor{blue}{\textbf{Relative Frequency \%}}\\ \hline 175 $\leq$ time $<$ 180 & \textcolor{white}{l} &\textcolor{white}{1} &\textcolor{white}{2} \\ 180 $\leq$ time $<$ 185 & \textcolor{white}{lll} &\textcolor{white}{3} &\textcolor{white}{6} \\ 185 $\leq$ time $<$ 190 & \textcolor{white}{lll} &\textcolor{white}{3} &\textcolor{white}{6} \\ 190 $\leq$ time $<$ 195 & \textcolor{white}{\sout{llll} l} & \textcolor{white}{6} & \textcolor{white}{12}\\ 195 $\leq$ time $<$ 200 & \textcolor{white}{l}\textcolor{white}{\sout{llll} \sout{llll}} & \textcolor{white}{10}& \textcolor{white}{20}\\ 200 $\leq$ time $<$ 205 & \textcolor{white}{\sout{llll} \sout{llll} ll} & \textcolor{white}{12} & \textcolor{white}{24}\\ 205 $\leq$ time $<$ 210 & \textcolor{white}{\sout{llll} lll} & \textcolor{white}{8}& \textcolor{white}{16}\\ 210 $\leq$ time $<$ 215 & \textcolor{white}{lll} & \textcolor{white}{3}& \textcolor{white}{6}\\ 215 $\leq$ time $<$ 220 & \textcolor{white}{lll} & \textcolor{white}{3}& \textcolor{white}{6}\\ 220 $\leq$ time $<$ 225 & \textcolor{white}{l} & \textcolor{white}{1}& \textcolor{white}{2}\\ \hline \textbf{Totals} & &\textcolor{white}{50} &\textcolor{white}{100} \\ \hline \end{tabular} \end{center} } \frame{ \frametitle{Solution to Example 3.2} For the call centre data, we might get: \begin{center} \begin{tabular}{|c|c|c|c|} \hline \textcolor{blue}{\textbf{Class Interval}} & \textcolor{blue}{\textbf{Tally}} & \textcolor{blue}{\textbf{Frequency}} & \textcolor{blue}{\textbf{Relative Frequency \%}}\\ \hline 175 $\leq$ time $<$ 180 & \textcolor{white}{l} &\textcolor{white}{1} &\textcolor{white}{2} \\ 180 $\leq$ time $<$ 185 & \textcolor{white}{lll} &\textcolor{white}{3} &\textcolor{white}{6} \\ 185 $\leq$ time $<$ 190 & \textcolor{white}{lll} &\textcolor{white}{3} &\textcolor{white}{6} \\ 190 $\leq$ time $<$ 195 & \textcolor{white}{\sout{llll} l} & \textcolor{white}{6} & \textcolor{white}{12}\\ 195 $\leq$ time $<$ 200 & \textcolor{white}{l}\textcolor{white}{\sout{llll} \sout{llll}} & \textcolor{white}{10}& \textcolor{white}{20}\\ 200 $\leq$ time $<$ 205 & \textcolor{white}{\sout{llll} \sout{llll} ll} & \textcolor{white}{12} & \textcolor{white}{24}\\ 205 $\leq$ time $<$ 210 & \textcolor{white}{\sout{llll} lll} & \textcolor{white}{8}& \textcolor{white}{16}\\ 210 $\leq$ time $<$ 215 & l\textcolor{white}{ll} & \textcolor{white}{3}& \textcolor{white}{6}\\ 215 $\leq$ time $<$ 220 & \textcolor{white}{lll} & \textcolor{white}{3}& \textcolor{white}{6}\\ 220 $\leq$ time $<$ 225 & \textcolor{white}{l} & \textcolor{white}{1}& \textcolor{white}{2}\\ \hline \textbf{Totals} & &\textcolor{white}{50} &\textcolor{white}{100} \\ \hline \end{tabular} \end{center} } \frame{ \frametitle{Solution to Example 3.2} For the call centre data, we might get: \begin{center} \begin{tabular}{|c|c|c|c|} \hline \textcolor{blue}{\textbf{Class Interval}} & \textcolor{blue}{\textbf{Tally}} & \textcolor{blue}{\textbf{Frequency}} & \textcolor{blue}{\textbf{Relative Frequency \%}}\\ \hline 175 $\leq$ time $<$ 180 & \textcolor{white}{l} &\textcolor{white}{1} &\textcolor{white}{2} \\ 180 $\leq$ time $<$ 185 & \textcolor{white}{lll} &\textcolor{white}{3} &\textcolor{white}{6} \\ 185 $\leq$ time $<$ 190 & \textcolor{white}{lll} &\textcolor{white}{3} &\textcolor{white}{6} \\ 190 $\leq$ time $<$ 195 & \textcolor{white}{\sout{llll} l} & \textcolor{white}{6} & \textcolor{white}{12}\\ 195 $\leq$ time $<$ 200 & \textcolor{white}{l}\textcolor{white}{\sout{llll} \sout{llll}} & \textcolor{white}{10}& \textcolor{white}{20}\\ 200 $\leq$ time $<$ 205 & \textcolor{white}{\sout{llll} \sout{llll} ll} & \textcolor{white}{12} & \textcolor{white}{24}\\ 205 $\leq$ time $<$ 210 & \textcolor{white}{\sout{llll} lll} & \textcolor{white}{8}& \textcolor{white}{16}\\ 210 $\leq$ time $<$ 215 & l\textcolor{white}{ll} & \textcolor{white}{3}& \textcolor{white}{6}\\ 215 $\leq$ time $<$ 220 & \textcolor{white}{lll} & \textcolor{white}{3}& \textcolor{white}{6}\\ 220 $\leq$ time $<$ 225 & l & \textcolor{white}{1}& \textcolor{white}{2}\\ \hline \textbf{Totals} & &\textcolor{white}{50} &\textcolor{white}{100} \\ \hline \end{tabular} \end{center} } \frame{ \frametitle{Solution to Example 3.2} For the call centre data, we might get: \begin{center} \begin{tabular}{|c|c|c|c|} \hline \textcolor{blue}{\textbf{Class Interval}} & \textcolor{blue}{\textbf{Tally}} & \textcolor{blue}{\textbf{Frequency}} & \textcolor{blue}{\textbf{Relative Frequency \%}}\\ \hline 175 $\leq$ time $<$ 180 & \textcolor{white}{l} &\textcolor{white}{1} &\textcolor{white}{2} \\ 180 $\leq$ time $<$ 185& \textcolor{white}{lll} &\textcolor{white}{3} &\textcolor{white}{6} \\ 185 $\leq$ time $<$ 190 & \textcolor{white}{lll} &\textcolor{white}{3} &\textcolor{white}{6} \\ 190 $\leq$ time $<$ 195 & \textcolor{white}{\sout{llll} l} & \textcolor{white}{6} & \textcolor{white}{12}\\ 195 $\leq$ time $<$ 200 & \textcolor{white}{l}\textcolor{white}{\sout{llll} \sout{llll}} & \textcolor{white}{10}& \textcolor{white}{20}\\ 200 $\leq$ time $<$ 205 & \textcolor{white}{\sout{llll} \sout{llll} ll} & \textcolor{white}{12} & \textcolor{white}{24}\\ 205 $\leq$ time $<$ 210 & \textcolor{white}{\sout{llll} lll} & \textcolor{white}{8}& \textcolor{white}{16}\\ 210 $\leq$ time $<$ 215 & l\textcolor{white}{ll} & \textcolor{white}{3}& \textcolor{white}{6}\\ 215 $\leq$ time $<$ 220 & l\textcolor{white}{ll} & \textcolor{white}{3}& \textcolor{white}{6}\\ 220 $\leq$ time $<$ 225 & l & \textcolor{white}{1}& \textcolor{white}{2}\\ \hline \textbf{Totals} & &\textcolor{white}{50} &\textcolor{white}{100} \\ \hline \end{tabular} \end{center} } \frame{ \frametitle{Solution to Example 3.2} For the call centre data, we might get: \begin{center} \begin{tabular}{|c|c|c|c|} \hline \textcolor{blue}{\textbf{Class Interval}} & \textcolor{blue}{\textbf{Tally}} & \textcolor{blue}{\textbf{Frequency}} & \textcolor{blue}{\textbf{Relative Frequency \%}}\\ \hline 175 $\leq$ time $<$ 180 & \textcolor{white}{l} &\textcolor{white}{1} &\textcolor{white}{2} \\ 180 $\leq$ time $<$ 185 & \textcolor{white}{lll} &\textcolor{white}{3} &\textcolor{white}{6} \\ 185 $\leq$ time $<$ 190 & \textcolor{white}{lll} &\textcolor{white}{3} &\textcolor{white}{6} \\ 190 $\leq$ time $<$ 195 & \textcolor{white}{\sout{llll} l} & \textcolor{white}{6} & \textcolor{white}{12}\\ 195 $\leq$ time $<$ 200 & l\textcolor{white}{\sout{llll} \sout{llll}} & \textcolor{white}{10}& \textcolor{white}{20}\\ 200 $\leq$ time $<$ 205 & \textcolor{white}{\sout{llll} \sout{llll} ll} & \textcolor{white}{12} & \textcolor{white}{24}\\ 205 $\leq$ time $<$ 210 & \textcolor{white}{\sout{llll} lll} & \textcolor{white}{8}& \textcolor{white}{16}\\ 210 $\leq$ time $<$ 215 & l\textcolor{white}{ll} & \textcolor{white}{3}& \textcolor{white}{6}\\ 215 $\leq$ time $<$ 220 & l\textcolor{white}{ll} & \textcolor{white}{3}& \textcolor{white}{6}\\ 220 $\leq$ time $<$ 225 & l & \textcolor{white}{1}& \textcolor{white}{2}\\ \hline \textbf{Totals} & &\textcolor{white}{50} &\textcolor{white}{100} \\ \hline \end{tabular} \end{center} } \frame{ \frametitle{Solution to Example 3.2} For the call centre data, we might get: \begin{center} \begin{tabular}{|c|c|c|c|} \hline \textcolor{blue}{\textbf{Class Interval}} & \textcolor{blue}{\textbf{Tally}} & \textcolor{blue}{\textbf{Frequency}} & \textcolor{blue}{\textbf{Relative Frequency \%}}\\ \hline 175 $\leq$ time $<$ 180 & \textcolor{white}{l} &\textcolor{white}{1} &\textcolor{white}{2} \\ 180 $\leq$ time $<$ 185 & \textcolor{white}{lll} &\textcolor{white}{3} &\textcolor{white}{6} \\ 185 $\leq$ time $<$ 190 & \textcolor{white}{lll} &\textcolor{white}{3} &\textcolor{white}{6} \\ 190 $\leq$ time $<$ 195& \textcolor{white}{\sout{llll} l} & \textcolor{white}{6} & \textcolor{white}{12}\\ 195 $\leq$ time $<$ 200 & l\textcolor{white}{\sout{llll} \sout{llll}} & \textcolor{white}{10}& \textcolor{white}{20}\\ 200 $\leq$ time $<$ 205 & \textcolor{white}{\sout{llll} \sout{llll} ll} & \textcolor{white}{12} & \textcolor{white}{24}\\ 205 $\leq$ time $<$ 210 & \textcolor{white}{\sout{llll} lll} & \textcolor{white}{8}& \textcolor{white}{16}\\ 210 $\leq$ time $<$ 215 & ll\textcolor{white}{l} & \textcolor{white}{3}& \textcolor{white}{6}\\ 215 $\leq$ time $<$ 220 & l\textcolor{white}{ll} & \textcolor{white}{3}& \textcolor{white}{6}\\ 220 $\leq$ time $<$ 225 & l & \textcolor{white}{1}& \textcolor{white}{2}\\ \hline \textbf{Totals} & &\textcolor{white}{50} &\textcolor{white}{100} \\ \hline \end{tabular} \end{center} } \frame{ \frametitle{Solution to Example 3.2} For the call centre data, we might get: \begin{center} \begin{tabular}{|c|c|c|c|} \hline \textcolor{blue}{\textbf{Class Interval}} & \textcolor{blue}{\textbf{Tally}} & \textcolor{blue}{\textbf{Frequency}} & \textcolor{blue}{\textbf{Relative Frequency \%}}\\ \hline 175 $\leq$ time $<$ 180 & l &\textcolor{white}{1} &\textcolor{white}{2} \\ 180 $\leq$ time $<$ 185 & lll &\textcolor{white}{3} &\textcolor{white}{6} \\ 185 $\leq$ time $<$ 190 & lll &\textcolor{white}{3} &\textcolor{white}{6} \\ 190 $\leq$ time $<$ 195 & \sout{llll} l & \textcolor{white}{6} & \textcolor{white}{12}\\ 195 $\leq$ time $<$ 200& \textcolor{white}{l}\sout{llll} \sout{llll} & \textcolor{white}{10}& \textcolor{white}{20}\\ 200 $\leq$ time $<$ 205 & \sout{llll} \sout{llll} ll & \textcolor{white}{12} & \textcolor{white}{24}\\ 205 $\leq$ time $<$ 210 & \sout{llll} lll & \textcolor{white}{8}& \textcolor{white}{16}\\ 210 $\leq$ time $<$ 215 & lll & \textcolor{white}{3}& \textcolor{white}{6}\\ 215 $\leq$ time $<$ 220 & lll & \textcolor{white}{3}& \textcolor{white}{6}\\ 220 $\leq$ time $<$ 225 & l & \textcolor{white}{1}& \textcolor{white}{2}\\ \hline \textbf{Totals} & &\textcolor{white}{50} &\textcolor{white}{100} \\ \hline \end{tabular} \end{center} } \frame{ \frametitle{Solution to Example 3.2} For the call centre data, we might get: \begin{center} \begin{tabular}{|c|c|c|c|} \hline \textcolor{blue}{\textbf{Class Interval}} & \textcolor{blue}{\textbf{Tally}} & \textcolor{blue}{\textbf{Frequency}} & \textcolor{blue}{\textbf{Relative Frequency \%}}\\ \hline 175 $\leq$ time $<$ 180 & l &1 &\textcolor{white}{2} \\ 180 $\leq$ time $<$ 185 & lll &3 &\textcolor{white}{6} \\ 185 $\leq$ time $<$ 190 & lll &3 &\textcolor{white}{6} \\ 190 $\leq$ time $<$ 195 & \sout{llll} l & 6 & \textcolor{white}{12}\\ 195 $\leq$ time $<$ 200 & \textcolor{white}{l}\sout{llll} \sout{llll} & 10& \textcolor{white}{20}\\ 200 $\leq$ time $<$ 205 & \sout{llll} \sout{llll} ll & 12 & \textcolor{white}{24}\\ 205 $\leq$ time $<$ 210 & \sout{llll} lll & 8& \textcolor{white}{16}\\ 210 $\leq$ time $<$ 215 & lll & 3& \textcolor{white}{6}\\ 215 $\leq$ time $<$ 220 & lll & 3& \textcolor{white}{6}\\ 220 $\leq$ time $<$ 225 & l & 1& \textcolor{white}{2}\\ \hline \textbf{Totals} & &50 &\textcolor{white}{100} \\ \hline \end{tabular} \end{center} } \frame{ \frametitle{Solution to Example 3.2} For the call centre data, we might get: \begin{center} \begin{tabular}{|c|c|c|c|} \hline \textcolor{blue}{\textbf{Class Interval}} & \textcolor{blue}{\textbf{Tally}} & \textcolor{blue}{\textbf{Frequency}} & \textcolor{blue}{\textbf{Relative Frequency \%}}\\ \hline 175 $\leq$ time $<$ 180 & l &1 &2 \\ 180 $\leq$ time $<$ 185 & lll &3 &6 \\ 185 $\leq$ time $<$ 190 & lll &3 &6 \\ 190 $\leq$ time $<$ 195 & \sout{llll} l & 6 & 12\\ 195 $\leq$ time $<$ 200 & \textcolor{white}{l}\sout{llll} \sout{llll} & 10& 20\\ 200 $\leq$ time $<$ 205 & \sout{llll} \sout{llll} ll & 12 & 24\\ 205 $\leq$ time $<$ 210 & \sout{llll} lll & 8& 16\\ 210 $\leq$ time $<$ 215& lll & 3& 6\\ 215 $\leq$ time $<$ 220 & lll & 3& 6\\ 220 $\leq$ time $<$ 225 & l & 1& 2\\ \hline \textbf{Totals} & &50 &100 \\ \hline \end{tabular} \end{center} } \normal \frame{ \frametitle{3.4 Graphical methods for presenting data} \noindent So far we have looked at \pause \vspl %\pause \begin{itemize} \item different types of \textcolor{purple}{\textbf{data}}, \pause %\pause %\item how to take \textcolor{red}{\textbf{surveys}} %\pause %\item how to \textcolor{red}{\textbf{sample}} from the \textcolor{red}{\textbf{population}} %\pause \item and how to summarise data in a \textcolor{purple}{\textbf{frequency table}} %\begin{itemize} %\item [--]frequency tables %\item [--]\textit{relative} frequency tables %\end{itemize} \end{itemize} \pause \vspace{0.5cm} %\pause We now look at how to display data \textcolor{purple}{\textbf{graphically}}. \pause \bigskip Allows us to see the \textcolor{purple}{\bf main features} of a dataset quickly and clearly. \pause \medskip \begin{itemize} \item Is the distribution symmetric or asymmetric? \pause \vspl \item Are the data spread out or tightly concentrated? \pause \vspl \item Do there appear to be groupings/clusters within the data? \end{itemize} } \frame{ \frametitle{3.4.1 Stem and leaf plots} \noindent \textcolor{purple}{\textbf{Stem and leaf plots}} are a quick and easy way of representing data graphically. \pause \vspace{0.3cm} \noindent They can be used with both \textcolor{purple}{\textbf{discrete}} and \textcolor{purple}{\textbf{continuous}} data. \pause \vspace{0.3cm} \noindent A bit like constructing a frequency table -- but carries visual information about the shape of the distribution. \pause \vspace{0.3cm} \noindent Best seen via an example! } \frame{ \frametitle{Example: Percentage returns on a share} \noindent The following numbers show the percentage returns on an ordinary share for 23 consecutive months: \begin{center} \begin{tabular}{c c c c c c c c } \textcolor{blue}{\textbf{0.2}} & \textcolor{blue}{\textbf{--2.1}} & \textcolor{blue}{\textbf{1.0}} & \textcolor{blue}{\textbf{0.1}} & \textcolor{blue}{\textbf{--0.5}} & \textcolor{blue}{\textbf{2.4}} & \textcolor{blue}{\textbf{--2.3}} & \textcolor{blue}{\textbf{1.5}} \\ \textcolor{blue}{\textbf{1.2}} & \textcolor{blue}{\textbf{--0.6}} & \textcolor{blue}{\textbf{2.4}} & \textcolor{blue}{\textbf{--1.2}} & \textcolor{blue}{\textbf{1.7}} & \textcolor{blue}{\textbf{--1.3}} & \textcolor{blue}{\textbf{--1.2}} & \textcolor{blue}{\textbf{0.9}} \\ \textcolor{blue}{\textbf{0.5}} & \textcolor{blue}{\textbf{0.1}} & \textcolor{blue}{\textbf{--0.1}} & \textcolor{blue}{\textbf{0.3}} & \textcolor{blue}{\textbf{--0.4}} & \textcolor{blue}{\textbf{0.5}} &\textcolor{blue}{\textbf{0.9}} & \\ \end{tabular} \end{center} %\pause \begin{itemize} \item The largest value is 2.4 and the smallest --2.3, and we have lots of decimal values in between \item It seems sensible to have a \textcolor{purple}{\textbf{stem unit}} of 1 and a \textcolor{purple}{\textbf{leaf unit}} of 0.1 \end{itemize} } \frame{ \frametitle{Example: Percentage returns on a share} A stem and leaf diagram for this set of returns might look like: \pause \begin{center} \begin{tabular}{c | c c c c c c c c } \textbf{--2} & 3 & 1 & & & & & & \\ \textbf{--1} & 3 & 2 & 2 & & & & &\\ \textbf{--0} & 6 & 5 & 4 & 1 & & & & \\ \textbf{0} & 1 & 1 & 2 & 3 & 5 & 5 & 9 & 9 \\ \textbf{1} & 0 & 2 & 5 & 7 & & & &\\ \textbf{2} & 4 & 4 & & & & & & \\ \noalign{\medskip} \multicolumn{1}{l}{\textcolor{purple}{\textbf{Stem}}}&\multicolumn{8}{l}{\textcolor{purple}{\textbf{Leaf}}}\\ \end{tabular} $n=23$,\quad $\text{stem unit}=1$,\quad $\text{leaf unit}= 0.1$. \end{center} } \frame{ \frametitle{Example: Production line data} \noindent If there is more than one significant figure in the data, the extra digits are \textcolor{purple}{\textbf{cut}}, not \textcolor{purple}{\textbf{rounded}} -- e.g. $2.97$ would be ``cut'' to 2.9. \vspace{0.3cm} \pause \noindent Consider the following data on lengths of items on a production line (in cm): \begin{center} \begin{tabular}{c c c c c} \textcolor{blue}{\textbf{2.97}} &\textcolor{blue}{\textbf{3.81}} &\textcolor{blue}{\textbf{2.54}} &\textcolor{blue}{\textbf{2.01}} &\textcolor{blue}{\textbf{3.49}} \\ \textcolor{blue}{\textbf{3.09}} &\textcolor{blue}{\textbf{1.99}} &\textcolor{blue}{\textbf{2.64}} &\textcolor{blue}{\textbf{2.31}} & \textcolor{blue}{\textbf{2.22}} \end{tabular} \end{center} \pause \noindent The stem and leaf plot for this is as follows: \begin{center} \begin{tabular}{l| c c ccccccc} \noalign{\medskip} \textbf{1} & 9\\ \textbf{2} & 0 & 2 & 3\\ \textbf{2} & 5 & 6 & 9\\ \textbf{3} & 0 & 4\\ \textbf{3} & 8\\ \noalign{\medskip} \multicolumn{10}{l}{$n=10, \quad \text{stem unit}=1$ cm,\quad $\text{leaf unit}=0.1$ cm.} \end{tabular} \end{center} } \sol \frame{ \frametitle{Example: Production line data} Why do you think we \textcolor{purple}{\bf cut} the extra digits? \pause \vsp So values such as 2.97 do not end up in the wrong row! } \normal \sol \frame{ \frametitle{Example 3.3} The observations in the table below are the recorded time it takes to get through to an operator at a telephone call centre (in seconds). \pause \begin{center} \begin{tabular}{| c c c c c c c c c c|} \hline 54 & 56 & 50 & 67 & 55 & 38 & 49 & 45 & 39 & 50\\ 45 & 51 & 47 & 53 & 29 & 42 & 44 & 61 & 51 & 50\\ 30 & 39 & 65 & 54 & 44 & 54 & 72 & 65 & 58 & 62\\ \hline \end{tabular} \end{center} \pause \begin{center} \begin{tabular}{c | c c c c c c c c c c c c} \noalign{\medskip} \textbf{2} &9 & & & & & & & & & & & \\ \textbf{3} &0 &8 &9 &9 & & & & & & & & \\ \textbf{4} &2 &4 &4 &5 &5 &7 &9 & & & & & \\ \textbf{5} &0 &0 &0 &1 &1 &3 &4 &4 &4 &5 &6 &8 \\ \textbf{6} &1 &2 &5 &5 &7 & & & & & & & \\ \textbf{7} &2 & & & & & & & & & & & \\ \noalign{\medskip} \multicolumn{1}{l}{\textcolor{purple}{\textbf{Stem}}}&\multicolumn{12}{l}{\textcolor{purple}{\textbf{Leaf}}}\\ \noalign{\medskip} \multicolumn{13}{l}{$n=30, \quad \text{stem unit}=10, \quad \text{leaf unit}=1$} \end{tabular} \end{center} } \normal \frame{ \frametitle{Example 3.4} The stem and leaf plot below represents the marks on a test for 52 students. Comment on the distribution of these marks. \begin{center} \begin{tabular}{l|ccccccccccccccccc} \noalign{\medskip} 1 & 4\\ 1 & 5&7&7\\ 2 & 1&1&2&3\\ 2 & 5&5&6&7&8&8 \\ 3 & 2&3&3&3&4&4&4\\ 3 & 5&5&6&7&7&8&8&9&9&9&9\\ 4 & 0&0&1&1&1&2&2&4&4&4\\ 4 & 5&7&7&8&8&8&9\\ 5 & 0&0&0\\ \noalign{\medskip} \multicolumn{17}{l}{$n=52, \quad \text{stem unit}=10$,\quad $\text{leaf unit}=1$.} \end{tabular} \end{center} } \sol \frame{ \frametitle{Solution to Example 3.4} \begin{itemize} \item \textcolor{purple}{\bf Asymmetric distribution} of marks \pause -- in fact, marks are \textcolor{purple}{\bf negatively skewed} \pause \vsp \item Majority of students achieve marks in the 30s/40s \pause \vsp \item The \textcolor{purple}{\bf modal class} is 35--39 \pause \vsp \item Large \textcolor{purple}{\bf spread} of marks \end{itemize} } \normal \frame{ \frametitle{3.4.2 Bar charts} \noindent \textcolor{purple}{\textbf{Bar charts}} are a commonly--used and clear way of presenting categorical data or ungrouped discrete frequency observations. \pause See \textcolor{orange}{\bf section 2.3.2} of the \textcolor{orange}{\bf summer revision booklet}. \vspl Recall the example on students' modes of transport: \begin{center} \begin{tabular}{|c | c|| c| c|| c| c|} \hline \textcolor{purple}{\textbf{Student}} & \textcolor{purple}{\textbf{Mode}} & \textcolor{purple}{\textbf{Student}} & \textcolor{purple}{\textbf{Mode}} & \textcolor{purple}{\textbf{Student}} & \textcolor{purple}{\textbf{Mode}}\\ \hline 1 & Car & 11 & Walk & 21 & Walk\\ 2 & Walk & 12 & Walk & 22 & Metro\\ 3 & Car & 13 & Metro & 23 & Car\\ 4 & Walk & 14 & Bus & 24 & Car\\ 5 & Bus & 15 & Train & 25 & Car\\ 6 & Metro & 16 & Bike & 26 & Bus\\ 7 & Car & 17 & Bus & 27 & Car\\ 8 & Bike & 18 & Bike & 28 & Walk\\ 9 & Walk & 19 & Bike & 29 & Car\\ 10 & Car & 20 & Metro & 30 & Car\\ \hline \end{tabular} \end{center} %\noindent \textbf{Draw a bar chart to represent these data} } \frame{ \frametitle{3.4.2 Bar charts} \begin{center} \includegraphics[angle=270,width=10.5cm]{transport2.ps} \end{center} } \frame{ \frametitle{3.4.3 Pie charts} These charts are commonly used within industry to communicate simple ideas, for example market share. \vsp They are used to show the proportions of a whole. \vsp They are best used when there are only a handful of categories to display. \vsp See \textcolor{orange}{Section 2.3.3} of the \textcolor{orange}{summer revision booklet} for more details. } \begin{comment} \frame{ \frametitle{Announcements} \begin{itemize} \item Last week: \textit{definitions}; how we \textit{get} data; how we \textit{look at} data using tables; \textit{relative frequency} \pause \vspl \item This week in seminars: \textit{stem-and-leaf plots}, and mentioned \textit{bar charts} and \textit{pie charts} \pause \vspl \item Today: More graphical techniques, and some numerical techniques, for summarising data \pause \vspl \item \textbf{CBA2} goes live in practice mode next week (assessed mode the week after) \end{itemize} } \normal \frame{ \frametitle{3.4.4 Histograms} \noindent Bar charts have their limitations, one of which is that they cannot be used to present \textcolor{purple}{\textbf{continuous data}}. \pause %\pause \vspace{0.3cm} \noindent When dealing with continuous random variables a different kind of graph is required -- one such graph is the \textcolor{purple}{\textbf{histogram}}. \vspace{0.3cm} %\pause } %\setbeamercolor{structure}{fg=red!80!black} \frame{ \frametitle{3.4.4 Histograms} \noindent Consider the following data, which show service times (in seconds) for a telephone call centre. \begin{center} \begin{tabular}{|c c c c c|} \hline 214.8412 & 220.6484 & 216.7294 & 195.1217 & 211.4795\\ 195.8980 & 201.1724 & 185.8529 & 183.4600 & 178.8625\\ 196.3321 & 199.7596 & 206.7053 & 203.8093 & 203.1321\\ 200.8080 & 201.3215 & 205.6930 & 181.6718 & 201.7461\\ 180.2062 & 193.3125 & 188.2127 & 199.9597 & 204.7813\\ 198.3838 & 193.1742 & 204.0352 & 197.2206 & 193.5201\\ 205.5048 & 217.5945 & 208.8684 & 197.7658 & 212.3491\\ 209.9000 & 197.6215 & 204.9101 & 203.1654 & 192.9706\\ 208.9901 & 202.0090 & 195.0241 & 192.7098 & 219.8277\\ 208.8920 & 200.7965 & 191.9784 & 188.8587 & 206.8912\\ \hline \end{tabular} \end{center} %\noindent \textbf{Produce a histogram to display these data graphically} } \frame{ \frametitle{3.4.4 Histograms} \noindent Producing a histogram is much like producing a bar chart. \pause \vspace{0.3cm} %\pause \noindent It is often best to produce a frequency table first which collects all the data together in an ordered format. \pause \vspace{0.3cm} %\pause \begin{center} \begin{tabular}{|c|c|} \hline \textcolor{purple}{\textbf{Service time}} & \textcolor{purple}{\textbf{Frequency}}\\ \hline $175\leq \textrm{\sf time} <180$ & 1\\ $180\leq \textrm{\sf time} <185$ & 3\\ $185\leq \textrm{\sf time} <190$ & 3\\ $190\leq \textrm{\sf time} <195$ & 6\\ $195\leq \textrm{\sf time} <200$ & 10\\ $200\leq \textrm{\sf time} <205$ & 12\\ $205\leq \textrm{\sf time} <210$ & 8\\ $210\leq \textrm{\sf time} <215$ & 3\\ $215\leq \textrm{\sf time} <220$ & 3\\ $220\leq \textrm{\sf time} <225$ & 1\\ \hline \textbf{Total} & 50\\ \hline \end{tabular} \end{center} } \frame{ \frametitle{3.4.4 Histograms} %\begin{center} \hspace*{-1cm} \includegraphics[height=7.3cm]{hist.eps}%\includegraphics[height=7.5cm]{callcentre.ps} %\end{center} } \begin{frame} \frametitle{3.4.4 Histograms} \noindent At first sight histograms look similar to bar charts. There are, however, some critical differences: %\pause \bigskip \begin{itemize} \item The horizontal ($x$-axis) is a \textcolor{purple}{\bf continuous scale}\pause \vspl \item As a result there are \textcolor{purple}{\bf no gaps between the bars}\pause \vspl \item Strictly speaking, the \textcolor{purple}{\bf area} of the bar is proportional to the frequency. \end{itemize} \end{frame} \frame{ \frametitle{Example 3.5} The \textit{Holiday Hypermarket} travel agency received 64 telephone calls yesterday morning. The table below gives information of the lengths, in minutes, of these telephone calls. \begin{center} \begin{tabular}{|c|c|c|} \hline \textbf{Length ($\boldsymbol{x}$) minutes} &\textbf{Frequency} &\textbf{Frequency density} \\ \hline $0\leq x < 5$ &4 &0.8 \\ $5\leq x <15$ &10 &1.0 \\ $15\leq x <30$ &24 & \\ $30\leq x <40$ &20 & \\ $40 \leq x<45$ &6 & \\ \hline \end{tabular} \end{center} Complete this table, and construct a histogram for these data. What is the \textit{modal class} here? } \sol \frame{ \frametitle{Solution to Example 3.5 (1/3)} First, the \textcolor{purple}{\bf frequency densities}, or bar heights. \pause \vsp Now we know that the area of the bar represents frequency. \pause \vsp The width of the first is 5, and the area must be 4, so \pause \begin{eqnarray*} \textrm{\sf Area} &=& \textrm{\sf Base} \times \textrm{\sf Height} \pause \\ 4 &=& 5 \times \textrm{\sf Height} \end{eqnarray*} \pause and so \pause \[ \textrm{\sf Height} = 4/5 = 0.8. \] } \sol \frame{ \frametitle{Solution to Example 3.5 (3/3)} \begin{center} \includegraphics[angle=-90,width=11cm]{histy1.ps} \end{center} } \frame{ \frametitle{Solution to Example 3.5 (3/3)} \begin{center} \includegraphics[angle=-90,width=11cm]{histy2.ps} \end{center} } \frame{ \frametitle{Solution to Example 3.5 (3/3)} \begin{center} \includegraphics[angle=-90,width=11cm]{histy3.ps} \end{center} } \frame{ \frametitle{Solution to Example 3.5 (3/3)} \begin{center} \includegraphics[angle=-90,width=11cm]{histy4.ps} \end{center} } \frame{ \frametitle{Solution to Example 3.5 (3/3)} \begin{center} \includegraphics[angle=-90,width=11cm]{histy5.ps} \end{center} } \normal \frame{ \frametitle{3.4.4 Histograms} \textbf{Summary} \vsp \begin{itemize} \item Like a bar chart, but for continous data \pause \vspl \item First step: Collect your data into a frequency table \pause \vspl \item If the width of all class intervals is equal, just have ``frequency'' on the $y$--axis \pause \vspl \item Otherwise, you will need to think about ``frequency density'' \end{itemize} } \frame{ \frametitle{3.4.5 Percentage relative frequency histograms} \noindent Instead of using \textcolor{purple}{\textbf{frequency}} on the vertical axis ($y$-axis), you \textit{could} use the \textcolor{purple}{\textbf{percentage relative frequency}}. \pause For example:\pause \begin{center} \begin{tabular}{|c|c|c|} \hline \textcolor{blue}{\textbf{Service time}} & \textcolor{blue}{\textbf{Frequency}} &\textcolor{blue}{\textbf{Relative Frequency (\%)}}\\ \hline $175\leq \textrm{\sf time}<180$ & 1 & 2\\ $180\leq \textrm{\sf time}<185$ & 3 & 6\\ $185\leq \textrm{\sf time}<190$ & 3 & 6\\ $190\leq \textrm{\sf time}<195$ & 6 & 12\\ $195\leq \textrm{\sf time}<200$ & 10 & 20\\ $200\leq \textrm{\sf time}<205$ & 12 & 24\\ $205\leq \textrm{\sf time}<210$ & 8 & 16\\ $210\leq \textrm{\sf time}<215$ & 3 & 6\\ $215\leq \textrm{\sf time}<220$ & 3 & 6\\ $220\leq \textrm{\sf time}<225$ & 1 & 2\\ \hline \textbf{Totals} & \textbf{50} & \textbf{100}\\ \hline \end{tabular} \end{center} %\pause } %\setbeamercolor{structure}{fg=red!80!black} \frame{ \frametitle{3.4.5 Percentage relative frequency histograms} \begin{center} \hspace*{-1cm} \includegraphics[height=7.3cm]{percrelfreqhist.eps}%\includegraphics[height=7.5cm]{callcentre2.ps} \end{center} } \frame{ \frametitle{3.4.5 Percentage relative frequency histograms} \begin{center} \hspace*{-0.8cm} \includegraphics[height=7.3cm]{hist.eps}%\includegraphics[height=7.5cm]{callcentre2.ps} \end{center} } \frame{ \frametitle{Why use percentage relative frequency?} \textbf{It is useful for comparing two or more histograms} \vsp \begin{itemize} \item [--] If one sample were larger than the other, the standard histograms would look different just because of the different sample sizes. \pause \vspl \item [--] Looking at percentages ``\textcolor{purple}{\textbf{puts both samples on the same scale}}'' and removes this difference.\pause \vspl \item [--] This enables us to look at \textcolor{purple}{\textbf{relative}} differences. \end{itemize} } \frame{ \frametitle{Why use percentage relative frequency?} \noindent The following (frequency) histograms were created from two samples, one of size 100 and one of size 400 ... \begin{center} \includegraphics[height=6.75cm]{overlay1.ps} \end{center} } \frame{ \frametitle{Why use percentage relative frequency?} \noindent ... and this is what happens when we use \textcolor{purple}{\textbf{\% relative frequency}} \begin{center} \includegraphics[height=6.75cm]{overlay2.ps} \end{center} } \frame{ \frametitle{3.4.6 Polygons} \noindent So percentage relative frequency histograms are useful for comparing two groups. \pause \vspace{0.3cm} \noindent However, can you imagine how \textcolor{purple}{\textbf{messy}} these ``superimposed'' histograms would get if we had three or more groups? \vspace{0.3cm} \pause \noindent To get round this \textcolor{purple}{\textbf{messy problem}} we use \textcolor{purple}{\textbf{polygons}}. } \frame{ \frametitle{3.4.6 Polygons} \begin{center} \includegraphics[angle=-90,width=11cm]{histy99.ps} \end{center} } \frame{ \frametitle{3.4.6 Polygons} \begin{center} \includegraphics[angle=-90,width=11cm]{histy999.ps} \end{center} } \frame{ \frametitle{3.4.6 Polygons} \begin{center} \includegraphics[angle=-90,width=11cm]{histy9999.ps} \end{center} } \frame{ \frametitle{3.4.6 Polygons} \begin{center} \includegraphics[angle=-90,width=11cm]{histy99999.ps} \end{center} } \frame{ \frametitle{3.4.6 Polygons} \begin{center} \includegraphics[height=7cm]{Income.ps} \end{center} } \frame{ \frametitle{3.4.7 Cumulative relative frequency polygons (Ogives)} %\noindent \textcolor{red}{\textbf{Cumulative}} percentage relative frequency is also a useful tool. %\vspace{0.3cm} \noindent The \textcolor{purple}{\textbf{cumulative}} percentage relative frequency is simply the sum of the percentage relative frequencies at the end of each class interval. \vspace{0.3cm} \noindent In other words, we add the frequencies up as we go along! \vspace{0.3cm} \pause \noindent \textbf{Consider the simple example:} \begin{center} \begin{tabular}{|c|c|c|} \hline \textcolor{red}{\textbf{Class Interval}} & \textcolor{red}{\textbf{ \% Relative Frequency}} & \textcolor{red}{\textbf{Cum. \% Rel. Freq.}}\\ \hline $0 \leq x < 10$ & 10 & 10\\ $10 \leq x < 20$ & 20 & 30\\ $20 \leq x < 30$ & 35 & 65\\ $30 \leq x < 40$ & 25 & 90\\ $40 \leq x < 50$ & 10 & 100\\ \hline \end{tabular} \end{center} } \frame{ \frametitle{3.4.7 Cumulative relative frequency polygons (Ogives)} The corresponding graph, or \textcolor{purple}{\textbf{ogive}}, is simple to produce by hand -- watch out though! For these plots we use the \textcolor{purple}{\textbf{end--points}} instead of the \textcolor{purple}{\textbf{mid--points}}! %Why?! } \frame{ \frametitle{3.4.7 Cumulative relative frequency polygons (Ogives)} \begin{center} \includegraphics[height=8cm]{simple-ogive}%\includegraphics[height=7cm]{ogive.ps} \end{center} } \frame{ \frametitle{3.4.7 Cumulative relative frequency polygons (Ogives)} \noindent For the income data on West Road and Jesmond Road, we get: \begin{center} \includegraphics[height=7cm]{Ogivesforincome.ps} \end{center} } \frame{ \frametitle{Other graphical summaries} There are many other forms of graphical summary, including: \vsp \pause \begin{itemize} \item Multiple Bar Charts \vspl \item Scatter plots \vspl \item Time series plots \end{itemize} \vsp These are widely used in business presentations. \vsp \pause \begin{center} \includegraphics[width=3cm]{bookworm.ps} \end{center} } \frame{ \frametitle{3.5 Numerical summaries of data} \noindent So far we have only considered graphical methods for presenting data -- \pause useful starting points. \pause \vsp As we shall see, however, for many purposes we might also require \textcolor{purple}{\bf numerical} methods for summarising data. \pause \vsp Before we introduce some ways of summarising data numerically, let us first think about some notation. } \frame{ \frametitle{3.5.1 Mathematical notation} \noindent Before we can talk more about numerical techniques we first need to define some basic notation. \vsp This will allow us to generalise all situations with a \textcolor{purple}{\bf simple shorthand}. } \frame{ \frametitle{3.5.1 Mathematical notation} Very often in statistics we replace actual numbers with letters in order to be able to write \textcolor{purple}{\bf general formulae}. \vsp We generally use a single \textcolor{purple}{\bf upper case} letter to represent our random variable and the \textcolor{purple}{\bf lower case} to represent sample data, with subscripts to distinguish individual observations in the sample. \vsp Amongst the most common letters to use is $x$, although $y$ and $z$ are frequently used as well. } \frame{ \frametitle{3.5.1 Mathematical notation} For example, suppose we ask a random sample of three people how many mobile phone calls they made yesterday. \vsp We might get the following data: \textcolor{brown}{\bf 1}, \textcolor{brown}{\bf 5}, \textcolor{brown}{\bf 7}. \pause \vsp If we take another sample we will most likely get different data, say \textcolor{brown}{\bf 2}, \textcolor{brown}{\bf 0}, \textcolor{brown}{\bf 3}. } \frame{ \frametitle{3.5.1 Mathematical notation} Using algebra we can represent the \textcolor{purple}{\bf general case} as $x_1$, $x_2$, $x_3$: \pause \begin{center} \begin{tabular}{lc c c} \textcolor{brown}{\bf 1st sample} & 1 & 5 & 7\\ \pause \textcolor{brown}{\bf 2nd sample} & 2 & 0 & 3\\ \hline \pause \textcolor{brown}{\bf typical sample} & $x_1$ & $x_2$ & $x_3$ \end{tabular} \end{center} } \frame{ \frametitle{3.5.1 Mathematical notation} \begin{center} \begin{tabular}{lc c c} \textcolor{brown}{\bf 1st sample} & 1 & 5 & 7\\ \pause \textcolor{brown}{\bf 2nd sample} & 2 & 0 & 3\\ \hline \pause \textcolor{brown}{\bf typical sample} & $x_1$ & $x_2$ & $x_3$ \end{tabular} \end{center} This can be generalised further by referring to the random variable \textcolor{purple}{\bf as a whole} as $X$ and the $i\text{th}$ observation in the sample as $x_i$. \pause \vsp Hence, in the first sample above, the second observation is $x_2=5$ whilst in the second sample it is $x_2=0$. \vsp The letters $i$ and $j$ are most commonly used as the index numbers for the subscripts. \vsp \noindent The total number of observations in a sample is usually referred to by the letter $n$. Hence in our simple example above $n=3$.} \frame{ \frametitle{3.5.1 Mathematical notation} \noindent The next important piece of notation to introduce is the symbol \LARGE \[ \boldsymbol{\sum} \] \normalsize \vsp This is the upper case of the Greek letter \textcolor{purple}{\bf sigma}. \vsp It is used to represent the phrase ``\textcolor{purple}{\bf sum the values}''. } \frame{ \frametitle{3.5.1 Mathematical notation} This symbol is used as follows: \pause \begin{eqnarray*} \sum_{i=1}^n x_i &=&\pause x_1 \pause + x_2 \pause + x_3 \pause + \dots + x_n. \end{eqnarray*} \pause \noindent This notation is used to represent the sum of all the values in our data (from the first $i=1$ to the last $i=n$). \pause \vsp This is often abbreviated to $\sum x$ when we sum over all the data in our sample. } \frame{ \frametitle{3.5.2 Measures of Location} \noindent These are also referred to as measures of \textcolor{purple}{\bf centrality} or, more commonly, \textcolor{purple}{\bf averages}. \vsp In general terms, they tell us the value of a ``\textcolor{purple}{\bf typical}'' observation. \vsp There are three measures which are commonly used: \pause \vspl \begin{itemize} \item the \textcolor{purple}{\bf mean}; \pause \vspl \item the \textcolor{purple}{\bf median}, and \pause \vspl \item the \textcolor{purple}{\bf mode}.\pause \end{itemize} \vsp We will consider these in turn.} \frame{ \frametitle{3.5.2 The Arithmetic Mean} \noindent The \textcolor{purple}{\bf arithmetic mean} is perhaps the most commonly used measure of location. \vsp We often refer to it as \underline{the} average or just the \textcolor{purple}{\bf mean}. \vsp The arithmetic mean is calculated by simply adding all our data together and dividing by the number of data we have. \vsp So if our data were \textcolor{brown}{\bf 10},\textcolor{brown}{\bf 12}, and \textcolor{brown}{\bf 14}, then our mean would be \pause \begin{align*} \frac{10 + 12 + 14}{3} &= \frac{36}{3} = 12. \end{align*} } \frame{ \frametitle{3.5.2 The Arithmetic Mean} We denote the mean of our sample, or sample mean, using the notation $\bar x$ (``$x$ bar''). \vsp In general, the mean is calculated using the formula \[ \bar x = \frac{1}{n}\sum_{i=1}^{n} x_i \] or equivalently as \pause \[ \bar x=\frac{\sum x}{n}. \] \pause \noindent For small data sets this is easy to calculate \textcolor{purple}{\bf by hand}, though this is simplified by using the \textcolor{purple}{\bf statistics mode} on a University approved calculator (see next week's tutorials). } \frame{ \frametitle{3.5.2 The Arithmetic Mean} \noindent Sometimes we might not have the raw data; instead, the data might be available in the form of a \textcolor{purple}{\bf table}. \vsp It is still possible to calculate the mean from such data. } \frame{ \frametitle{3.5.2 The Arithmetic Mean} Let us first consider the case where we have some \textcolor{purple}{\bf ungrouped discrete data}. \pause \vsp Previously we have seen the data: \begin{center} \begin{tabular}{|c|c||c|c|} \hline \textbf{Date} & \textbf{Cars Sold} & \textbf{Date} & \textbf{Cars Sold}\\ \hline 01/07/12 & 9 & 08/07/12 & 10\\ 02/07/12 & 8 & 09/07/12 & 5\\ 03/07/12 & 6 & 10/07/12 & 8\\ 04/07/12 & 7 & 11/07/12 & 4\\ 05/07/12 & 7 & 12/07/12 & 6\\ 06/07/12 & 10 & 13/07/12 & 8\\ 07/07/12 & 11 & 14/07/12 & 9\\ \hline \end{tabular} \end{center} \noindent The mean number of cars sold per day is \pause \begin{align*} \bar x = \frac{9 + 8 + \ldots + 8 + 9}{14} = 7.71. \end{align*}} \frame{ \frametitle{3.5.2 The Arithmetic Mean} \noindent These data can be presented as the \textcolor{purple}{\bf frequency table}: {\normalsize \begin{center} \begin{tabular}{|c|c c c c c c c c|} \hline \textbf{Cars Sold $(x_{(j)})$} &4 &5 &6 &7 &8 &9 &10 &11 \\ \hline \textbf{Frequency $(f_j)$} &1 &1 &2 &2 &3 &2 &2 &1 \\ \hline \end{tabular} \end{center}} \pause \noindent The sample mean can be calculated from these data as \begin{eqnarray*} \bar x &=& \frac{4+5+\overbrace{6+6}^{\times 2}+\overbrace{7+7}^{\times 2}+\overbrace{8+8+8}^{\times 3}+\overbrace{9+9}^{\times 2}+\overbrace{10+10}^{\times 2}+11}{14}\\ \pause & & \\ & & \\ &=&\frac{(\textbf{4} \times 1) + (\textbf{5} \times 1) + (\textbf{6} \times 2) + (\textbf{7}\times 2)+ \cdots +(\textbf{11} \times 1)}{14} \\ \pause & & \\ & & \\ &=& 7.71. \end{eqnarray*} } \frame{ \frametitle{3.5.2 The Arithmetic Mean} \noindent We can express this calculation of the sample mean from discrete tabulated data as \pause \[ \bar x = \frac{1}{n}\sum_{j=1}^{k}f_j x_{(j)} . \] \pause \noindent Here the different values of $X$ which occur in the data are $ x_{(1)}, x_{(2)}, \ldots , x_{(k)}$. \pause \vsp In this example $ x_{(1)}=4,\ x_{(2)}=5, \ldots, x_{(k)}=11 $ and $ k=8. $ } \frame{ \frametitle{3.5.2 The Arithmetic Mean} \noindent If we only have \textcolor{purple}{\bf grouped frequency data}, it is still possible to \emph{approximate} the value of the sample mean. \pause \vsp Consider the following (ordered) data: \begin{center} \begin{tabular}{rrrrrrrrrr} 8.4 & 8.7 & 9.0 & 9.0 & 9.2 & 9.3 & 9.3 & 9.5 & 9.6 & 9.6 \\ 9.6 & 9.7 & 9.7 & 9.9 & 10.3 & 10.4 & 10.5 & 10.7 & 10.8 & 11.4 \end{tabular} \pause \end{center} \noindent The sample mean of these data is \textbf{9.73}.} % s=0.75818} \frame{ \frametitle{3.5.2 The Arithmetic Mean} Grouping these data into a frequency table gives \begin{center} \begin{tabular}{|c|c|c|} \hline \textbf{Class Interval} & \textbf{mid--point $(m_j)$} & \textbf{Frequency $(f_j)$}\\ \hline $8.0 \leq x <8.5$ & 8.25 & 1\\ $8.5 \leq x <9.0$ & 8.75 & 1\\ $9.0 \leq x <9.5$ & 9.25 & 5\\ $9.5 \leq x <10.0$ & 9.75 & 7\\ $10.0 \leq x <10.5$ & 10.25 & 2\\ $10.5 \leq x <11.0$ & 10.75 & 3\\ $11.0 \leq x <11.5$ & 11.25 & 1\\ \hline \textbf{Total ($n$)} & & 20\\ \hline \end{tabular} \end{center} } \frame{ \frametitle{3.5.2 The Arithmetic Mean} \noindent Therefore, the (approximate) sample mean is calculated using the the frequencies ($f_j$) and the mid--points ($m_j$) as \[ \bar x = \frac{1}{n}\sum_{j=1}^{k} f_j m_j. \] \pause \noindent For the grouped data above, we obtain \pause \[ \bar x = \frac{1}{20}\left(1\times 8.25 + 1\times 8.75 + \cdots + 3\times 10.75 + 1\times 11.25\right)= 9.775. \] \pause \noindent This value is fairly close to the correct sample mean and is a reasonable approximation given the partial information we have in the table.} \frame{ \frametitle{3.5.2 The Median} \noindent The \textcolor{purple}{\bf median} is occasionally used instead of the mean, particularly when the data have an \textcolor{purple}{\bf asymmetric profile}. \pause \vsp The median is the \textcolor{purple}{\bf middle value} of the observations when they are listed in ascending order. \pause \vsp It is straightforward to determine the median for small data sets, particularly via a stem and leaf plot. \pause \vsp For larger data sets, the calculation is more easily done using \texttt{Minitab} (see Semester 2). } \frame{ \frametitle{3.5.2 The Median} \noindent The median is that value that has half the observations above it and half below. \vsp If the sample size ($n$) is an odd number, we have: \pause \[ \text{\sf median}= \left(\frac{n + 1}{2}\right)^{th} \text{\sf largest observation}. \] } \frame{ \frametitle{3.5.2 The Median} \noindent For example, if our data were \begin{center} \textcolor{brown}{\bf 2}, \textcolor{brown}{\bf 3}, \textcolor{brown}{\bf 3}, \textcolor{brown}{\bf 5}, \textcolor{brown}{\bf 6}, \textcolor{brown}{\bf 7}, \textcolor{brown}{\bf 9}, \pause \end{center} then the sample size ($n=7$) is an odd number and therefore the median is the \pause \begin{align*} \frac{7 + 1}{2} &= 4^{th} \text{ \sf largest observation}, \end{align*} \pause that is, the median is the fourth largest (or smallest) ranked observation. \pause \vsp For these data the $\text{\sf median}=5$. } \frame{ \frametitle{3.5.2 The Median} \noindent If the sample size ($n$) is an even number the process is slightly more complicated: \begin{eqnarray*} \text{\sf median}&= &\text{\sf average of the } \left(\frac{n}{2}\right)^{th}\\ & &\\ & & \hspace{1cm} \textrm{\sf and the } \left(\frac{n}{2} +1\right)^{th} \textrm{\sf largest observations}. \end{eqnarray*} } \frame{ \frametitle{3.5.2 The Median} For example, if our data were \begin{center} \textcolor{brown}{\bf 2}, \textcolor{brown}{\bf 3}, \textcolor{brown}{\bf 3}, \textcolor{brown}{\bf 5}, \textcolor{brown}{\bf 6}, \textcolor{brown}{\bf 7}, \textcolor{brown}{\bf 9}, \textcolor{brown}{\bf 10}, \end{center} \pause then the sample size ($n=8$) is an even number and therefore \pause \begin{eqnarray*} \text{\sf median}&=&\textrm{\sf average of the } \left(\frac{8}{2}\right)^{th}\\ & & \\ & & \hspace{1cm} \textrm{\sf and the } \left(\frac{8}{2} +1\right)^{th} \text{\sf largest observations}\\ \pause & & \\ &=& \frac{5 + 6}{2}\\ & & \\ &=& 5.5. \end{eqnarray*} } \frame{ \frametitle{3.5.2 The Median} \noindent It is possible to estimate the median value from an \textcolor{purple}{\bf ogive} as it is half way through the ordered data and hence is at the 50\% level of the cumulative frequency. \pause \vsp The accuracy of this estimate will depend on the accuracy of the ogive drawn. } \frame{ \frametitle{3.5.2 The Mode} \noindent This is the final measure of location we will look at. \vsp It is the value of the random variable in the sample which \textcolor{purple}{\bf occurs with the highest frequency}. \vsp It is usually found by inspection (no formula). } \frame{ \frametitle{3.5.2 The Mode} For discrete data this is easy -- the mode is simply the most common value. \vsp So, on a bar chart, it would be the category with the \textcolor{purple}{\bf highest bar}. \pause \vsp For example, consider the following data: \pause \begin{center} \textcolor{brown}{\bf 2}, \textcolor{brown}{\bf 2}, \textcolor{brown}{\bf 2}, \textcolor{brown}{\bf 3}, \textcolor{brown}{\bf 3}, \textcolor{brown}{\bf 4}, \textcolor{brown}{\bf 5}.\pause \end{center} Quite obviously the mode is 2 as it occurs most often. } \frame{ \frametitle{3.5.2 The Mode} \noindent It is possible to refer to \textcolor{purple}{\bf modal classes} with grouped data. \vsp This is simply the class with the greatest frequency of observations. \pause \vsp For example, the model class of \begin{center} \begin{tabular}{|c|c|} \hline \textbf{Class} & \textbf{Frequency}\\ \hline $10 \leq x < 20$ & 10\\ $20 \leq x < 30$ & 15\\ $30 \leq x < 40$ & 30\\ \hline \end{tabular} \end{center} is obviously $30 \leq x < 40$. } \frame{ \frametitle{3.5.3 Measures of Spread} \noindent A measure of location is insufficient in itself to summarise data as it only describes the value of a \textcolor{purple}{\bf typical outcome} and not how much \textcolor{purple}{\bf variation} there is in the data. For example: \begin{center} \begin{tabular}{c|c c c} \textbf{Dataset 1} &\textcolor{brown}{\bf 6} & \textcolor{brown}{\bf 22} &\textcolor{brown}{\bf 38} \\ \textbf{Dataset 2} &\textcolor{brown}{\bf 21} &\textcolor{brown}{\bf 22} &\textcolor{brown}{\bf 23}\end{tabular} \end{center} \pause \vspl For both of these datasets, $\bar{x}=\textsf{median}=22$. However,\pause \vspl \begin{itemize} \item dataset 1 ranges considerably from these average values; \pause \vspl \item dataset 2 is very tightly concentrated around these averages. \end{itemize} } \frame{ \frametitle{3.5.3 Measures of Spread} The mean or the median does not fully represent the data. \pause \vsp There are three basic \textcolor{purple}{\bf measures of spread} which we will consider: \pause \vsp \begin{itemize} \item the \textcolor{purple}{\bf range} \pause \vspl \item the \textcolor{purple}{\bf inter--quartile range}, and \pause \vspl \item the \textcolor{purple}{\bf sample variance}/\textcolor{purple}{\bf standard deviation}. \end{itemize} } \frame{ \frametitle{3.5.3 The Range} \noindent This is the \textcolor{purple}{\bf simplest measure of spread}.\pause \vsp It is simply the difference between the largest and smallest observations. \pause \vsp In our simple example above: \pause \begin{eqnarray*} \textsf{Range for dataset 1} &=&38 - 6 = 32 \pause \\ & & \\ \textsf{Range for dataset 2} &=&23 - 21 = 2 \end{eqnarray*} \pause These clearly describe very different data sets. The first set has a much wider range (data more spread out) than the second. } \frame{ \frametitle{3.5.3 The Range} \noindent There are two problems with the range as a measure of spread. \pause \vsp \begin{itemize} \item Uses the two most extreme data points -- value can be unduly influenced by one particularly large/small value (\textcolor{purple}{\bf outliers}) \pause \vspl \item Only suitable for comparing (roughly) equally--sized samples \end{itemize} } \frame{ \frametitle{3.5.3 The Inter--Quartile Range} \noindent The inter--quartile range describes the range of the \textcolor{purple}{\bf middle half of the data} and so is less prone to the influence of the extreme values. \pause \vsp \noindent To calculate the inter--quartile range (IQR) we simply divide the the ordered data into \textcolor{purple}{\bf four quarters}.\pause \vsp The three values that split the data into these quarters are called the \textcolor{purple}{\bf quartiles}. } \frame{ \frametitle{3.5.3 The Inter--Quartile Range} \begin{itemize} \item The first quartile (\textcolor{purple}{\bf lower quartile}, $Q1$) has 25\% of the data below it \pause \vspl \item The second quartile (\textcolor{purple}{\bf median}, $Q2$) has 50\% of the data below it \pause \vspl \item The third quartile (\textcolor{purple}{\bf upper quartile}, $Q3$) has 75\% of the data below it. \pause \end{itemize}} \frame{ \frametitle{3.5.3 The Inter--Quartile Range} We already know how to find the median, and the other quartiles are calculated as follows: \pause \begin{eqnarray*} Q1 &=& \frac{(n+1)}{4}\text{th}\text{ \sf smallest observation}\\ & & \\ Q3 &=& \frac{3(n+1)}{4}\text{th}\text{ \sf smallest observation}. \end{eqnarray*} \pause \noindent Just as with the median, these quartiles might not correspond to actual observations. \vsp For example, in a dataset with $n=20$ values, the lower quartile is the \[ \frac{21}{4} = 5\,\frac{1}{4}^{\sf th} \textsf{ largest observation,} \] that is, a quarter of the way between the 5th and 6th largest observations.} \frame{ \frametitle{3.5.3 The Inter--Quartile Range} Consider again the data $(n=20)$: \pause \begin{center} \begin{tabular}{rrrrrrrrrr} 8.4 & 8.7 & 9.0 & 9.0 & 9.2 & 9.3 & 9.3 & 9.5 & 9.6 & 9.6 \\ 9.6 & 9.7 & 9.7 & 9.9 & 10.3 & 10.4 & 10.5 & 10.7 & 10.8 & 11.4 \end{tabular} \end{center} \noindent Here the 5th and 6th smallest observations are 9.2 and 9.3 respectively. Therefore, the lower quartile is $Q1=9.225$. \pause \vsp Similarly, the upper quartile is the \[ \frac{3 \times 21}{4}=15\frac{3}{4}^{\sf th} \textsf{ smallest observation,} \] that is, three quarters of the way between 10.3 and 10.4; so $Q3=10.375$. } \frame{ \frametitle{3.5.3 The Inter--Quartile Range} \noindent The inter--quartile range is simply the difference between the upper and lower quartiles, that is \pause \begin{eqnarray*} IQR &=& Q3 - Q1 \\ & & \\ &=&10.375-9.225 \\ & & \\ &=&1.15. \end{eqnarray*} } \frame{ \frametitle{3.5.3 The Inter--Quartile Range} \noindent The interquartile range can also be \emph{estimated} from the ogives in a similar manner to the median. \pause \vsp Simply draw the ogive and then read off the values for 75\% and 25\% and calculate the difference between them. This is especially useful if you only have grouped data. \vsp Again the accuracy depends on the quality of your graph. \pause \vsp \noindent The inter--quartile range is useful as it allows us to make comparisons between the ranges of two data sets, without the problems caused by \textcolor{purple}{\bf outliers} or \textcolor{purple}{\bf unequal sample sizes}. } \frame{ \frametitle{3.5.3 The Sample Variance and Standard Deviation} \noindent The \textcolor{purple}{\bf sample variance} is the standard measure of spread used in statistics. \vsp It is usually denoted by $s^2$ and is simply the ``average'' of the squared distances of the observations from the sample mean. \pause \vsp We use the formula \pause \[ s^2 = \frac{\left( x_1 - \bar x \right)^2 + \left( x_2-\bar x \right)^2 + \ldots + \left( x_n - \bar x \right)^2}{n-1}; \]\pause the $(n-1)$ divisor will be explained in semester 2. } \frame{ \frametitle{3.5.3 The Sample Variance and Standard Deviation} \noindent We can rewrite this using more condensed mathematical notation: \pause \begin{align*} s^2 &= \frac{1}{n-1} \sum_{i=1}^n \left(x_i - \bar x \right)^2,\\ \intertext{\sf or equivalently as} \\ s^2 &= \frac{1}{n-1}\left\{ \sum_{i=1}^n x_i^2 - n \left(\bar x \right)^2 \right\}. \end{align*} \pause \noindent Note that the notation $x_i^2$ represents the squared value of the observation $x_i.$ That is, $x_i^2=(x_i)^2$. \pause \vsp \noindent The \textcolor{purple}{\bf sample standard deviation} $s$ is the positive square root of the sample variance. } \frame{ \frametitle{3.5.3 The Sample Variance and Standard Deviation} \noindent A different calculation is needed when the data are given in the form of a grouped frequency table with frequencies ($f_i$) in intervals with mid--points ($m_i$). \pause \vsp First the sample mean $\bar x$ is approximated (as described earlier) and then the sample variance is approximated as \[ s^2 = \frac{1}{n-1}\left\{ \sum_{i=1}^k f_i m_i^2 - n \left(\bar x \right)^2 \right\}. \] } \frame{ \frametitle{Example 3.6} Consider again the data \begin{center} \begin{tabular}{rrrrrrrrrr} 8.4 & 8.7 & 9.0 & 9.0 & 9.2 & 9.3 & 9.3 & 9.5 & 9.6 & 9.6 \\ 9.6 & 9.7 & 9.7 & 9.9 & 10.3 & 10.4 & 10.5 & 10.7 & 10.8 & 11.4 \end{tabular} \end{center} Calculate the sample variance and hence the sample standard deviation. } \sol \frame{ \frametitle{Example 3.6: Solution (1/2)} The \textcolor{purple}{\bf sample mean} is \[ \bar{x} = 9.73. \]\pause If we use the second version of the formula \[ s^2 = \frac{1}{n-1}\left\{ \sum_{i=1}^n x_i^2 - n \left(\bar x \right)^2 \right\}, \] we also need $\sum x^{2}$, which is: \[ 8.4^{2}+8.7^{2}+9.0^{2} + \ldots + 11.4^{2} = 1904.38. \] } \frame{ \frametitle{Example 3.6: Solution (2/2)} So \pause \begin{eqnarray*} s^2 &=& \frac{1}{\textcolor{blue}{\bf 20}-1}\left\{\textcolor{blue}{\bf 1904.38} - \textcolor{blue}{\bf 20} \left(\textcolor{blue}{\bf 9.73}\right)^2 \right\} \\ & & \\ &=& 0.5748. \end{eqnarray*} \pause The standard deviation is just the square root of this, i.e. \pause \[ s = \sqrt{0.5748} = 0.7582. \] } \normal \frame{ \frametitle{3.5.4 Box--and--whisker plots} \noindent Box and whisker plots are another graphical method for displaying data and are particularly useful in highlighting differences between groups. \pause \vsp These plots use some of the key summary statistics we have looked at earlier -- the \textcolor{purple}{\bf quartiles} -- and also the \textcolor{purple}{\bf maximum} and \textcolor{purple}{\bf minimum} observations. } \frame{ \frametitle{3.5.4 Box--and--whisker plots} \noindent The plot is constructed as follows: \pause \vsp \begin{itemize} \item Lay out an $x$--axis for the full range of the data \pause \vspl \item Draw a rectangle with ends at the the upper and lower quartiles\pause \vspl \item Split the rectangle into two at the median: You now have the ``\textcolor{purple}{\bf box}'' \pause \vspl \item Finally, draw lines from the box to the minimum and maximum values -- these are the ``\textcolor{purple}{\bf whiskers}'' \end{itemize} } \frame{ \frametitle{Example 3.7} \noindent Suppose that, from our data, we obtain the following summary statistics: \begin{center} \begin{tabular}{l c l} Minimum & & $min=10$\\ Lower quartile & & $Q1=40$\\ Median & & $Q2=43$\\ Upper quartile & & $Q3=45$\\ Maximum & & $max=50$ \end{tabular} \end{center} \noindent Complete the associated box--and--whisker plot, and comment. } \frame{ \frametitle{Example 3.7: Solution} \begin{center} \includegraphics[width=8cm,angle=-90]{box1.ps} \end{center} } \sol \frame{ \frametitle{Example 3.7: Solution} \begin{center} \includegraphics[width=8cm,angle=-90]{box2.ps} \end{center} } \frame{ \frametitle{Example 3.7: Solution} \begin{center} \includegraphics[width=8cm,angle=-90]{box3.ps} \end{center} } \frame{ \frametitle{Example 3.7: Solution} \begin{center} \includegraphics[width=8cm,angle=-90]{box4.ps} \end{center} } %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% \normal \frame{ \frametitle{Announcements} \begin{itemize} \item \textbf{CBA 2:} \pause \vspl \begin{itemize} \item [--] Now live in practice mode \pause \vspl \item [--] Assessed mode opens on Monday \pause \vspl \item [--] Deadline: \textbf{23:59, Sunday 20th December} \pause \vspl \item [--] Some questions relate to material in chapter 4 \end{itemize} \pause \vsp \item \textbf{Prize question winners:} Apex Tshisekedi, Arjun Jain, Rahil Gandhi, Zhong Tong Tong, Mingrui Li, Sarah Koullapi, Sharon Tan! And Abigail Farnworth from last time... \end{itemize} } \frame{ \frametitle{Announcements} \begin{itemize} \item Case study 2: \pause \vspl \begin{itemize} \item [--] Due in by \textbf{4pm, Friday 15th January} -- hand-in arrangements as before...\pause \vspl \item [--]After today's lecture, \textit{should} be OK to complete \underline{\textit{most}} of the work \pause \vspl \item [--]Assessed questions in next week's workshops \pause \vspl \item [--] Graphical summaries: be selective! Use the most appropriate graph for the data. \textbf{Must be hand-drawn!} \pause \vspl \item [--] Numerical summaries: usually \textbf{one} measure of average AND \textbf{one} measure of spread for each dataset: be selective! \pause \vspl \item [--] Comments: shape of the distribution (symmetric/asymmetric); outliers; comparisons between groups (percentage scale for graphs?) \pause \vspl \item [--] Two pages - graphs separate (in an Appendix at the end, correctly referenced) \end{itemize} \end{itemize} } %\end{comment} \normal \frame{ \vspace{1cm} \huge \textbf{Chapter 4} \vspace{1cm} \Large \textbf{Probability} } \frame{ \frametitle{What we'll cover...} \begin{itemize} \item \textcolor{purple}{\bf Definitions}: Experiments, outcomes, sample space... \pause \vspl \item \textcolor{purple}{\bf Interpretations} of probability: \textit{Classical}, \textit{Frequentist}, \textit{Bayesian} \pause \vspl \item \textcolor{purple}{\bf Laws} of probability \pause \vspl \item Probability \textcolor{purple}{\bf trees} \pause \vspl \item The \textcolor{purple}{\bf binomial} distribution \pause \vspl \item The \textcolor{purple}{\bf Poisson} distribution \end{itemize} } \normal \frame{ \frametitle{4. Probability} \noindent \textcolor{purple}{\bf Probability} is the language we use to model uncertainty. \pause \vspl \begin{itemize} \item Few things in life are certain \pause \vspl \item Usually an element of uncertainty around outcomes of our choices \pause \vspl \item This can make all the difference between a good investment and a poor one! \pause \vspl \item Hence the importance of probability! \end{itemize} } \frame{ \frametitle{4.1 Definitions} \vsp \noindent We often use the letter $P$ to represent a probability. \pause For example, $P(Rain)$ would be the probability that it rains. \pause \vsp An \textcolor{purple}{\bf Experiment} is an activity where we do not know for certain what will happen, but we will \textit{observe} what happens. For example: \pause \begin{itemize} \item We will ask someone whether or not they have used our product. \pause \item We will observe the temperature at midday tomorrow. \pause \item We will toss a coin and observe whether it shows ``heads'' or ``tails''. \end{itemize} } \frame{ \frametitle{4.1 Definitions} An \textcolor{purple}{\bf outcome}, or \textcolor{purple}{\bf elementary event}, is one of the possible things that can happen. \pause \vsp For example, suppose that we are interested in the (UK) shoe size of the next customer to come into a shoe shop. Possible outcomes: \pause \vspl \begin{itemize} \item ``eight'' \item ``twelve'' \item ``nine and a half'' \end{itemize} \vspl In any experiment, \textbf{one}, \textbf{and only one}, outcome occurs. } \frame{ \frametitle{4.1 Definitions} The \textcolor{purple}{\bf sample space} is the set of all possible outcomes. For example, it could be the set of all shoe sizes. \pause \vspace{0.5cm} An \textcolor{purple}{\bf event} is a set of outcomes. \pause \vsp For example ``the shoe size of the next customer is less than 9'' is an event. \vsp It is made up of all of the outcomes where the shoe size is less than 9. } \frame{ \frametitle{4.1 Definitions} \noindent Probabilities are usually expressed in terms of fractions or decimal numbers or percentages. \vsp Therefore we could express the probability of it raining today as \[ P(\textrm{\sf Rain}) = \frac{1}{20} = 0.05 = 5\%. \] \pause \noindent All probabilities are measured on a scale ranging from \textcolor{purple}{\bf 0} $\rightarrow$ \textcolor{purple}{\bf 1}. \vspl \begin{itemize} \item An event with probability zero is an \textcolor{purple}{\bf impossible} event \pause \vspl \item An event with probability one is a \textcolor{purple}{\bf certain} event. \pause \vspl \item The collection of all possible outcomes, that is the \textcolor{purple}{\bf sample space}, has a probability of 1. \vspl For example, if an event consists of only two outcomes -- \textit{success} or \textit{failure} -- then $P(success\text{ or }failure)=1$. \end{itemize} } \frame{ \frametitle{4.1 Definitions} \noindent Two events are said to be \textcolor{purple}{\bf mutually exclusive} if both cannot occur simultaneously: \pause In the example above, the outcomes \textit{success} and \textit{failure} are mutually exclusive. \pause \vsp \noindent Two events are said to be \textcolor{purple}{\bf independent} if the occurrence of one does not affect the probability of the second occurring. \pause \vsp For example, if you toss a coin and look out of the window, it would be reasonable to suppose that the events ``get heads'' and ``it is raining'' would be independent.} \frame{ \frametitle{4.2 How do we measure Probability?} \noindent There are three main ways in which we can measure probability: \pause \vspl \begin{itemize} \item The \textcolor{purple}{\bf classical} interpretation \pause \vspl \item The \textcolor{purple}{\bf frequentist} interpretation \pause \vspl \item The \textcolor{purple}{\bf subjective} or \textcolor{purple}{\bf Bayesian} inteperetation \end{itemize} \pause \vsp All three obey the basic rules described above. \vsp Different people argue in favour of the different views of probability and some will argue that each kind has its uses depending on the circumstances. } \frame{ \frametitle{4.2.1 The classical interpretation} \noindent If all possible outcomes are ``\textcolor{purple}{\bf equally likely}'' then we can adopt the \textcolor{purple}{\bf classical} approach to measuring probability. \pause \vsp For example, if we tossed a fair coin, there are only two possible outcomes -- a head or a tail -- both of which are equally likely, and hence \pause \[ P(Head)=\frac{1}{2}\quad\text{\sf and}\quad P(Tail)=\frac{1}{2}. \] } \frame{ \frametitle{4.2.1 The classical interpretation} \noindent The underlying idea behind this view of probability is \textcolor{purple}{\bf symmetry}. \pause \vsp In this example, there is no reason to think that the outcome \textit{Head} and the outcome \textit{Tail} have different probabilities and so they should have the same probability. \pause \vsp Since there are two outcomes and one of them must occur, both outcomes must have probability 1/2. \pause \vsp \noindent Generally, \[ P(\textsf{\sf Event})=\frac{\textsf{\sf Total number of outcomes in which event occurs}}{\textrm{\sf Total number of possible outcomes}}. \] } %\frame{ %\frametitle{4.2.1 The classical interpretation} %\noindent Another commonly--used example is rolling dice. % %\vsp% % %There are six possible %outcomes (1, 2, 3, 4, 5, 6) when a die is rolled and each of them should have an equal chance of %occurring. \pause % %\vsp % %Hence %\[ %P(1) = \frac{1}{6}, P(2) = \frac{1}{6}, \ldots~. \pause %\] %\pause % %\noindent Other calculations can be made such as \pause %\[ %P(\textrm{\sf Even Number}) = \frac{3}{6} = \frac{1}{2}. \pause %\] %This follows from the formula \pause %\[ %P(\textrm{\sf Event})=\frac{\textrm{\sf Total number of outcomes in which event occurs}}{\textrm{\sf Total number of possible outcomes}}. %\] %} %\noindent Note that this formula only works when all possible events are equally %likely -- not a practical assumption for most real life situations. \frame{ \frametitle{4.2.2 The frequentist interpretation} When the outcomes of an experiment are \textbf{not} equally likely, we can \textcolor{purple}{\bf conduct experiments} to give us some idea of how likely the different outcomes are. \pause \vsp For example, suppose we were interested in measuring the probability of producing a defective item in a manufacturing process. \pause \vspl \begin{itemize} \item Monitor the process over a long period \pause \vspl \item Calculate the proportion of defective items \pause \end{itemize} \vspl \textbf{What constitutes a ``long period of time''?} \pause \vspl Generally, \[ P(\textrm{\sf Event})=\frac{\textrm{\sf Number of times an event occurs}}{\textrm{\sf Total number of times experiment done}}. \] } %\frame{ %\frametitle{4.2.2 The frequentist interpretation} % In a more simple case, if we %did not believe that a coin was fair, we could toss the coin a large %number of times and see how often we obtained a head. \pause % %\vsp % %In both cases we %perform the same experiment a large number of times and observe the %outcome. \pause % %\vsp % %This is the basis of the \textcolor{purple}{\bf frequentist} view. \pause % %\vsp % %By conducting experiments, the probability of an event can easily be %estimated using the following formula: \pause %\[ %P(\textrm{\sf Event})=\frac{\textrm{\sf Number of times an event %occurs}}{\textrm{\sf Total number of times experiment done}}. %\] \pause % %\noindent The larger the experiment, the closer this probability is to the %``true'' probability. %} %\frame{ %\frametitle{4.2.2 The frequentist interpretation} %\begin{itemize} %\item Probability is a ``property of nature'' \pause %\vspl %\item In practice, we cannot conduct \textcolor{purple}{\bf infinite} sequences of trials... \pause %\vspl %\item ... therefore, we never really know the ``true'' values of probabilities \pause %\vspl %\item We have to be able to imagine a long sequence of \textcolor{purple}{\bf identical trials}... \pause %\vspl %\item ... not appropriate for \textcolor{purple}{\bf one--off} experiments like the launch of a new product. %\end{itemize} %} \frame{ \frametitle{4.2.3 The Subjective/Bayesian interpretation} We are probably all intuitively familiar with this method of assigning probabilities: \pause \vspl \begin{itemize} \item When we board a plane, we judge the probability of it crashing to be sufficiently small that we are happy to undertake the journey. \pause \vspl \item Similarly, the odds given by bookmakers on a horse race reflect people's beliefs about which horse will win. \end{itemize} \pause \vspl This probability does not fit within the frequentist definition as the same race cannot be run a large number of times. } %\frame{ %\frametitle{4.2.3 The Subjective/Bayesian interpretation} %\noindent One potential difficulty with using subjective probabilities is that it {\em is} subjective. \pause % %\vsp % %So the probabilities which two people assign to the same event can be different. \pause % %\vsp % %This becomes important if these probabilities are to be used in decision %making. For example: \pause %\vspl %\begin{itemize} %\item Should we launch a new product? \pause %\vspl %\item Two people have very different ideas about how likely success/failure will be \pause %\vspl %\item Decision to go ahead is controversial! \pause %\vspl %\item Need a way to reconcile these differences! %\end{itemize} %} %\frame{ %\frametitle{4.2.3 The Subjective/Bayesian interpretation} % Subjective probability is still subject to the same rules as the other %forms of probability: \pause %\vspl %\begin{itemize} %\item All probabilities $\geq 0$ \pause %\vspl %\item The probability of all outcomes should sum to one %\end{itemize} %} \frame{ \frametitle{Example 4.1} \noindent A fast--food chain with 700 outlets describes the geographic location of its restaurants with the following table: {\small \begin{center} \begin{tabular}{c c |c c c c} & &\multicolumn{4}{c}{\textbf{Region}} \\ & &NE &SE &SW &NW \\ \hline &Under 10,000 &35 &42 &21 &70 \\ \textbf{Population}&10,000--100,000&70 &105 &84 &35 \\ &Over 100,000 &175 &28 &35 &0 \\ \hline \end{tabular} \end{center} } \noindent A health and safety organisation selects a restaurant at random for a hygiene inspection. \begin{itemize} \item [\textbf{(a)}] Which of the three approaches to probability would you use to find: \begin{itemize} \item [(i)] $P(\textrm{\sf NE restaurant chosen})$, %\item [(ii)] $P(\textrm{\sf Restaurant chosen from a city with a population $>$ 100,000})$, and \item [(ii)]$P(\textrm{\sf SW and city with a population $<$ 100,000})$? \end{itemize} \item [\textbf{(b)}] Now use this approach to find the probabilities given above. \end{itemize} } \sol \frame{ \frametitle{Example 4.1: Solution to part (a)} A restaurant is chosen at random -- all outcomes are equally likely -- so \textbf{classical}. \pause \vsp \textit{[Not that important, really...]} } \frame{ \frametitle{Example 4.1: Solution to part (b)(i)} {\small \begin{center} \begin{tabular}{c c |c c c c} & &\multicolumn{4}{c}{\textbf{Region}} \\ & &NE &SE &SW &NW \\ \hline &Under 10,000 &35 &42 &21 &70 \\ \textbf{Population}&10,000--100,000&70 &105 &84 &35 \\ &Over 100,000 &175 &28 &35 &0 \\ \hline \end{tabular} \end{center} } \begin{eqnarray*} P(\textsf{NE restaurant chosen}) &=& \pause \frac{35+70+175}{700} \pause \\ & & \\ &=& \frac{280}{700} \\ & & \\ &=& 0.4 \end{eqnarray*} } \frame{ \frametitle{Example 4.1: Solution to part (b)(ii)} {\small \begin{center} \begin{tabular}{c c |c c c c} & &\multicolumn{4}{c}{\textbf{Region}} \\ & &NE &SE &SW &NW \\ \hline &Under 10,000 &35 &42 &21 &70 \\ \textbf{Population}&10,000--100,000&70 &105 &84 &35 \\ &Over 100,000 &175 &28 &35 &0 \\ \hline \end{tabular} \end{center} } \begin{eqnarray*} P(\textrm{\sf SW and city with a population $<$ 100,000}) &=& \pause \frac{21+84}{700} \pause \\ & & \\ &=& \frac{105}{700} \\ \pause & & \\ &=&0.15 \end{eqnarray*} } \normal %\frame{ %\frametitle{Another example} %The spinner shown below is spun once. %\begin{center} %\includegraphics[width=7cm]{spinner.ps} %\end{center} %} %\frame{ %\frametitle{Another example} %\begin{itemize} %\item [(a)] Which of the three approaches to probability would you use to find: %\begin{itemize} %\item [(i)] $P(\textrm{\sf lands red})$, %\item [(ii)] $P(\textrm{\sf lands triangle})$, and %\item [(iii)] $P(\textrm{\sf lands quadrilateral})$? %\end{itemize} %\item [(b)] Now use this approach to find the probabilities given above. %\end{itemize} %} %\sol %\frame{ %\frametitle{Solution to part (a)} %We can use the \pause %\begin{center} %\huge %\textbf{Classical} approach %\end{center} %\normalsize %to probability since all outcomes are equally likely. %} %\frame{ %\frametitle{Solution to part (b)} %\includegraphics[width=4cm]{spinner.ps} % %\begin{eqnarray*} %P(\textsf{lands red}) &=& \frac{1}{6} \\ \pause %& & \\ %P(\textsf{lands triangle}) &=& \frac{1}{6} \\ \pause %& & \\ %P(\textsf{lands quadrilateral}) &=& \frac{4}{6} = \frac{2}{3} %\end{eqnarray*} %} \normal %\frame{ %\frametitle{Another example!} % %\noindent On the probability scale, how likely do you think it is that Newcastle United will finish in a top--six position this season? Which approach to probability would you use to estimate this? %} %\normal \frame{ \frametitle{4.3.1 Laws of Probability: Multiplication law} The probability of two \textcolor{purple}{\bf independent} events $E_1$ and $E_2$ both occurring can be written as \[ P(E_1 \text{ \textbf{and} } E_2) = P(E_1) \times P(E_2), \]\pause \noindent and this is known as the \textcolor{purple}{\bf multiplication law} of probability. \pause \vsp For example, the probability of throwing a six followed by another six on two rolls of a die is calculated as follows:\pause \vspl \begin{itemize} \item The outcomes of the two rolls of the die are \textcolor{purple}{\bf independent} \pause \vspl \item Let $E_1$ denote a six on the first roll and $E_2$ a six on the second roll. \pause \vspl \item \begin{eqnarray*} P(\text{\sf two sixes})&=&P(E_1 \text{ \sf and } E_2) \\ \pause &=& P(E_1) \times P(E_2) \\ \pause &=& \frac{1}{6} \times \frac{1}{6} =\frac{1}{36}. \end{eqnarray*} \end{itemize} } %\frame{ %\frametitle{4.3.1 Laws of Probability: Multiplication law} %\noindent This method of calculating probabilities extends to when there are %many \textcolor{purple}{\bf independent} events: \pause %\[ %P(E_1\text{ and } E_2 \text{ and }\cdots\text{ and } E_n) %=P(E_1)\times P(E_2)\times\cdots\times P(E_n). %\] } \frame{ \frametitle{4.3.2 Laws of probability: Addition Law} The multiplication law is concerned with the probability of two or more independent events occurring. \pause \vsp The \textcolor{purple}{\bf addition law} describes the probability of any of two \textbf{or} more events occurring. \pause \vsp The addition law for two events $E_1$ and $E_2$ is \[ P(E_1\text{ \textbf{or} }E_2)=P(E_1)+P(E_2)-P(E_1\text{ and }E_2). \] This describes the probability of \textit{either} event $E_1$ \textit{or} event $E_2$ happening. } %\frame{ %\frametitle{Example 4.2} %\noindent Consider the following information: \pause %\vspl %\begin{itemize} %\item 50 percent of families in a city subscribe to the morning newspaper %\vspl %\item 65 percent subscribe to the afternoon newspaper %\vspl %\item 30 percent of the families subscribe to \textbf{both} newspapers %\end{itemize} \pause %\vsp %What proportion of families subscribe to at least one newspaper? %} %\frame{ %\frametitle{Example 4.2: Solution} %\noindent We are told $P(\text{\sf Morning})=0.5$, $P(\text{\sf Afternoon})=0.65$ and $P(\text{Morning and Afternoon})=0.3$. \pause % %\vsp %We could show this information on a \textcolor{purple}{\bf Venn diagram}: \pause % % %\begin{center} %\includegraphics[width=7cm,angle=-90]{venn.ps} %\end{center} \pause % %Clearly, %\[ %P(\text{\sf at least one paper})=P(\text{\sf Morning or Afternoon}) = 0.85. %\] %} %\frame{ %\frametitle{Example 4.2: Solution} %Also, using the \textcolor{purple}{\bf addition law}: %\begin{eqnarray*} %P(\text{\sf at least one paper})&=&P(\text{\sf Morning or Afternoon})\\ %& &\\ %&=&P(\text{\sf Morning})+P(\text{\sf Afternoon})\\ % & &-P(\text{\sf Morning and Afternoon})\\ %& &\\ %&=& 0.5+0.65-0.3\\ %&=& 0.85. %\end{eqnarray*} %} \frame{ \frametitle{4.3.2 Laws of probability: Addition Law} \noindent A more basic version of the rule works where events are \textcolor{purple}{\bf mutually exclusive}. \pause \vsp If events $E_1$ and $E_2$ are mutually exclusive then \[ P(E_1\text{ or }E_2)=P(E_1)+P(E_2). \] This simplification occurs because when two events are mutually exclusive they cannot happen together and so $P(E_1\text{ \sf and }E_2)=0$. } %\frame{ %\frametitle{4.4 Tree Diagrams} %\noindent \textcolor{purple}{\bf Tree diagrams} or \textcolor{purple}{\bf probability trees} are simple clear ways of presenting %probabilistic information. \pause % %\vsp % %Let us first consider a simple example in which a die is rolled twice. \pause % %\vsp % %Suppose we are interested in the probability that we score %a six on both rolls. } %frame{ %\frametitle{4.4 Tree Diagrams} %This probability can be calculated as %\begin{eqnarray*} %P(\text{ \sf Six and Six})&=&P(\text{ \sf Six on 1st throw})\times P(\text{ \sf Six on 2nd throw})\\ %& & \\ %&=&\frac{1}{6}\times \frac{1}{6}\\ %& & \\ %&=&\frac{1}{36}. %\end{eqnarray*} %} \frame{ \frametitle{4.4 Tree Diagrams} \noindent This example can be represented as a tree diagram in which experiments are represented by circles (called \textcolor{purple}{\bf nodes}) and the outcomes of the experiments as \textcolor{purple}{\bf branches}: \pause \begin{center} \includegraphics[width=8cm,height=5cm]{probtree-coin.eps} \end{center} } \frame{ \frametitle{Example 4.3} A machine is used to produce components. Each time it produces a component there is a chance that the component will be defective. \vsp When the machine is working correctly the probability that a component is defective is 0.05. Sometimes, though, the machine requires adjustment and, when this is the case, the probability that a component is defective is 0.2. \vsp At the time in question there is a probability of 0.1 that the machine requires adjustment. } \sol \frame{ \frametitle{Solution to Example 4.3} \begin{center} \includegraphics[width=11.7cm]{tree1.eps} \end{center} } \frame{ \frametitle{Solution to Example 4.3} \begin{center} \includegraphics[width=11.7cm]{tree2.eps} \end{center} } \frame{ \frametitle{Solution to Example 4.3} \begin{center} \includegraphics[width=11.7cm]{tree3.eps} \end{center} } \normal \frame{ \frametitle{Example 4.3} Components produced by the machine are tested and either accepted or rejected. A component which is not defective is accepted with probability 0.97 or (falsely) rejected with probability 0.03. \vsp A defective component is (falsely) accepted with probability 0.15 and rejected with probability 0.85. } \sol \frame{ \frametitle{Solution to Example 4.3} \begin{center} \includegraphics[width=11.7cm]{tree4.eps} \end{center} } \frame{ \frametitle{Solution to Example 4.3} \begin{center} \includegraphics[width=11.7cm]{tree5.eps} \end{center} } \announce \frame{ \frametitle{Welcome back!} \begin{itemize} \item \textbf{My name:} Lee Fawcett \pause \vspl \item \textbf{Where you can find me:} 2.07, 2nd floor Herschel \pause \vspl \item \textbf{What I do:} Teach Stats \pause \vspl \item \textbf{This semester:} \pause \vspl \begin{itemize} \item [--] Lecture every Friday, 12--1, Herschel LT1 \pause \vspl \item [--] Workshops the following week -- new days/times! \pause \vspl \item [--] All compulsory; bring your notes, stationary and calculators! \pause \vspl \item [--] Four assignments: Two CBAs, one case study and one computing assignment using Minitab \pause \vspl \item [--] See webpage: \texttt{www.mas.ncl.ac.uk/$\sim$nlf8/teaching/acc1012} \end{itemize} \end{itemize} } \normal \frame{ \frametitle{4.5 The binomial distribution} Suppose we are interested in the \textcolor{purple}{\bf number of sixes} we get from 3 rolls of a die. \vsp Each roll of the die is an experiment or trial which gives a ``\textcolor{purple}{\bf six}'' (success, or $\textcolor{orange}{\it s}$) or ``\textcolor{purple}{\bf not a six}'' (failure, or $\textcolor{blue}{\it f}$). \vsp The probability of a success is $p=P(\textrm{\sf six})=1/6$. \vsp We have $n=3$ \textcolor{purple}{\bf independent} experiments or trials (rolls of the die). } \frame{ \frametitle{4.5 The binomial distribution} Let $X$ be the number of sixes obtained. \pause \vsp We can now obtain the full \textcolor{purple}{\bf probability distribution} of $X$; \pause a probability distribution is a list of all the possible outcomes for $X$ with along with their associated probabilities. } \frame{ \frametitle{4.5 The binomial distribution} \noindent For example, suppose we want to work out the probability of obtaining \textbf{three} sixes: (three ``\textcolor{purple}{\bf successes}'' --- i.e. $\textcolor{orange}{\it sss}$ --- or $P(X=3)$). \pause \vsp Since the rolls of the die can be considered independent, we get (using the \textcolor{purple}{\bf multiplication law}): \pause \[ P(\textcolor{orange}{\it sss})= P(\textcolor{orange}{\it s}) \times P(\textcolor{orange}{\it s}) \times P(\textcolor{orange}{\it s}) = \frac{1}{6}\times \frac{1}{6} \times \frac{1}{6}=\left(\frac{1}{6}\right)^{3} \] } \frame{ \frametitle{4.5 The binomial distribution} \noindent That one's easy! \vsp What about the probability that we get \textbf{two} sixes --- i.e. $P(X=2)$? \pause \vsp This one's a bit more tricky, because that means we need two $\textcolor{orange}{\it s}$'s and one $\textcolor{blue}{\it f}$... \pause \vsp ...but the $\textcolor{blue}{\it f}$ (``not six'') could appear on the first roll, or the second roll, or the third! \pause \vsp Thinking about it, there are actually \textbf{eight} possible outcomes for the three rolls of the die: } \normal \frame{ \frametitle{4.5 The binomial distribution} \medskip \begin{center} \psfragscanon \psfrag{s}[][][1.3][0]{\tiny $\textcolor{orange}{\it s}$} \psfrag{f}[][][1.3][0]{\tiny$\textcolor{blue}{\it f}$} \psfrag{1/6}[][][1.3][0]{\tiny$\frac{1}{6}$} \psfrag{5/6}[][][1.3][0]{\tiny$\frac{5}{6}$} \psfrag{(1/6)^3}[][][1.3][0]{\tiny$(\frac{1}{6})^3$} \psfrag{(1/6)^2*(5/6)}[][][1.3][0]{\tiny$(\frac{1}{6})^2(\frac{5}{6})$} \psfrag{(1/6)*(5/6)^2}[][][1.3][0]{\tiny$(\frac{1}{6})(\frac{5}{6})^2$} \psfrag{(5/6)^3}[][][1.3][0]{\tiny$(\frac{5}{6})^3$} \includegraphics[width=11cm,height=7cm]{probtree-3dice.eps} \end{center} } \frame{ \frametitle{4.5 The binomial distribution} \noindent So, for $P(X=2)$, we \textit{could} have: \pause \begin{eqnarray*} P(\textcolor{blue}{\it f}\textcolor{orange}{\it ss}) &=& \frac{5}{6}\times \frac{1}{6} \times \frac{1}{6}= \left(\frac{1}{6}\right)^{2}\times\frac{5}{6}, \end{eqnarray*} \pause or we \textit{could} have:\pause \begin{eqnarray*} P(\textcolor{orange}{\it s}\textcolor{blue}{\it f}\textcolor{orange}{\it s}) &=& \frac{1}{6}\times \frac{5}{6} \times \frac{1}{6}= \left(\frac{1}{6}\right)^{2}\times\frac{5}{6}, \end{eqnarray*} \textit{or even}: \pause \begin{eqnarray*} P(\textcolor{orange}{\it ss}\textcolor{blue}{\it f}) &=& \frac{1}{6}\times \frac{1}{6} \times \frac{5}{6}= \left(\frac{1}{6}\right)^{2}\times\frac{5}{6}, \end{eqnarray*} } \frame{ \frametitle{4.5 The binomial distribution} \noindent Can you see that we therefore get: \pause \begin{eqnarray*} P(X=2) &=& 3 \times \left(\frac{1}{6}\right)^{2}\times\frac{5}{6}. \end{eqnarray*} \pause \noindent Which takes the form: \begin{eqnarray*} P(X=2)&=&\text{\sf Number of ways to get two sixes} \\ & &\\ & & \times P(\text{\sf 2 sixes})\times P(\text{\sf 1 ``not six''}). \end{eqnarray*} } \frame{ \frametitle{4.5 The binomial distribution} \noindent Using the same argument as above we can calculate the other probabilities: \pause \begin{eqnarray*} P(X=0) &=& \left(\frac{5}{6}\right)^{3} = 0.579\\ \pause & & \\ P(X=1) &=& 3 \times \left(\frac{1}{6}\right)\times \left(\frac{5}{6}\right)^{2} = 0.347 \pause \\ & & \\ P(X=2) &=& 3 \times \left(\frac{1}{6}\right)^{2}\times\frac{5}{6} = 0.069 \\ \pause & & \\ P(X=3) &=& \left(\frac{1}{6}\right)^{3} = 0.005... \end{eqnarray*} } \frame{ \frametitle{4.5 The binomial distribution} ... and so the full \textcolor{purple}{\bf probability distribution} for $X$ is: \pause \begin{center} \begin{tabular}{|c|c c c c|} \hline $x$ &0 &1 &2 &3 \\ \hline $P(X=x)$ &0.579 &0.347 &0.069 &0.005 \\ \hline \end{tabular} \end{center} \pause \vsp \noindent This probability distribution shows that most of the time we would get either 0 or 1 sixes and, for example, 3 sixes would be quite rare. \pause \vsp Try your own experiment! \vsp \texttt{Minitab} } \frame{ \frametitle{4.5 The binomial distribution} \noindent Now this is a bit long--winded $\ldots$ and that was just for \textbf{three} rolls of the die! \vsp Imagine what it would be like to calculate for \textbf{100} rolls of the die! \vsp \noindent We would like a more concise way of working these probabilities out without having to list all the possible outcomes as we did above. } \frame{ \frametitle{4.5 The binomial distribution} You should see from the tree diagram that we can construct a \textcolor{purple}{\bf general formula}, taking the form: \pause \begin{eqnarray*} P(X=r)&=&\text{\sf \# ways to get $r$ successes out of $n$ trials} \\ & & \\ & & \times P(\text{\sf $r$ successes})\times P(\text{\sf $n-r$ failures}) \end{eqnarray*}} \frame{ \frametitle{4.5.1 The binomial distribution: notation} \noindent In many surveys and experiments data is collected in the form of \textcolor{purple}{\bf counts}. For example: \pause \vspl \begin{itemize} \item The number of people in the survey who bought a CD in the past month \pause \vspl \item The number of people who said they would vote Labour at the next election \pause \vspl \item The number of defective items in a sample taken from a production line \end{itemize} } \frame{ \frametitle{4.5.1 The binomial distribution: notation} All these variables have common features: \pause \vspl \begin{itemize} \item [1.]Each person/item has only two possible (exclusive) responses (Yes/No, Defective/Not defective etc)\\ --- this is referred to as a \textcolor{purple}{\bf trial} which results in a \textcolor{purple}{\bf success} or \textcolor{purple}{\bf failure}. \pause \vspl \item [2.]The survey/experiment takes the form of a random sample\\ --- the responses are \textcolor{purple}{\bf independent}. \pause \vspl \item [3.]The probability of a success in each trial is $p$ (in which case the probability of a failure is $1-p$). \pause \vspl \item [4.] We are interested in the random variable $X$, the total number of successes out of $n$ trials. \end{itemize} } \frame{ \frametitle{4.5.1 The binomial distribution: notation} If these conditions are met, then $X$ has a \textcolor{purple}{\bf binomial distribution} with index $n$ and probability $p$. \pause \vsp We write this as $X\sim Bin(n,p)$, which reads as ``$X$ has a binomial distribution with index $n$ and probability $p$''. \pause \vsp Here, $n$ and $p$ are known as the \textcolor{purple}{\bf parameters} of the binomial distribution. } \frame{ \frametitle{4.5.1 The binomial distribution: notation} Returning to the dice--rolling experiment earlier: we were interested in the number of sixes obtained from three rolls of a six--sided die. \pause \vsp Treating each roll of the die as a \textcolor{purple}{\bf trial}, with a six representing a \textcolor{purple}{\bf success} and ``not a six'' representing a \textcolor{purple}{\bf failure}, we can see that: \pause \vspl \begin{itemize} \item We have $n=3$ independent trials; \pause \vspl \item Each trial has probability of success $p=1/6$. \end{itemize} \pause \vsp Thus if $X$ represents the number of sixes on the three rolls, we have that \[ X \sim Bin(3,1/6). \] } \frame{ \frametitle{4.5.2 Probability calculations} Recall our derived formula for working out probabilities in the die example: \pause \begin{eqnarray*} P(X=r)&=&\text{\sf \# ways to get $r$ successes out of $n$ trials} \\ & & \\ & &\times P(\text{\sf $r$ successes}) \times P(\text{\sf $n-r$ failures}). \end{eqnarray*} \pause We can write this more succinctly as \begin{center} \framebox[7cm]{ \parbox[t]{6cm}{ \vspace{0.1cm} \begin{align*} P(X=r)&=\comb{n}{r}\times p^{r}\times(1-p)^{n-r}. \end{align*}}} \end{center} } \frame{ \frametitle{4.5.2 Probability calculations} \begin{center} \framebox[7cm]{ \parbox[t]{6cm}{ \vspace{0.1cm} \begin{align*} P(X=r)&=\comb{n}{r}\times p^{r}\times(1-p)^{n-r}. \end{align*}}} \end{center} The \textcolor{purple}{\bf binomial coefficient} $\comb{n}{r}$ works out how many ways we can choose $r$ objects out of $n$, and so is commonly read as ``$n$ choose $r$''. \pause \vsp There is a formula for this, but it is more easily obtained using the $nCr$ button on the calculator!} \frame{ \frametitle{4.5.2 Probability calculations} \begin{center} \framebox[7cm]{ \parbox[t]{6cm}{ \vspace{0.1cm} \begin{align*} P(X=r)&=\comb{n}{r}\times p^{r}\times(1-p)^{n-r}. \end{align*}}} \end{center} The expression for $P(X=2)$ in the die rolling example can be calculated using this formula. \pause \vsp In this example, \begin{itemize} \item $n=3$ \pause \item $r=2$ \pause \item The probability of success is $p=\frac{1}{6}$ \pause \item The probability of failure is $1-p=\frac{5}{6}$. \end{itemize} \pause } \frame{ \frametitle{4.5.2 Probability calculations} So, \begin{eqnarray*} P(X=2) &=&\comb{3}{2} \times \left(\frac{1}{6}\right)^2 \times \left(1-\frac{1}{6}\right)^{3-2}\\ \pause & & \\ &=&3 \times \left(\frac{1}{6}\right)^{2}\times\frac{5}{6} \\ \pause & & \\ &=& 0.069, \end{eqnarray*} \pause which matches up exactly with what we calculated before (as it should)! \pause \vsp You should try this formula for $r$ = 0, 1 and 3 to make sure you can recover the other values in the table on \st{page 112} page 108. } \frame{ \frametitle{4.5.3 Mean and variance} \noindent If we have the probability distribution for $X$ rather than the raw observations: \pause \vspl \begin{itemize} \item We denote the mean for $X$ not by $\bar{x}$ but by $E[X]$ (``expectation of $X$'') \pause \vspl \item We denote the variance by $Var(X)$, and the standard deviation by $SD(X)$ \end{itemize} \pause \vspl If $X$ is a random variable with Binomial $Bin(n,p)$ distribution, then its mean and variance are \[ E[X] = n\times p \qquad Var(X) = n\times p \times (1-p). \] } \frame{ \frametitle{4.5.3 Mean and variance} For example, in the die--rolling experiment, \pause \[ E[X] = 3\times \frac{1}{6} = 0.5 \] \pause and \begin{eqnarray*} Var(X) &=& 3 \times \frac{1}{6}\times \frac{5}{6} = 0.417 \qquad \textrm{\sf and so}\\ \pause SD(X)&=&\sqrt{0.417} = 0.645. \end{eqnarray*}} \frame{ \frametitle{Example 4.4} A salesperson has a 40\% chance of making a sale on a customer visit and she arranges 6 visits in a day. Suppose that the salesperson's visits result in sales \textit{independently}. \vspl \begin{itemize} \item [\textbf{(a)}] What are the probabilities of her making 0, 1, 2, 3, 4, 5 and 6 sales? \vspl \item [\textbf{(b)}] Find the salesperson's expected number of sales. What is the standard deviation? \end{itemize}} \sol \frame{ \frametitle{Solution to Example 4.4 (a)} Let $X$: number of sales. Then we have that \pause \[ X \sim Bin(6,0.4) \] \pause Thus \pause \begin{eqnarray*} P(X=0) &=& \comb{6}{0}\times 0.4^0 \times 0.6^6 = \pause 0.047 \pause\\ P(X=1) &=& \comb{6}{1}\times 0.4^1 \times 0.6^5 = \pause 0.187 \pause\\ P(X=2) &=& \comb{6}{2}\times 0.4^2 \times 0.6^4 = \pause 0.311 \pause \\ & & \\ \vdots &\vdots& \vdots \\ & & \\ P(X=6) &=& \comb{6}{6} \times 0.4^6 \times 0.6^0 = 0.004. \end{eqnarray*} } \frame{ \frametitle{Solution to Example 4.4 (b)} If $X\sim Bin(n,p)$, \vspl \begin{itemize} \item$E[X] = n\times p$ and \vspl \item $Var(X) = n\times p \times (1-p)$, \end{itemize} \vspl giving \pause \begin{eqnarray*} E[X] &=& 6 \times 0.4 = 2.4 \textrm{ \sf sales.} \end{eqnarray*} \pause Also, \begin{eqnarray*} Var(X) &=& 6 \times 0.4 \times 0.6 = 1.44 \qquad \textrm{\sf so} \\ SD(X) &=&\sqrt{1.44} = 1.2 \textrm{ \sf sales}. \end{eqnarray*} } \begin{comment} \frame{ \frametitle{Welcome back!} \begin{itemize} \item \textcolor{purple}{\bf New lecture time:} Friday \textbf{12 noon}, Bedson Building LT1\pause \vspl \item \textcolor{purple}{\bf New workshop times:} including some computer practical sessions in the CHART cluster (every third week, you'll be reminded when it's a computer week) \pause \vspl \item Your \textcolor{purple}{\bf marked case studies} will be returned soon, with feedback \pause \end{itemize} } \frame{ \frametitle{Welcome back!} \begin{itemize} \item This semester: \pause \vspl \begin{itemize} \item [--] Two more CBAs -- deadlines: \textbf{1st March}, \textbf{10th May} \pause \vspl \item [--] One more case study -- deadline: \textbf{8th May} \pause \vspl \item [--] Computer practical work -- deadline: \textbf{13th March} \end{itemize} \pause \vspl \item This week: \pause \vspl \begin{itemize} \item [--] We will probably finish Chapter 4 in this Friday's lecture \pause \vspl \item [--] Please collect your notes for Chapter 5 from the Maths General Office at some point this week -- any time from Wednesday lunchtime onwards \end{itemize} \end{itemize} } \end{comment} \begin{comment} \normal \frame{ \frametitle{Recap: The binomial distribution} If, for our discrete random variable $X$, we have: \vspl \begin{itemize} \item A \textcolor{purple}{\bf finite number of "trials"}; \\ \textcolor{gray}{\textit{[e.g. roll a fair die three times -- number of trials $n=3$]}}\pause \vspl \item and each trial has \textcolor{purple}{\bf two outcomes} -- "success" and "failure"; \\ \textcolor{gray}{\textit{[e.g. "get a six" and "do not get a six"]}}\pause \vspl \item if we know the \textcolor{purple}{\bf probability of "success"}; \\ \textcolor{gray}{\textit{[e.g. here, $p=1/6$]}} \pause \vspl \item and if the outcome of each trial is \textcolor{purple}{\bf independent} of every other trial... \\ \textcolor{gray}{\textit{[e.g. the outcome of each roll has no influence over all the other outcomes]}}\pause \end{itemize} \vspl ... then we can say that \[ X\sim \textsf{Bin}(n,p) \] } \frame{ \frametitle{Recap: The binomial distribution} In the last lecture, we "derived" the following key formula for a binomial random variable, where $X\sim \textsf{Bin}(n,p)$: \vsp \begin{itemize} \item $P(X=r) = \comb{n}{r} \times p^{r} \times (1-p)^{n-r}$ \pause \vsp \item $E[X] = n\times p $ \pause \vsp \item $\textit{\sf Var}(X) = n \times p \times (1-p)$ \end{itemize} } \frame{ \frametitle{Recap: The binomial distribution} \noindent \textbf{Example from last lecture:} Suppose \[ X: \textsf{Number of sixes on three rolls of a die}. \pause \] You should see that we can say: \[ X \sim \textsf{Bin}(3,1/6), \] \pause and that, for example, \[ P(X=2) = \comb{3}{2}\times\left(\frac{1}{6}\right)^{2} \times \left(\frac{5}{6}\right) = 0.069. \] \pause Also, \begin{eqnarray*} E[X] &=& 3 \times \frac{1}{6} = 0.5 \qquad \textsf{and} \\ \textsf{Var}(X) &=& 3 \times \frac{1}{6} \times \frac{5}{6} = 0.417. \end{eqnarray*} } \frame{ \frametitle{Your turn...} Now try \vsp \begin{itemize} \item \textbf{Question 6} -- think carefully about part (d) \vspl \item \textbf{Question 8 }(Alicia, Robyn, Josh -- sorry if you don't like McDonald's!) \vspl \item \textbf{Question 9} -- think carefully about part (c) \vspl \item \textbf{Question 11} -- hard! \end{itemize} \pause \vsp Don't forget to ask if you have any questions, and please sign the attendance list! } \end{comment} \normal \frame{ \frametitle{4.6 The Poisson distribution} The \textcolor{purple}{\bf Poisson distribution} is a very important discrete probability distribution which arises in many different contexts. \vsp Typically, Poisson random quantities are used in place of binomial random quantities in situations where $n$ is large, $p$ is small, and both $np$ and $n(1-p)$ exceed~5. \pause \vsp In general, it is used to model data which are counts of (random) events in a certain area or time interval, \underline{without a known/fixed upper limit} but with a known \textcolor{purple}{\bf rate} of occurrence. } \frame{ \frametitle{4.6 The Poisson distribution} \noindent For example, consider the number of calls made in a 1 minute interval to an Internet service provider (ISP). \pause \vsp The ISP has thousands of subscribers, but each one will call with a very small probability. \pause \vsp If the ISP knows that on average 5 calls will be made in the interval, the actual number of calls will be a Poisson random variable, with mean 5. } \frame{ \frametitle{4.6.1 Notation and probability calculations} \noindent If $X$ is a random variable with a Poisson distribution with parameter~$\lambda$ (Greek lower case ``lambda''), then the probability it takes different values is \pause \[ P(X=r)=\frac{\lambda^r\,e^{-\lambda}}{r!}, \quad r=0,1,2,\ldots~. \]\pause We write this as $X\sim Po(\lambda)$. The parameter $\lambda$ has a very simple interpretation as the \textcolor{purple}{\bf rate} at which events occur. } \frame{ \frametitle{4.6.2 Mean and variance} The Poisson distribution has mean and variance \[ E(X)=\lambda,\qquad Var(X)=\lambda. \] } \frame{ \frametitle{4.6 The Poisson distribution} \noindent Returning to the ISP example, suppose we want to know the probabilities of different numbers of calls made to the ISP. \pause \vsp Let $X$ be the number of calls made in a minute. Then $X\sim Po(5)$ and, for example, the probability of receiving \textcolor{blue}{\bf 4} calls is \pause \[ P(X=\textcolor{blue}{\bf 4})=\frac{5^{\textcolor{blue}{\bf 4}}e^{-5}}{\textcolor{blue}{\bf 4}!}= 0.1755. \] \pause Similarly, the probability of receiving \textcolor{blue}{\bf 7} calls is \pause \[ P(X=\textcolor{blue}{\bf 7}) = \frac{5^{\textcolor{blue}{\bf 7}}e^{-5}}{\textcolor{blue}{\bf 7}!} = 0.1044. \] } \frame{ \frametitle{4.6 The Poisson distribution} \noindent We can use the formula for Poisson probabilities to calculate the probability of all possible outcomes: {\small \begin{center} \begin{tabular}{c c c} $r$ & $P(X=r)$ & $P(X\leq r)$ \\ \hline 0 & \textcolor{white}{0.0067} & \textcolor{white}{0.0067}\\ 1 & \textcolor{white}{0.0337} & \textcolor{white}{0.0404}\\ 2 & \textcolor{white}{0.0843}& \textcolor{white}{0.1247}\\ 3 & \textcolor{white}{0.1403} & \textcolor{white}{0.2650}\\ 4 & 0.1755 & \textcolor{white}{0.4405}\\ 5 & \textcolor{white}{0.1755} & \textcolor{white}{0.6160}\\ 6 & \textcolor{white}{0.1462} & \textcolor{white}{0.7622}\\ 7 & 0.1044 & \textcolor{white}{0.8666}\\ 8 & \textcolor{white}{0.0653} & \textcolor{white}{0.9319}\\ $\vdots$ & \textcolor{white}{$\vdots$} & \textcolor{white}{$\vdots$} \\ \hline sum & \textcolor{white}{1.000000} & \\ \end{tabular} \end{center}} {\color{white} \noindent Therefore the probability of receiving between 2 and 8 calls is \[ P(2\leq X\leq 8)=P(X\leq 8)-P(X\leq 1)=0.9319-0.0404=0.8915 \]} } \frame{ \frametitle{4.6 The Poisson distribution} \noindent We can use the formula for Poisson probabilities to calculate the probability of all possible outcomes: {\small \begin{center} \begin{tabular}{c c c} $r$ & $P(X=r)$ & $P(X\leq r)$ \\ \hline 0 & 0.0067 & \textcolor{white}{0.0067}\\ 1 & 0.0337 & \textcolor{white}{0.0404}\\ 2 & 0.0843 & \textcolor{white}{0.1247}\\ 3 & 0.1403 & \textcolor{white}{0.2650}\\ 4 & 0.1755 & \textcolor{white}{0.4405}\\ 5 & 0.1755 & \textcolor{white}{0.6160}\\ 6 & 0.1462 & \textcolor{white}{0.7622}\\ 7 & 0.1044 & \textcolor{white}{0.8666}\\ 8 & 0.0653 & \textcolor{white}{0.9319}\\ $\vdots$ & $\vdots$ & \textcolor{white}{$\vdots$} \\ \hline sum & 1.000000 & \\ \end{tabular} \end{center}} {\color{white} \noindent Therefore the probability of receiving between 2 and 8 calls is \[ P(2\leq X\leq 8)=P(X\leq 8)-P(X\leq 1)=0.9319-0.0404=0.8915 \]} } \frame{ \frametitle{4.6 The Poisson distribution} \noindent We can use the formula for Poisson probabilities to calculate the probability of all possible outcomes: {\small \begin{center} \begin{tabular}{c c c} $r$ & $P(X=r)$ & $P(X\leq r)$ \\ \hline 0 & 0.0067 & 0.0067\\ 1 & 0.0337 & 0.0404\\ 2 & 0.0843 & 0.1247\\ 3 & 0.1403 & 0.2650\\ 4 & 0.1755 & 0.4405\\ 5 & 0.1755 & 0.6160\\ 6 & 0.1462 & 0.7622\\ 7 & 0.1044 & 0.8666\\ 8 & 0.0653 & 0.9319\\ $\vdots$ & $\vdots$ & $\vdots$ \\ \hline sum & 1.000000 & \\ \end{tabular} \end{center}} {\color{white} \noindent Therefore the probability of receiving between 2 and 8 calls is \[ P(2\leq X\leq 8)=P(X\leq 8)-P(X\leq 1)=0.9319-0.0404=0.8915 \]} } \frame{ \frametitle{4.6 The Poisson distribution} \noindent We can use the formula for Poisson probabilities to calculate the probability of all possible outcomes: {\small \begin{center} \begin{tabular}{c c c} $r$ & $P(X=r)$ & $P(X\leq r)$ \\ \hline 0 & 0.0067 & 0.0067\\ 1 & 0.0337 & 0.0404\\ 2 & \textcolor{orange}{\bf 0.0843} & 0.1247\\ 3 & \textcolor{orange}{\bf 0.1403} & 0.2650\\ 4 & \textcolor{orange}{\bf 0.1755} & 0.4405\\ 5 & \textcolor{orange}{\bf 0.1755} & 0.6160\\ 6 & \textcolor{orange}{\bf 0.1462} & 0.7622\\ 7 & \textcolor{orange}{\bf 0.1044} & 0.8666\\ 8 & \textcolor{orange}{\bf 0.0653} & 0.9319\\ $\vdots$ & $\vdots$ & $\vdots$ \\ \hline sum & 1.000000 & \\ \end{tabular} \end{center}} \noindent Therefore the probability of receiving between 2 and 8 calls is {\color{white} \[ P(2\leq X\leq 8)=P(X\leq 8)-P(X\leq 1)=0.9319-0.0404=0.8915 \]} } \frame{ \frametitle{4.6 The Poisson distribution} \noindent We can use the formula for Poisson probabilities to calculate the probability of all possible outcomes: {\small \begin{center} \begin{tabular}{c c c} $r$ & $P(X=r)$ & $P(X\leq r)$ \\ \hline 0 & 0.0067 & 0.0067\\ 1 & 0.0337 & \textbf{0.0404}\\ 2 & \textcolor{orange}{\bf 0.0843} & 0.1247\\ 3 & \textcolor{orange}{\bf 0.1403} & 0.2650\\ 4 & \textcolor{orange}{\bf 0.1755} & 0.4405\\ 5 & \textcolor{orange}{\bf 0.1755} & 0.6160\\ 6 & \textcolor{orange}{\bf 0.1462} & 0.7622\\ 7 & \textcolor{orange}{\bf 0.1044} & 0.8666\\ 8 & \textcolor{orange}{\bf 0.0653} & \textbf{0.9319}\\ $\vdots$ & $\vdots$ & $\vdots$ \\ \hline sum & 1.000000 & \\ \end{tabular} \end{center}} \noindent Therefore the probability of receiving between 2 and 8 calls is \pause \[ P(2\leq X\leq 8)=P(X\leq 8)-P(X\leq 1)=0.9319-0.0404=0.8915 \] } \frame{ \frametitle{4.6 The Poisson distribution} Probability calculations such as this enable ISPs to calibrate the likely demand for their service and hence the resources they need to provide the service. \vsp Using such a model we can also account for ``extreme'' situations. \vsp For example, suppose that, for this ISP, we observed the following number of calls per minute over a five minute period: \textcolor{brown}{\bf 6}, \textcolor{brown}{\bf 3}, \textcolor{brown}{\bf 5}, \textcolor{brown}{\bf 4}, \textcolor{brown}{\bf 6}. \pause \vsp Using simple frequentist reasoning, we would have \begin{eqnarray*} P(\textrm{\sf 7 calls made}) &=& \frac{0}{5} = 0, \end{eqnarray*} i.e. we will \textit{never} observe seven calls in any one minute period! } \frame{ \frametitle{4.6 The Poisson distribution} However, using the Poisson model, we have \begin{eqnarray*} P(X=7) &=& 0.1044, \end{eqnarray*} which is probably more realistic. \pause \vsp You can see from the previous table of probabilities that, although when $r$ is large the associated probabilities are small, at least they are accounted for and are, more realistically, non--zero. } \normal \frame{ \frametitle{Review: Binomial and Poisson distributions} \textbf{Binomial:} $X \sim \textsf{Bin}(n,p)$ \pause \vsp \begin{itemize} \item $n$ is the number of ``trials'' \pause \vspl \item $p$ is the ``success'' probability \pause \vspl \item Each trial must have two possible outcomes (``success'' and ``failure'') \pause \vspl \item Each trial must be independent of every other trial \pause \vspl \item Since $n$ is known, there is a \textcolor{purple}{\bf known upper limit to the random variable} \end{itemize} } \frame{ \frametitle{Review: Binomial and Poisson distributions} \textbf{Poisson:} $X\sim \textsf{Po}(\lambda)$ \pause \vsp \begin{itemize} \item $\lambda$ is the ``rate of occurrence'' \pause \vspl \item Often used when we are counting something over a fixed time interval \pause \pause \item The value of $\lambda$ is constant across time \pause \vspl \item We usually \textcolor{purple}{\bf don't know the upper limit of the random variable} \end{itemize} } \frame{ \frametitle{Review: Binomial and Poisson distributions} \textbf{Key formulae} \pause \begin{center} \begin{tabular}{|c|c|c|c|} \hline &$P(X=r)$ &Expectation &Variance \\ \hline & & & \\ Binomial &$\comb{n}{r}p^{r}(1-p)^{n-r}$ &$np$ &$np(1-p)$ \\ & & & \\ \hline & & & \\ Poisson &$\displaystyle \frac{e^{-\lambda}\lambda^{r}}{r!}$ &$\lambda$ &$\lambda$ \\ & & & \\ \hline \end{tabular} \end{center} } \normal \frame{ \frametitle{Example 4.5} \textit{Pronto Pizzeria} have recently opened a new restaurant in Newcastle. They have noticed that, during any one hour period on a Saturday night, they usually see about eight parties of customers arrive. \vspl \begin{itemize} \item [\textbf{(a)}] What probability distribution might be appropriate for $X$, the number of parties that arrive between 7pm and 8pm this Saturday evening? \vspl \item [\textbf{(b)}] What is the probability that, between 7pm and 8pm this Saturday, at least three parties of customers arrive at the restaurant? \vspl \item [\textbf{(c)}] The management are considering a Half Price Happy Hour Menu between 6pm and 7pm every Saturday night. Why might the model you proposed in part (a) no longer be valid? \end{itemize} } \sol \frame{ \frametitle{Solution to Example 4.5(a)} \textbf{Poisson} distribution -- we have a rate (8 per hour), and we do not know the upper limit for the number of parties. \pause \vsp If $X$: Number of parties arriving in a one hour period. then \[ X \sim Po(8). \] } \frame{ \frametitle{Solution to Example 4.5(b)} \begin{eqnarray*} P(X\geq 3) &=& \pause 1-P(X\leq 2) \pause \\ &=& 1-\left\{P(X=0)+P(X=1)+P(X=2) \right\} \\ \pause & & \\ &=& 1 -\left\{\frac{e^{-8}8^{0}}{0!}+\frac{e^{-8}8^{1}}{1!}+\frac{e^{-8}8^{2}}{2!}\right\} \pause \\ & & \\ &=& 1- \left\{0.0003+0.0027+0.0107 \right\} \pause \\ & & \\ &=&0.9863. \end{eqnarray*} } \frame{ \frametitle{Solution to Example 4.5(c)} We might expect more people to come to the restaurant during the Happy Hour -- therefore we cannot assume the same rate of occurrence. \vsp Specifically, we might expect $\lambda>8$. } \normal \frame{ \frametitle{Example 4.6} Consider the following two scenarios: \pause \vsp \textbf{Scenario 1} \vspl \noindent \textit{RyanJet} currently leads all European airlines in on--time arrivals, with 82.5\% of it's flights arriving on time or earlier. Of the next 6 scheduled \textit{RyanJet} flights, $X$ is the number that will land on time. \pause \vsp \textbf{Scenario 2} \vspl \noindent Customer bookings on the \textit{RyanJet} website occur at a rate of 20 per hour. Due to a technical problem, the \textit{RyanJet} website will be down for maintenance between 17:30 and 18:00 this evening. We are interested in $Y$, the number of bookings we can expect \textit{RyanJet} to lose, because of this website down--time. } \frame{ \frametitle{Example 4.6} \begin{itemize} \item [\textbf{1.}] Would you use the binomial distribution or the Poisson distribution to model the scenarios outlined above? \vsp In each case explain why, and state fully the most appropriate distribution, including the values of the associated parameters. \end{itemize} } \sol \frame{ \frametitle{Example 4.6: Solution (1/2)} \textbf{Scenario 1} -- \pause \textcolor{purple}{\bf Binomial distribution} \pause \vspl \begin{itemize} \item Finite number of trials ($n=6$) \pause \vspl \item Each trial has two outcomes (success = land on time, failure = dosen't land on time) \pause \vspl \item We know that $p=0.825$ \pause \vspl \item Assuming independence, we have $X\sim \textsf{Bin}(6,0.825)$. \end{itemize} } \frame{ \frametitle{Example 4.6: Solution (2/2)} \textbf{Scenario 2} -- \pause \textcolor{purple}{\bf Poisson distribution} \pause \vspl \begin{itemize} \item Don't know the upper bound for $Y$ \pause \vspl \item We have a rate of occurrence: $\lambda=20$ per hour \pause \vspl \item Assuming a constant rate, we have $Y\sim \textsf{Po}(10)$ \end{itemize} } \normal \frame{ \frametitle{Example 4.6} \begin{itemize} \item [\textbf{2.}] Find the expectation, and standard deviation, of the number of on--time flights of the next 6 scheduled \textit{RyanJet} flights. \vsp Also find the expected number of lost \textit{RyanJet} bookings between 17:30 and 18:00, with the associated standard deviation. \end{itemize} } \sol \frame{ \frametitle{Example 4.6: Solution (1/1)} Since $X\sim \textsf{Bin}(6,0.825)$, then \pause \vspl \begin{itemize} \item $E[X] = n \times p = 6\times 0.825 = 4.95 \textsf{ flights}$. \pause \vspl \item $Var(X) = n \times p \times (1-p) = 6\times 0.825 \times 0.175 = 0.86625$. Thus, $SD(X) = \sqrt{0.86625} = 0.931 \textsf{ flights}$. \pause \end{itemize} \vsp Since $Y\sim \textsf{Po}(10)$, then \pause \vspl \begin{itemize} \item $E[Y] = \lambda = 10 \textsf{ bookings}.$ \pause \vspl \item $Var(Y) = \lambda = 10$. Thus $SD(Y) = \sqrt{10} = 3.162 \textsf{ bookings}.$ \end{itemize} } \normal \frame{ \frametitle{Example 4.6} \begin{itemize} \item [\textbf{3.}] Find the probability that, of \textit{RyanJet}'s next six flights, exactly 4 arrive on time. \end{itemize} } \sol \frame{ \frametitle{Example 4.6: Solution (1/1)} We have $X\sim\textsf{Bin}(6,0.825)$. So \pause \begin{eqnarray*} P(X=4) &=&\comb{6}{4}\times 0.825^{4} \times 0.175^{2} \\ \pause & & \\ &=& 15 \times 0.4632504 \times 0.030625 \pause \\ && \\ &=& 0.2128. \end{eqnarray*} } \normal \frame{ \frametitle{Example 4.6} \begin{itemize} \item [\textbf{4.}] Find the probability that \textit{RyanJet} will lose \vspl \begin{itemize} \item [\textbf{(a)}] exactly 8 bookings; \vspl \item [\textbf{(b)}] more than 2 bookings, \end{itemize} \vspl during their website down--time. \end{itemize} } \sol \frame{ \frametitle{Example 4.6: Solution (1/2)} We know that $Y \sim \textsf{Po}(10)$. So\pause \begin{eqnarray*} P(Y=8) &=& \frac{e^{-10} \times 10^{8}}{8!} \pause \\ & & \\ &=&0.1126 \end{eqnarray*} } \frame{ \frametitle{Example 4.6: Solution (2/2)} Similarly, \pause \begin{eqnarray*} P(Y>2) &=& P(Y=3) + P(Y=4) + \cdots \pause \\ &=&1-\left\{P(Y=0)+P(Y=1)+P(Y=2)\right\} \\ \pause && \\ &=&1-\left\{\frac{e^{-10}10^{0}}{0!} + \frac{e^{-10}10^{1}}{1!} + \frac{e^{-10}10^{2}}{2!} \right\}\pause \\ & & \\ &=& 1-\left\{0.000045 + 0.00045 + 0.0023\right\} \pause \\ && \\ &=& 0.9972. \end{eqnarray*} } \normal \frame{ \frametitle{Example 4.6} \begin{itemize} \item [\textbf{5.}] It is decided to move the website down--time to 03:00--03:30. \vsp Why might the model you proposed in question 1 no longer be appropriate? \end{itemize} } \sol \frame{ \frametitle{Example 4.6: Solution (1/1)} The rate will probably change -- surely, the number of bookings made in the middle of the night will not be the same as at other times of the day. } \normal \frame{ \frametitle{Example 4.7} When a customer places an order with \textit{Staples Online Office Supplies}, a computerised accounting information system automatically checks to see if the customer has exceeded his or her credit limit. \vsp Let $X$ be the number of customers who have exceeded their credit limit. \vsp In a random sample of $n$ orders, we can expect 1 customer to have exceeded their credit limit, with a variance of 0.95. } \frame{ \frametitle{Example 4.7} \begin{itemize} \item [\textbf{1.}] Would the binomial or the Poisson distributions be most appropriate to model $X$? \end{itemize} } \sol \frame{ \frametitle{Example 4.7: Solution (1/1)} The \textcolor{purple}{\bf Binomial distribution} -- as we have a finite number of orders ($n$), and each trial has two outcomes: \pause \vspl \begin{itemize} \item success (exceeded credit limit) \pause \vspl \item failure (does not exceed credit limit) \end{itemize} \pause \vspl We can also assume independence between trials, as we have a random sample of orders. } \normal \frame{ \frametitle{Example 4.7} \begin{itemize} \item [\textbf{2.}] Find the parameters of the model you proposed in question 1, and hence find the probability that at least 1 customer, of the next $n$ orders placed, exceeds their credit limit. \end{itemize} } \sol \frame{ \frametitle{Example 4.7: Solution (1/3)} We are told that $E[X]=1$; therefore, \pause \begin{eqnarray*} n \times p &=& 1, \qquad \textsf{and so} \\ \pause n &=&\frac{1}{p}. \end{eqnarray*} \pause We are also told that $Var(X) = 0.95$; therefore, \pause \begin{eqnarray*} n \times p \times (1-p) &=&0.95 \pause \qquad \textsf{and so} \\ \pause \frac{1}{p} \times p \times (1-p) &=&0.95 \pause \\ 1-p &=&0.95 \pause \\ p &=&0.05. \pause \end{eqnarray*} } \frame{ \frametitle{Example 4.7: Solution (2/3)} Substituting $p=0.05$ into the expression for the expectation gives \pause \[ n = \frac{1}{p} = \frac{1}{0.05} = 20. \] \pause So we have $X\sim\textsf{Bin}(20,0.05)$. } \frame{ \frametitle{Example 4.7: Solution (3/3)} Now we can find \pause \begin{eqnarray*} P(X\geq 1) &=& 1- P(X=0) \\ \pause &=& 1-\comb{20}{0} \times 0.05^{0} \times 0.95^{20} \pause \\ &=& 1-0.3585 \pause \\ &=&0.6415. \end{eqnarray*} } \announce \frame{ \frametitle{Announcements} \begin{itemize} \item CBA3 -- the first CBA of semester 2 -- goes live in practice mode this coming \textcolor{purple}{\bf Monday, 20th February} \pause \vspl \item Next week is a \textcolor{purple}{\bf computer practical week} -- see course webpage/your own personal timetable for details \pause \vspl \item Marked case study 2 reports available to collect from the ACC1012/1053 box upstairs \pause \vspl \item Course webpage now up-to-date, including one page summary of learning outcomes for Chapter 4 \pause \vspl \item ReCap recordings -- there have been problems locating these, but the ReCap team are sorting this out! \pause \vspl \item Today we will start Chapter 5, but first... \end{itemize} } \normal \frame{ \frametitle{Recap: Probability models} \begin{itemize} \item Is your variable \textcolor{purple}{\bf discrete} or \textcolor{purple}{\bf continuous}? \pause \vspl \item If \textcolor{purple}{\bf discrete} (e.g. \# customers arriving in a ten minute interval; \# of successful sales made on the next 6 visits) : \pause \vspl \begin{itemize} \item [--] Do you have a fixed number of independent trials, each with two outcomes (``\textcolor{orange}{\bf success}'' and ``\textcolor{orange}{\bf failure}")? \pause \vspl \item [--] Do you know the probability of success? \pause \vspl \item [--] Do you have a fixed rate of occurrence per unit time/per page/per mile cycled etc...? \end{itemize} \pause \vspl \item If \textcolor{purple}{\bf continuous}... \end{itemize} } \frame{ \vspace{1cm} \huge \textbf{Chapter 5} \vspace{1cm} \Large \textbf{Continuous probability models} } \frame{ \frametitle{What we'll cover...} \begin{itemize} \item \textcolor{purple}{\bf Probability density functions} \pause \vspl \item The \textcolor{purple}{\bf Normal distribution} \pause \vspl \item The \textcolor{purple}{\bf Uniform distribution} \pause \vspl \item The \textcolor{purple}{\bf exponential distribution} \pause \vspl \item \textcolor{purple}{\bf Poisson processes} \end{itemize} } \normal \frame{ \frametitle{5. Continuous probability models} \noindent We have seen how \textcolor{purple}{\textbf{discrete}} random variables can be modelled by discrete probability distributions such as the \textcolor{purple}{\textbf{binomial}} and \textcolor{purple}{\textbf{Poisson}} distributions. \pause \vspace{0.3cm} \noindent We now consider how to model \textcolor{purple}{\textbf{continuous}} random variables. } \frame{ \frametitle{5.1 Introduction} \noindent A variable is \textcolor{purple}{\textbf{discrete}} if it takes a \textcolor{purple}{\textbf{countable}} number of values. \vspace{0.3cm} \noindent \textbf{For example}, \begin{itemize} \item [--] the number of \textcolor{blue}{\textbf{blue}} cars that I count in a 5 minute period \item [--] the number of \textbf{heads} observed when I flip a coin ten times \item [--] Shoe sizes: $1, \ldots, 12, 13, 1, 2, \ldots$ \item [--] $r=0,0.1,0.2,\ldots,0.9,1.0$ \end{itemize} \pause \vspace{0.3cm} \noindent In contrast, the values which a \textcolor{purple}{\textbf{continuous variable}} can take form a \textcolor{purple}{\textbf{continuous scale}}, with no ``jumps''. \vspace{0.3cm} \noindent \textbf{For example}, \begin{itemize} \item [--] Height \item [--] Weight \item [--] Temperature \end{itemize} } \frame{ \frametitle{5.1 Introduction} \noindent Think about \textcolor{purple}{\textbf{height}}. \pause \vsp \begin{itemize} \item In practice, we might only record height to the \textcolor{purple}{\textbf{nearest cm}} \vspl \item If we could measure height \textit{exactly} we'd find that everyone had a \textcolor{purple}{\textbf{different height}} \pause \vspl \item This is the essential difference between discrete and continuous variables \pause \vspl \item If there are $n$ people on the planet, the probability that someone's height is $x$ would be $\frac{1}{n}$ \pause \vspl \item As $n$ gets bigger and bigger, this probability tends to zero!! \end{itemize} } \frame{ \frametitle{5.1 Introduction} \noindent Consider taking a sample of values from: \pause \[ X: \textsf{Weight of 10oz burger}. \] \begin{center} \includegraphics[width=6cm]{burger.ps} \end{center} } \frame{ \frametitle{5.1 Introduction} This is what we might observe as our sample got bigger: \begin{center} %\resizebox{\textwidth}{!}{\includegraphics{hist1.pdf}} %\resizebox{\textwidth}{!}{\includegraphics{hist2.pdf}} %\resizebox{\textwidth}{!}{\includegraphics{hist3.pdf}} %\resizebox{\textwidth}{!}{\includegraphics{histinf.pdf}} \psfrag{f}[][]{\tiny \sf Frequency} \psfrag{W}[][]{\tiny \sf oz } \psfrag{A}[][]{\tiny $n=100$} \psfrag{B}[][]{\tiny $n=1000$} \psfrag{C}[][]{\tiny $n=10,000$} \psfrag{D}[][]{\tiny $n=100,000$} \psfrag{E}[][]{\tiny $n=1000,000$} \psfrag{F}[][]{\tiny $n=10,000,000$} \psfrag{9.0}[][]{\tiny \sf 9.0} \psfrag{9.5}[][]{\tiny \sf 9.5} \psfrag{10.0}[][]{\tiny \sf 10.0} \psfrag{10.5}[][]{\tiny \sf 10.5} \psfrag{11.0}[][]{\tiny \sf 11.0} \psfrag{0}[][]{\tiny \sf 0} \psfrag{5}[][]{\tiny \sf 5} \psfrag{10}[][]{\tiny \sf 10} \psfrag{15}[][]{\tiny \sf 15} \psfrag{20}[][]{\tiny \sf 20} \psfrag{50}[][]{\tiny \sf } \psfrag{100}[][]{\tiny \sf 100} \psfrag{150}[][]{\tiny \sf } \psfrag{200}[][]{\tiny \sf 200} \psfrag{250}[][]{\tiny \sf } \psfrag{500}[][]{\tiny \sf } \psfrag{1000}[][]{\tiny \sf 1000} \psfrag{1500}[][]{\tiny \sf } \psfrag{2000}[][]{\tiny \sf 2000} \psfrag{2500}[][]{\tiny \sf } \psfrag{4000}[][]{\tiny \sf} \psfrag{6000}[][]{\tiny \sf } \psfrag{8000}[][]{\tiny \sf 8000} \psfrag{10000}[][]{\tiny \sf } \psfrag{12000}[][]{\tiny \sf } \psfrag{20000}[][]{\tiny \sf } \psfrag{40000}[][]{\tiny \sf } \psfrag{60000}[][]{\tiny \sf 60000} \psfrag{80000}[][]{\tiny \sf } \psfrag{100000}[][]{\tiny \sf } \psfrag{120000}[][]{\tiny \sf 120000} \psfrag{200000}[][]{\tiny \sf 200000} \psfrag{400000}[][]{\tiny \sf 400000} \psfrag{600000}[][]{\tiny \sf 600000} \psfrag{800000}[][]{\tiny \sf 800000} \psfrag{1000000}[][]{\tiny \sf 1000000} \psfrag{1200000}[][]{\tiny \sf 1200000} \psfrag{0e+00}[][]{\tiny \sf 0e+00} \psfrag{1e+05}[][]{\tiny \sf } \psfrag{2e+05}[][]{\tiny \sf } \psfrag{3e+05}[][]{\tiny \sf } \psfrag{4e+05}[][]{\tiny \sf 4e+05} \psfrag{5e+05}[][]{\tiny \sf } \psfrag{6e+05}[][]{\tiny \sf } \includegraphics[width=10cm, height=7cm]{cdfbigger.ps} \end{center} } \frame{ \frametitle{5.1 Introduction} \begin{itemize} \item As the sample size gets bigger, the \textcolor{purple}{\textbf{interval widths get smaller}} \item the jagged profile of the histogram \textcolor{purple}{\textbf{smooths out}} to become a curve \item When the sample size is infinitely large, this curve is known as the \textcolor{purple}{\textbf{probability density function}} (pdf) \end{itemize} } \frame{ \frametitle{5.1 Introduction} \frametitle{Features of the probability density function} \noindent \textbf{The key features of pdfs are:} \vspl \begin{enumerate} \item pdfs never take negative values \vspl \item the area under a pdf is one: $P(-\infty1.5)$. Now our tables give ``less than'' probabilities, and here we want a ``greater than'' probability. \pause \begin{center} \includegraphics[width=7.5cm,height=5cm]{blank1.ps} \end{center} \pause \noindent So we find $P(Z<1.5) = 0.9332$ and subtract this from 1 to give \textbf{0.0668}. \end{enumerate} } \frame{ \frametitle{5.2.3 The \textcolor{yellow}{\textit{standard}} Normal distribution} \begin{enumerate} \item [\textbf{5.}] What about the probability that $Z$ lies between $-1.2$ and $1.5$? It often helps to think about this graphically. \pause \begin{center} \includegraphics[width=7.5cm,height=5cm]{blank2i.ps} \end{center} \noindent \textcolor{white}{Doing so, gives:} \begin{align*} \textcolor{white}{P(-1.2110) &=& 1-P(X<110) \\ \pause & & \\ &=& 1-0.7486 \\ \pause & & \\ &=& 0.2514. \end{eqnarray*} } \frame{ \frametitle{Example: Solution to (iii)} \begin{eqnarray*} P(X>125) &=& 1-P(X<125) \\ \pause & & \\ &=& 1-P\left(Z<\frac{125-100}{15}\right) \\ \pause & & \\ &=& 1-P(Z<1.67) \\ \pause & & \\ &=& 1-0.9525 \\ \pause & & \\ &=& 0.0475. \end{eqnarray*} } \frame{ \frametitle{Example: Solution to (iv)} \begin{eqnarray*} P(958.5) &=& P\left(Z>\frac{8.5-7}{2}\right) \\ & & \\ \pause &=&P(Z>0.75) \\\pause & & \\ &=&1-P(Z<0.75) \\ \pause & & \\ &=&1-0.7734 \pause = 0.2266. \end{eqnarray*} } \frame{ \frametitle{Example 5.1(b): Solution} \begin{eqnarray*} P(X<2.5) &=&P\left(Z<\frac{2.5-7}{2}\right) \\\pause & & \\ &=&P(Z<-2.25) \\ \pause & & \\ &=&0.0122. \end{eqnarray*} } \frame{ \frametitle{Example 5.1(c): Solution} \begin{eqnarray*} P(2.5b. \end{cases} \] } \frame{ \frametitle{5.3.1 The Uniform distribution} \noindent Therefore, for example, the \textcolor{purple}{\textbf{probability that the inspectors visit the hotel in the morning}} (within 180 minutes after 9am) is \[ P(X\leq 180)=\pause\dfrac{180-0}{540-0}=\pause\frac{1}{3}. \] } \frame{ \frametitle{5.3.1 The Uniform distribution} \noindent The \textcolor{purple}{\textbf{probability of a visit during the lunch hour}} (12.30 to 13.30) is \begin{eqnarray*} P(210\leq X\leq 270)&=&P(X\leq 270)-P(X<210)\\ \pause &=&\frac{270-0}{540-0}~-\frac{210-0}{540-0}\\ \pause & & \\ &=&\frac{270-210}{540}\\ & & \\ \pause &=&\frac{60}{540}\\ & & \\ &=&\frac{1}{9}. \end{eqnarray*} } \frame{ \frametitle{5.3.1 Uniform distribution: Mean and Variance} \noindent \textbf{Recall that:} \vsp \begin{itemize} \item If $X \sim \textsf{bin}(n, p)$, then \vspl \begin{itemize} \item [--] $E(X) = n \times p$ and \vspl \item [--] $\textsf{Var}(X) = n \times p \times (1-p)$ \end{itemize} \vspace{0.3cm} \pause \item If $X \sim \textsf{Po}(\lambda)$, then \vspl \begin{itemize} \item [--] $E(X) = \lambda$ and \vspl \item [--] $\textsf{Var}(X) = \lambda$ \end{itemize} \end{itemize} } \frame{ \frametitle{5.3.1 Uniform distribution: Mean and Variance} \noindent We have equivalent formulae for $X \sim U(a,b)$: \begin{eqnarray*} E(X) &=& \frac{a+b}{2} \\ & & \\ \textsf{Var}(X) &=& \frac{(b-a)^2}{12}. \end{eqnarray*} } \frame{ \frametitle{5.3.1 Uniform distribution: Mean and Variance} \noindent In the above example, we have \begin{align*} E(X)&=\frac{a+b}{2}=\frac{0+540}{2}=270, \end{align*} \noindent so that the mean arrival of the inspectors is 9am + 270 minutes = 13.30. \pause \vspace{0.3cm} Also \begin{align*} \textsf{Var}(X)&=\frac{{(540-0)}^2}{12}=24300, \end{align*} and therefore \pause $SD(X)=\sqrt{Var(X)}=\sqrt{24300}=155.9$ minutes. } \frame{ \frametitle{5.3.2 The Exponential distribution} \noindent The \textcolor{purple}{\textbf{exponential distribution}} is another common distribution that is used to describe continuous random variables. \vspace{0.3cm} \pause \noindent It is often used to model lifetimes of products and times between ``random'' events, for example: \vsp \begin{itemize} \item Lifetime of light bulbs \vspl \item Arrival of customers in a queueing system \vspl \item Arrival of orders \end{itemize} \pause \vspl \noindent The distribution has one parameter, $\lambda$. If our random variable $X$ follows an \textcolor{purple}{\textbf{exponential distribution}}, then we say \begin{eqnarray*} X &\sim& \textsf{exp}(\lambda). \end{eqnarray*} } \frame{ \frametitle{5.3.2 The Exponential distribution} \noindent Its \textcolor{purple}{\textbf{probability density function}} is \[ f(x)= \begin{cases} \lambda e^{- \lambda x}& \text{for }x\geq 0,\\ 0&\text{otherwise} \end{cases} \] \pause and probabilities can be calculated using \[ P(X\leq x)= \begin{cases} 0&\text{for }x<0\\ 1-e^{- \lambda x}& \text{for }x>0. \end{cases} \] } \frame{ \frametitle{5.3.2 The Exponential distribution} \noindent The main \textcolor{purple}{\textbf{features}} of this distribution are: \vsp \begin{enumerate} \item an exponentially distributed random variable can only take \textbf{positive} values \vspl \item larger values are \textbf{increasingly unlikely} -- ``\textcolor{purple}{\textbf{exponential decay}}'' \vspl \item the value of $\lambda$ fixes the \textbf{rate of decay} -- \textit{larger values correspond to more rapid decay}. \end{enumerate} } \sol \frame{ \frametitle{5.3.2 The Exponential distribution} \begin{center} \psfrag{Density}[][]{\tiny \sf Density} \psfrag{lambda=1}[][]{\tiny \sf $\lambda=1$} \psfrag{lambda=5}[][]{\tiny \sf $\lambda=5$} \psfrag{lambda=10}[][]{\tiny \sf $\lambda=10$} \psfrag{lambda=20}[][]{\tiny \sf $\lambda=20$} \includegraphics[width=11cm,height=6.5cm]{exp.ps} \end{center} } \normal \frame{ \frametitle{5.3.2 The Exponential distribution} \noindent Consider an example in which the time (in minutes) between successive users of a self--service supermarket checkout can be modelled by an exponential distribution with $\lambda = 0.3$. \vspace{0.3cm} \pause \noindent The probability of the gap between phone users being less than 5 minutes is \begin{eqnarray*} P(X<5)&=&1-e^{-0.3\times 5}\\ \pause &=&1-0.223 \pause =0.777. \end{eqnarray*} } \frame{ \frametitle{5.3.2 The Exponential distribution} \noindent Also the probability that the gap is more than 10 minutes is \begin{eqnarray*} P(X>10)&=&1-P(X< 10)\\ \pause &=&1-\left(1-e^{-0.3\times 10}\right)\\ \pause &=&e^{-0.3\times 10} \pause =0.050 \end{eqnarray*} \pause and the probability that the gap is between 5 and 10 minutes is \begin{eqnarray*} P(55) &=&1 - P(X<5) \pause \\ & & \\ &=&1-\frac{5-3}{6-3} = \pause \frac{1}{3} = 0.3333. \pause \end{eqnarray*} Also, $P(X>7) = \pause 0$... why? } \frame{ \frametitle{Example 5.2(d): Solution} If $X \sim U(a,b)$, \[ E[X] = \frac{a+b}{2} \qquad \textsf{and} \qquad \textsf{Var}(X) = \frac{(b-a)^{2}}{12}. \] \pause So we have \[ E[X] = \frac{3+6}{2} = 4.5 \textsf{ minutes} \] and \[ \textsf{SD}(X) = \sqrt{\frac{(6-3)^{2}}{12}} = 0.866 \textsf{ minutes} \] } \normal \frame{ \frametitle{Example 5.3} Customers arrive at the drive--through window of a fast food restaurant at a rate of 2 per minute during the lunch hour. \begin{itemize} \item [\textbf{(a)}] What is the probability that the next customer will arrive within 1 minute? \item [\textbf{(b)}] What is the probability that the next customer will arrive within 20 seconds? \item [\textbf{(c)}] What is the mean time between arrivals at the drive--through window? What about the standard deviation? \end{itemize} } \sol \frame{ \frametitle{Example 5.3(a),(b): Solution} Let $X$: Time between arrivals at the drive--through window (minutes); $X \sim \textsf{exp}(2)$. \pause Then \begin{eqnarray*} P(X<1) &=& 1-e^{-2\times 1} \pause \\ &=& 0.865. \end{eqnarray*} Similarly, \begin{eqnarray*} P\left(X<\frac{1}{3}\right) &=& 1-e^{-2 \times \frac{1}{3}} \pause \\ &=&0.487 \end{eqnarray*} } \frame{ \frametitle{Example 5.3(c): Solution} If $X \sim \textsf{exp}(\lambda)$, then \[ E[X] = \frac{1}{\lambda} \qquad \textsf{and} \qquad \textsf{Var}(X) = \frac{1}{\lambda^{2}}. \] \pause Thus, in our example, \[ E[X] = \frac{1}{2} = 0.5 \textsf{ minutes} = 30 \textsf{ seconds}, \] \pause and \[ \textsf{SD}(X) = \sqrt{\frac{1}{2^{2}}} = \sqrt{\frac{1}{4}} = 30 \textsf{ seconds}. \] } \normal \frame{ \vspace{1cm} \huge \textbf{Chapter 6} \vspace{1cm} \Large \textbf{Statistical inference} \textbf{for the population mean} } \frame{ \frametitle{What we'll cover...} \begin{itemize} \item \textcolor{purple}{\bf Confidence intervals} \pause \vspl \item One sample \textcolor{purple}{\bf hypothesis tests} for the mean \pause \vspl \item Two sample \textcolor{purple}{\bf hypothesis tests} for the mean \pause \end{itemize} } \frame { \frametitle{6.1 Introduction} \noindent \textbf{What does our sample tell us about the population?} \pause \vsp \begin{itemize} \item We can rarely observe the entire population, so the \textbf{\textcolor{purple}{population mean}} and \textbf{\textcolor{purple}{population variance}} are hardly ever known \textit{exactly}; \pause \item These unknown quantities are called \textbf{\textcolor{purple}{parameters}}; \pause \item We use Greek letters to denote them -- $\mu$ for the mean, and $\sigma^{2}$ for the variance (and so $\sigma$ for the standard deviation); \pause \item We hope that the \textbf{\textcolor{purple}{sample mean}} ($\bar{x}$) will be quite close to the true mean ($\mu$); \pause \item \textbf{But how do we know if it is?} \end{itemize} } \frame{ \frametitle{Example: \textit{The Vintage Clothing Co.}} \noindent The \textit{Vintage Clothing Co.} are a large retailer of bespoke and retro clothing. \pause \vsp They have 1,000 branches across the U.K., and all of their branches are open on Sundays. \pause \vsp However, they are considering whether or not it is worthwhile staying open on Sundays. \pause \vsp The table overleaf shows the number of transactions at each of their shops on Sunday 24th February 2013. } \frame{ \frametitle{Example: \textit{The Vintage Clothing Co.}} %\begin{table}[!h] \begin{center} \tiny \hspace*{-5mm} \begin{tabular}{|rrrrrrrrrr|rrrrrrrrrr|} \hline 282&258&399&271&343&285&247&513&171&123&168&327&430&240&410&341& 90&512&245&336\\ 290&263&446&185&330&111&243&376&139&351&311&389&546&321&393&487&287&514&149&315\\ 264&320&217&257&349&640& 97&298&393&454&363&354&360&326&199&502&154&273&213&413\\ 293&407&362&270&344&263&290&263& 50&253&345&581&229&264&304&394&246&235&417&452\\ 499&276&412&323&310&177&248&178&409&275&278&307&495&515&232&432&577&269&370&248\\ 339&404&371&262&336&218&274&483&211&245&316&381&432&233&223&447&412&250&262&337\\ 202&133&356&408&224&379&197&278&235&509&171&232&429&315&326&602& 63&290&230&121\\ 242&389&219&206&393&437&306&152&294&271&230&398&346&344&379&347&468&300&325&237\\ 305&174&291&261&214&532&335& 63&100&357&190&347&208&420&322&463&203&216&356&504\\ 389&236&445&378&255&301&308&150&289&453&464&273&211&450&222&250&214&259&296&356\\ \hline 320&420&357&160&372& 99&316&218&248&322&145&399&433&393&403&361&241&234&388&255\\ 261&279&369&342&168&322&304&254& 99&503&303&212&105&166&257&422&460&331&288&410\\ 346&370&235&355& 65&340&420&338&568&644&164&288&319&159&324&208&452&297&305&259\\ 268&340&305&361&319&519&293&380&286&431&402&329&363&330&612&248&302&592&589&349\\ 446&588&304&454&164&240&293&478&540&339&245&257&222&471&469&273&244&126&174&183\\ 277&216&555&401&380&338&212&476& 77&363&140&451&329& 66&217&461&435&380&314&324\\ 522&111&119&316&116&471&142&336&277&101&518&264&226&256&539&324&320&292&476&324\\ 333&332&404&362&202&204&341& 80&333&267&439&136&343&389&244&370&268&362&317&400\\ 372&595&314&182&470&192&555&374&368&192&225&321&435&403&316&312&307&368&236&452\\ 192& 63&407&125&253& 89& 70&186&491&342&122&367&106&334&161&177&180&355&356&317\\ \hline 454&122&286& 39&361&262&316&272&285&201&191&162&229&334&278&231&154&290&277&392\\ 644&297&398&118&246&148&478&167&337&344&395&334&255&401&504&304&408&204&673&126\\ 192&507& 41&457&405&306&282&446&195&512&252&510&557&191&321&404&542&438&291&449\\ 377&240&441&308&346&265&375&332&580&130&353&426& 95&588&332&109&467&333&388&309\\ 263&529&172&529&315&257&481&260&297&382&438& 64&226&185&369&275&320&126&321&375\\ 190&340&337&224&363&212&371&229&175&388&332&315&389&452&266&393&323&253&280&420\\ 219&400&378&241&616&551&359&489&314&450&645&224&320&405&182&251&370&341&318&232\\ 240&471&293&240&184&296&617&565&206&147&169&401&140&462&389&310&262&334&263&269\\ 323&351&187&544&387&425&353&175&378&484&205&295&413&189&559&251&480&283&262&304\\ 213&574&579&325&246&206&419&306&471&264&270&300&278&131&561&328&440&514&280&391\\ \hline 281&403&256&348&183&161&444&482&338&268&313&252&179&414&444&266&400&435&433&506\\ 203&269&450&322&459&183&212&242&144&406&401&174&605&270&487&494&235&316&368&319\\ 260&254&157&377&145&284&401&220&452& 59&335&467&251&192&371&298&317&382&363&397\\ 282&303&328&378&363&636&374&143&495&239&423&496&411&462&282&411&203&395&590&388\\ 278&272&417&666&233&316&287&268&186&247&339&397&276&291&324& 81&271&399&129&325\\ 247&152&315&224&130&323&352&276&398&338&231&258&310&421&215& 85&237&356&439&348\\ 507&277&240&188&321&419&370&374&211&224&340&264&441&226&563&279&297&114&546&277\\ 281&196&498&375&348&234&469&103&324&643&315&293&444&109&408&100&477&293&138&206\\ 485&279&494&513& 97&293&669&312&425& 70&181&210&241&187&448& 55&253&564&404&382\\ 31&496&234&200&411&386&218&382&483&405&435&414&379&360&194&291&393&247&314&285\\ \hline 204&188&444&416&106&485&276&250&248&200&352&463&251&197&197&456&293&333&373&240\\ 295&297&271&141&319&256&197&110&338&237&249&291&393&437&432&274&202&182&176&212\\ 482& 96&272&296&323&289&285&160&203&336&217&321&202&266&253&436&390&259&596&383\\ 236&291&226&250&270&439&360&310&326&415&447&336&354&273&243&390&213&318&346&599\\ 637&255& 61&393&324&492&484&259&271&150&550&185&224&352&387&441&232&261&313&410\\ 246&529& 97&448&369&199&140&498&287&293&258&431&267&396&217&340&278&297&387&281\\ 162&237&305&239&246&412&632&385&342&340&673&414&298&383&152&438&408&452&492&603\\ 439&223&404&466&380&214&155&410&291&234&248&325&391&338&416&262&361&358&484&129\\ 152&363& 90&383&365&500&362&190&343&138&233&179&200&476&128&308&221&649&278&152\\ 525&275&355&585&394&183&488&323&312&595&257&434&160&375&478&353&239&331&426&477\\ \hline \end{tabular} \end{center} } \frame{ \frametitle{Example: \textit{The Vintage Clothing Co.}} Suppose the marketing department of \textit{The Vintage Clothing Co.} are interested in the \textcolor{purple}{\bf average number of transactions} across all their stores on Sunday 24th February 2013. \pause \vsp Can we work this out \textit{exactly}? \pause \vsp \noindent The answer is ``yes'' \pause --- we have data from every single branch! \pause \vsp Actually, the table of results shows we have taken a \textcolor{purple}{\bf census} -- every single branch has been asked to provide us with data. \pause \vsp So in this case, it is possible to work out the \textcolor{purple}{\bf population mean} $\mu$: \pause \[ \mu =\frac{282+258+399+271+ \ldots + 426 + 477}{1000} = 320\textrm{ \sf transactions.} \] } \frame{ \frametitle{Example: \textit{The Vintage Clothing Co.}} \noindent Now, let's suppose the company don't have the \textcolor{purple}{\bf time}/\textcolor{purple}{\bf resources} to take a census. \pause \vsp In fact, a week before the 24th February, just \textcolor{purple}{\bf five stores} are selected at random --- \pause we work out the mean number of transactions using the data from these five stores only. \pause \vsp Let's suppose the top left--hand block of the table is for stores 1--100, the next block is for stores 101--200, etc. \pause \vsp We put the numbers 1--1000 into a bag and draw, without replacement, \textcolor{purple}{\bf 5 numbers at random}. } \frame{ \frametitle{Example: \textit{The Vintage Clothing Co.}} \begin{center} \begin{tabular}{c| c} Store &No. of transactions, $X$ \\ \hline 637 &$x_{1}=374$ \\ 327 &$x_{2}=452$ \\ 849 &$x_{3}=271$ \\ 666 &$x_{4}=419$ \\ 680 &$x_{5}=643$ \\ \end{tabular} \end{center} \pause \noindent Let's suppose this is the only information we have. \pause \vsp It is no longer possible to work out the true population mean, as we don't have information from every single shop; \pause we can now only work out the \textcolor{purple}{\bf sample mean} $\bar{x}$: \pause \[ \bar{x} = \frac{374+452+271+419+643}{5} = 431.8 \approx 432 \textrm{ \sf transactions.} \] } \frame{ \frametitle{Example: \textit{The Vintage Clothing Co.}} \noindent Obviously, the marketing team are not just interested in what goes on in these five shops. \pause \vsp However, this is the only information they have, and so they use this information to draw conclusions about all 1,000 shops as a whole. \pause \vsp This is known as the process of \textcolor{purple}{\bf statistical inference} -- we are trying to \textit{infer} things about the population, based on the limited information in our sample. \pause \vsp Hopefully, provided we don't have a \textcolor{purple}{\bf biased sample}, $\bar{x}$ will do a good job at estimating $\mu$. Has it done a good job here? } \sol \frame{ \frametitle{Example: \textit{The Vintage Clothing Co.}} The true \textcolor{purple}{\bf population mean} is $\mu=320$. \pause \vsp Our \textcolor{purple}{\bf sample mean} is $\bar{x}\approx 432$. \pause \vsp We have \textcolor{purple}{\bf over--estimated} the population mean by quite a bit \pause -- our estimate is 112 too high! \pause \vsp Perhaps we have a \textcolor{purple}{\bf biased sample} -- we seem to have selected a number of stores with a higher--than--average number of transactions -- especially store 680!} \normal \frame{ \frametitle{Example: \textit{The Vintage Clothing Co.}} This begs the question: \pause \begin{center}``\textit{How accurate can our sample mean be in estimating the population mean?}'' \end{center} \pause Let's take another random sample by drawing another five numbers from the bag: \pause \begin{center} \begin{tabular}{c| c} Store &No. of transactions, $X$ \\ \hline 558 &$x_{1}=253$ \\ 428 &$x_{2}=446$ \\ 903 &$x_{3}=251$ \\ 364 &$x_{4}=256$ \\ 14 &$x_{5}=185$ \\ \end{tabular} \end{center} \pause \noindent This gives \[ \bar{x} = \frac{253+446+251+256+185}{5} = 278.2 \approx 278 \textrm{ \sf transactions.} \] } \frame{ \frametitle{Example: \textit{The Vintage Clothing Co.}} This sample mean is much closer to the population mean, but still not \textit{very} close. \pause \vsp Also, it is quite different from the mean of the previous sample. \pause \vsp You could repeat this procedure yourself (filling in the table on page 146) to select three more random samples of size 5, and calculate the sample means. \pause \vsp In fact, consider this \textcolor{purple}{\bf homework} to do before the next practical session! } \frame{ \frametitle{Example: \textit{The Vintage Clothing Co.}} \begin{center} \begin{tabular}{ccc} \multicolumn{1}{l}{\textbf{Your random sample 1}}\hspace*{1cm}\\ \noalign{\medskip} $u=$\hspace*{1.75cm}&$store=$\hspace*{1.75cm}&$x_1=$\hspace*{1.75cm}\\ \noalign{\medskip} $u=$\hspace*{1.75cm}&$store=$\hspace*{1.75cm}&$x_2=$\hspace*{1.75cm}\\ \noalign{\medskip} $u=$\hspace*{1.75cm}&$store=$\hspace*{1.75cm}&$x_3=$\hspace*{1.75cm}\\ \noalign{\medskip} $u=$\hspace*{1.75cm}&$store=$\hspace*{1.75cm}&$x_4=$\hspace*{1.75cm}\\ \noalign{\medskip} $u=$\hspace*{1.75cm}&$store=$\hspace*{1.75cm}&$x_5=$\hspace*{1.75cm}\\ \noalign{\medskip} \cline{3-3} \noalign{\medskip} &&$\bar x=$\hspace*{1.75cm} \end{tabular} \end{center} \pause How close are these sample means to the correct population value $\mu=320 \textrm{ \sf transactions}$? } \frame{ \frametitle{Example: \textit{The Vintage Clothing Co.}} \noindent In fact, we could take many samples, and it's very likely that we'll get a \textcolor{purple}{\bf different value} for $\bar{x}$ each time. \pause \vsp It's also very \textit{unlikely} that any of our $\bar{x}$'s will be exactly the same as the \textcolor{purple}{\bf true population mean} $\mu$. \pause \vsp The graph on the next slide shows histograms of $\bar{x}$'s taken from 100 samples from our population of 1,000 stores. } \frame{ \frametitle{Example: \textit{The Vintage Clothing Co.}} \begin{center} \psfrag{0}[][]{\sf \tiny 0} \psfrag{5}[][]{\sf \tiny 5} \psfrag{10}[][]{\sf \tiny 10} \psfrag{15}[][]{\sf \tiny 15} \psfrag{20}[][]{\sf \tiny 20} \psfrag{150}[][]{\sf \tiny 150} \psfrag{200}[][]{\sf \tiny 200} \psfrag{250}[][]{\sf \tiny 250} \psfrag{300}[][]{\sf \tiny 300} \psfrag{350}[][]{\sf \tiny 350} \psfrag{400}[][]{\sf \tiny 400} \psfrag{450}[][]{\sf \tiny 450} \psfrag{500}[][]{\sf \tiny 500} \psfrag{25}[][]{\sf \tiny 25} \psfrag{30}[][]{\sf \tiny 30} \psfrag{40}[][]{\sf \tiny 40} \psfrag{xbar}[][]{\sf \scriptsize $\bar{x}$} \psfrag{n=5}[][]{\sf \small $n=5$} \psfrag{n=10}[][]{\sf \small $n=10$} \psfrag{n=50}[][]{\sf \small $n=50$} \psfrag{n=100}[][]{\sf \small $n=100$} \psfrag{n=250}[][]{\sf \small $n=250$} \psfrag{n=500}[][]{\sf \small $n=500$} \psfrag{Frequency}[][]{\sf \tiny Frequency} \includegraphics[width=10cm,height=7.5cm]{CLT.ps} \end{center} } \frame{ \frametitle{Example: \textit{The Vintage Clothing Co.}} \noindent \textbf{You should notice two things:} \pause \vspl \begin{itemize} \item [\textbf{1.}] The distribution of $\bar{x}$'s looks like a \textcolor{purple}{\bf Normal distribution}; \pause \vspl \item [\textbf{2.}] As we increase the sample size ($n$), the distribution for $\bar{x}$ gets more and more \textcolor{purple}{\bf concentrated} -- around the \textcolor{purple}{\bf true population value} $\mu=320$! \end{itemize} \pause \vsp \noindent In fact, what we can see in action in this graph is known as the \textcolor{purple}{\bf Central Limit Theorem}. \pause \vsp This is a very powerful result in Statistics which tells us about the distribution of the sample mean $\bar{x}$. We now state this formally. } \frame{ \frametitle{The Central Limit Theorem} \noindent Suppose $x_{1}, x_{2}, \ldots, x_{n}$ are a random sample from \textit{any} population, with mean $\mu$ and variance $\sigma^{2}$. \pause \vsp If $n$ is large, then \begin{eqnarray*} \bar{x} \sim N\left(\mu, \frac{\sigma^{2}}{n}\right) \qquad \textrm{\sf approximately};\pause \end{eqnarray*} if $x_{1}, x_{2}, \ldots, x_{n}$ come from a Normal distribution themselves, then this result holds for \textit{any} $n$. } \frame{ \frametitle{The Central Limit Theorem} In other words, if we were to take many samples of size $n$ and: \pause \vspl \begin{itemize} \item For each sample calculate the mean $\bar{x}$; \pause \vspl \item Put all of our $\bar{x}$'s together and make a histogram of them; \end{itemize} \pause \vspl then our histogram of $\bar{x}$'s will \textcolor{purple}{\bf always} be Normally distributed around the true population mean $\mu$. \pause \vsp What's more, we also know about the \textcolor{purple}{\bf variability} of $\bar{x}$: \pause \[ \textsf{var}(\bar{x}) = \frac{\sigma^{2}}{n}. \] \pause Thus, the standard deviation of $\bar{x}$ is $\sigma/\sqrt{n}$, and we call this the \textcolor{purple}{\bf standard error}. } \frame{ \frametitle{6.2 Confidence intervals} \noindent The values we calculate for sample means and variances are \textcolor{purple}{\bf point estimates}. \pause \vsp They are single values based on a limited sample of the whole population. \pause \vsp Suppose that we wish to estimate the mean $\mu$ of a population. The natural estimate for $\mu$ is the sample mean $\bar{x}$. \pause \vsp However, as we have seen, $\bar{x}$ is never exactly equal to $\mu$; all we really hope is that $\bar{x}$ will be close to $\mu$. } \frame{ \frametitle{6.2 Confidence intervals} One way of improving our inference is to construct \textcolor{purple}{\bf interval estimates}, more commonly known as \textcolor{purple}{\bf confidence intervals}. \pause \vsp We simply place an interval over the point estimate for $\mu$ which allows us to say (with a certain level of confidence) within what range the population mean lies. \pause \vsp The calculation of these intervals depends on the size of our sample ($n$), the level of confidence we choose, and whether or not the population variance ($\sigma^{2}$) is known. } \frame{ \frametitle{6.2.1 Case 1: Known variance $\sigma^{2}$} \noindent We know from the CLT that, if our random sample is drawn from a Normal distribution, or if $n$ is large (i.e. $n \geq 30$), then \pause \begin{eqnarray*} \bar{x} \sim N\left(\mu, \frac{\sigma^{2}}{n}\right). \end{eqnarray*} \pause If we initially assume we know the population variance $\sigma^{2}$, we can \textcolor{purple}{\bf standardise} $\bar{x}$ as we did in Chapter 5; i.e. \pause \begin{eqnarray*} Z &=& \frac{\bar{x}-\mu}{\sqrt{\sigma^{2}/n}}. \end{eqnarray*} \pause Recall that the standard Normal distribution is $Z\sim N(0,1)$, i.e. $Z$ has zero mean and variance (and so standard deviation) 1. \pause \vsp Also recall that approximately 95\% of the standard normal distribution lies between $-1.96$ and $1.96$, i.e: } \frame { \frametitle{6.2.1 Case 1: Known variance $\sigma^{2}$} \begin{itemize} \item We know that (from tables) \pause \begin{eqnarray*} \textsf{Pr}(\textsf{\sf --1.96} $), and so our values are: \pause \begin{center} \begin{tabular}{|c|c c c|} \hline Significance level &10\% &5\% &1\% \\ \hline Critical value &1.645 &1.96 &2.576 \\ \hline \end{tabular} \end{center} \pause \noindent Our test statistic $z=2.33$ lies between the critical values of 1.96 and 2.576, and so our $p$--value lies \textcolor{purple}{\bf between 1\% and 5\%}. \pause \vsp We can see this more clearly on a diagram. } \frame{ \frametitle{6.3.2 Case 1: Population variance known: Example} \noindent \textbf{Step 5} \textit{(conclusion)} \pause \vsp \noindent Using table 6.2 to interpret our $p$--value, we see that: \pause \vspl \begin{itemize} \item There is \textcolor{purple}{\bf moderate} evidence against $H_{0}$ \pause \item Thus, we should \textcolor{purple}{\bf reject} $H_{0}$ in favour of $H_{1}$ \pause \item It appears that the population mean transaction size \textcolor{purple}{\bf is not equal to} \pounds 130 \end{itemize} } \normal \frame{ \frametitle{6.3.2 Case 1: Population variance known: Example} \noindent Alternatively, since our sample mean $\bar{x}=\textrm{\sf \pounds} 123$ is smaller than the proposed value of $\textrm{\sf \pounds} 130$, we could have set up a \textcolor{purple}{\bf one--tailed} alternative hypothesis: \pause \begin{eqnarray*} H_{0} &:& \mu = \textrm{\sf \pounds} 130 \qquad \textrm{\sf against} \\ H_{1} &:& \mu < \textrm{\sf \pounds} 130. \end{eqnarray*} \pause \noindent This is now a one--tailed test and the critical values from tables are: \pause \begin{center} \begin{tabular}{|c| c c c|} \hline Significance level &10\% &5\% &1\% \\ \hline Critical value &1.282 &1.645 &2.326 \\ \hline \end{tabular} \end{center} } \frame{ \frametitle{6.3.2 Case 1: Population variance known: Example} \noindent The test statistic is (as before) $2.33$, which now lies ``to the right'' of the last critical value in the table (2.326). \pause \vsp Thus, our $p$--value is now smaller than 1\%, and so, using table 6.2, we see that \pause \vsp \begin{itemize} \item there is \textcolor{purple}{\bf strong} evidence against $H_{0}$ \pause \item We would \textcolor{purple}{\bf reject} $H_{0}$ in favour of $H_{1}$ \pause \item There is evidence to suggest that the population mean \textcolor{purple}{\bf is less than} \pounds 130 \end{itemize} } \frame{ \frametitle{6.3.3 Case 2: Population variance unknown: Example} \noindent The batteries for a fire alarm system are required to last for 20000 hours before they need replacing. \vsp 16 batteries were tested; they were found to have an average life of 19500 hours and a standard deviation of 1200 hours. \vsp Perform a hypothesis test to see if the batteries do, on average, last for 20000 hours. } \frame{ \frametitle{6.3.3 Case 2: Population variance unknown: Example} \noindent \textbf{Steps 1 and 2} \textit{(hypotheses)} \pause \vsp \noindent Using a \textcolor{purple}{\bf one--tailed test}, our null and alternative hypotheses are: \begin{eqnarray*} H_{0} &:& \mu = 20000 \qquad{\textrm{\sf versus}} \\ H_{1} &:& \mu < 20000. \end{eqnarray*} \pause \noindent We use a one--tailed test because we are interested in whether the batteries are effective or not; there is no problem if they last longer than 20000 hours. } \frame{ \frametitle{6.3.3 Case 2: Population variance unknown: Example} \noindent \textbf{Step 3} \textit{(calculating the test statistic)} \pause \vsp \noindent Unlike the previous example, the population variance $\sigma^{2}$ is unknown.\pause \vsp However, the \textcolor{purple}{\bf sample} standard deviation is given, based on a sample of size 16, and so we need to use a slightly different test statistic. \pause \vsp In fact, we do what we did last week when we were constructing confidence intervals -- i.e. we replace $\sigma^{2}$ with $s^{2}$ and then use tables of values from Student's $t$ distribution instead of the standard Normal distribution. \pause } \frame{ \frametitle{6.3.3 Case 2: Population variance unknown: Example} Thus, the test statistic is given by \pause \begin{eqnarray*} t &=& \frac{|\bar{x} - \mu |}{\sqrt{s^2 / n}}\\ \pause & & \\ &=& \frac{|19500-20000|}{\sqrt{1200^{2}/16}} \\ \pause & & \\ &=& \frac{500}{\sqrt{1440000/16}} \\ \pause & & \\ &=& 1.667. \end{eqnarray*} } \frame{ \frametitle{6.3.3 Case 2: Population variance unknown: Example} \noindent \textbf{Step 4} \textit{(finding the $p$--value)} \pause \vsp \noindent Since $\sigma^{2}$ is unknown, we use $t$--distribution tables to obtain a range for our $p$--value. \pause \vsp The degrees of freedom, $\nu=n-1=16-1=15$, and under a one--tailed test this gives the following critical values: \pause \begin{center} \begin{tabular}{|c| c c c|} \hline Significance level &10\% &5\% &1\% \\ \hline Critical value &1.341 &1.753 &2.602 \\ \hline \end{tabular} \end{center} \pause \noindent Our test statistic of $t=1.667$ lies between the critical values of 1.341 and 1.753, and so the corresponding $p$--value lies between 5\% and 10\%. } \frame{ \frametitle{6.3.3 Case 2: Population variance unknown: Example} \noindent \textbf{Step 5} \textit{(conclusion)} \pause \vsp \noindent Using table 6.2 to interpret our $p$--value, we see that there is \pause \vspl \begin{itemize} \item only \textcolor{purple}{\bf slight} evidence against the null hypothesis \pause \item this is not enough grounds to reject it, so we \textcolor{purple}{\bf retain} $H_{0}$ \pause \item There is insufficient evidence to suggest there is a problem with these batteries. \end{itemize} } \frame { \frametitle{6.4 Testing two means} \noindent If we have \textbf{two} independent random samples from \textbf{two} populations, we can compare the two sample means (\textit{c.f.} comparing one sample mean to a \textit{proposed value} in the one--sample case). \pause \vspace{0.5cm} \noindent We use the same framework for hypothesis testing as for the one--sample test; however, the calculations required for the test statistic are slightly different. There are two situations to consider here: \pause \begin{itemize} \item Are \textbf{both} population variances ($\sigma_{1}^{2}$ \textbf{and} $\sigma_{2}^{2}$) \textcolor{purple}{\textbf{known}}? \pause \item Are \textbf{both} population variances \textcolor{purple}{\textbf{unknown}}? \end{itemize} } \frame { \frametitle{6.4.1 Both population variances ($\boldsymbol{\sigma_{1}^{2}}$ and $\boldsymbol{\sigma_{2}^{2}}$) known} \pause \begin{itemize} \pause \item[\textbf{1.}] \textcolor{purple}{\textbf{State the null hypothesis}} \\ \noindent This time, the null hypothesis is \begin{eqnarray*} H_{0} &:& \mu_{1} = \mu_{2}, \end{eqnarray*} i.e. the two population means are equal. \pause \vspl \item [\textbf{2.}] \textcolor{purple}{\textbf{State the alternative hypothesis}} \\ \noindent We usually test against the (\textcolor{purple}{\bf two--tailed}) alternative: \begin{eqnarray*} H_{1} &:& \mu_{1} \neq \mu_{2}, \end{eqnarray*} i.e. the population means \textit{are not} equal. However, we might use the \textcolor{purple}{\bf one--tailed} alternatives: \pause \begin{eqnarray*} H_{1} &:& \mu_{1} > \mu_{2}, \qquad \textsf{\sf or} \\ \pause H_{1} &:& \mu_{1} < \mu_{2}. \end{eqnarray*} \end{itemize} } \frame{ \frametitle{6.4.1 Both population variances ($\boldsymbol{\sigma_{1}^{2}}$ and $\boldsymbol{\sigma_{2}^{2}}$) known} \begin{itemize} \item [\textbf{3.}] \textcolor{purple}{\textbf{Calculate the test statistic}} \\ \noindent The test statistic for a two--sample test (when both population variances are known) is \begin{eqnarray*} z &=& \frac{|\bar{x}_{1}-\bar{x}_{2}|}{\sqrt{\frac{\sigma_{1}^{2}}{n_{1}} + \frac{\sigma_{2}^{2}}{n_{2}}}} \end{eqnarray*} \end{itemize} } \frame{ \frametitle{Both population variances ($\boldsymbol{\sigma_{1}^{2}}$ and $\boldsymbol{\sigma_{2}^{2}}$) known} \begin{itemize} \item [\textbf{4.}]\textcolor{purple}{\textbf{Find the $\boldsymbol{p}$--value}} \\ \noindent This is found from statistical tables; since, in this case, both population variances $\sigma_{1}^{2}$ and $\sigma_{2}^{2}$ are \textit{known}, we refer to standard normal tables. As before, we find a range for our $p$--value by comparing our test statistic to the 10\%, 5\% and 1\% critical values. \pause \vspl \item [\textbf{5.}] \textcolor{purple}{\textbf{Form a conclusion}} \\ \noindent Exactly the same again! Use table 6.2 to help you decide what to do! Word your conclusions in the context of the original question. \pause \end{itemize} } \frame{ \frametitle{Example 6.3} Before a training session for call centre employees a sample of 50 calls to the call centre had an average duration of 5 minutes, whereas after the training session a sample of 45 calls had an average duration of 4.5 minutes. \pause \vsp The population variance is \textbf{known} to have been 1.5 minutes before the course and 2 minutes afterwards. \pause \vsp Has the course been effective? } \sol \frame{ \frametitle{Example 6.3: Solution} \textbf{Steps 1 \& 2} \textit{(hypotheses)} \pause \vsp \begin{eqnarray*} H_{0}&:& \mu_{B}=\mu_{A} \qquad \textrm{\sf versus} \\ \pause & & \\ H_{1} &:& \mu_{B}\neq \mu_{A} \end{eqnarray*} } \frame{ \frametitle{Example 6.3: Solution} \textbf{Step 3} \textit{(Test statistic)} \pause \vsp We use \pause \begin{eqnarray*} z &=& \frac{|\bar{x}_{B}-\bar{x}_{A}|}{\sqrt{\frac{\sigma_{B}^{2}}{n_{B}}+\frac{\sigma_{A}^{2}}{n_{A}}}} \pause \\ && \\ &=& \frac{|5 - 4.5|}{\sqrt{\frac{1.5}{50}+\frac{2}{45}}} \pause \\ & &\\ &=& \frac{0.5}{\sqrt{0.03+0.0444}} \\ \pause & & \\ &=& 1.833 \end{eqnarray*} } \frame{ \frametitle{Example 6.3: Solution} \textbf{Step 4} \textit{(p--value)} \pause \vsp Since this is case 1 (both population variances known), we use Normal tables (page 158). \pause \vsp Recall that we have a \textcolor{purple}{\bf two--tailed} alternative, and so we get \pause \begin{center} \begin{tabular}{|c| c c c|} \hline Significance level &10\% &5\% &1\% \\ \hline Critical value &1.645 &1.96 &2.576 \\ \hline \end{tabular} \end{center} \pause Our test statistic $z=1.833$ shows that our $p$--value is between 5\% and 10\%.} \frame{ \frametitle{Example 6.3: Solution} \textbf{Step 5} \textit{(Conclusion)} \pause \vsp From Table 6.2, \pause \vsp \begin{itemize} \item We have \textcolor{purple}{\bf slight} evidence against $H_{0}$ \pause \vspl \item This is not enough to reject and so we \textcolor{purple}{\bf retain} $H_{0}$ \pause \vspl \item There is insufficient evidence to suggest the training course has been effective! \end{itemize} } \normal \frame { \frametitle{6.4.2 $\boldsymbol{\sigma_{1}^{2}}$ and $\boldsymbol{\sigma_{2}^{2}}$ unknown} \noindent In the more likely situation where the population variances are unknown, the test statistic becomes \pause \begin{eqnarray*} t &=& \frac{|\bar{x}_{1}-\bar{x}_{2}|}{s \times \sqrt{\frac{1}{n_{1}}+\frac{1}{n_{2}}}}, \end{eqnarray*} where $s$ is a ``pooled standard deviation'', and is found as \begin{eqnarray*} s &=& \sqrt{\frac{(n_{1}-1)s_{1}^{2}+(n_{2}-1)s_{2}^{2}}{n_{1}+n_{2}-2}}. \end{eqnarray*} \pause \noindent Like before, we have to use $t$--tables to obtain our critical value; the degrees of freedom is now found as $\nu=n_{1}+n_{2}-\textsf{\sf 2}$. } \normal \frame{ \frametitle{Example 6.4} \noindent A company is interested in knowing if two branches have the same level of average transactions. \vsp The company sample a small number of transactions and calculates the following statistics: \begin{center} \begin{tabular}{c|ccc} Shop 1 & $\bar{x}_{1} = 130$ & $s_{1}^{2} = 700$ & $n_{1}=12$\\ Shop 2 & $\bar{x}_{2} = 120$ & $s_{2}^{2} = 800$ & $n_{2}=15$ \end{tabular} \end{center} \noindent Test whether or not the two branches have (on average) the same level of transactions. } \frame{ \frametitle{Example 6.4: Solution} \textbf{Steps 1 and 2} \textit{(hypotheses)} \pause \vspl \noindent Our null and alternative hypotheses are: \pause \begin{eqnarray*} H_{0} &:& \mu_{1} = \mu_{2} \qquad \textrm{\sf versus} \\ \pause H_{1} &:& \mu_{1} \neq \mu_{2}. \end{eqnarray*} \vsp \textbf{Step 3} \textit{(Test statistic)} \pause \vspl Since both population variances are unknown (only the \textit{sample} values are given), the test statistic is \pause \begin{eqnarray*} t &=& \frac{|\bar{x}_{1}-\bar{x}_{2}|}{s \times \sqrt{\frac{1}{n_{1}}+\frac{1}{n_{2}}}}. \end{eqnarray*} } \sol \frame{ \frametitle{Example 6.4: Solution} First the \textcolor{purple}{\bf pooled standard deviation}: \pause \begin{eqnarray*} s &=& \sqrt{\frac{(n_{1}-1)s_{1}^{2}+(n_{2}-1)s_{2}^{2}}{n_{1}+n_{2}-2}} \\ \pause & & \\ &=& \sqrt{\frac{(12-1)\times 700 + (15-1)\times 800}{12+15-2}} \\ \pause & & \\ &=& \sqrt{\frac{11\times 700 + 14 \times 800}{25}} \\ \pause & & \\ &=& \sqrt{\frac{7700 + 11200}{25}} \\ \pause & & \\ &=& \sqrt{\frac{18900}{25}} \\ \pause & & \\ &=& \sqrt{756} = 27.495. \end{eqnarray*}} \frame{ \frametitle{Example 6.4: Solution} Now we put this into our formula for $t$: \pause \begin{eqnarray*} t &=& \frac{|130-120|}{27.495 \times \sqrt{\frac{1}{12}+\frac{1}{15}}} \\ \pause & &\\ &=& \frac{10}{27.495 \times \sqrt{0.15}} \\ \pause & &\\ &=& \frac{10}{10.649} \\ \pause & &\\ &=& 0.939. \end{eqnarray*}} \normal \frame{ \frametitle{Example 6.4: Solution} \textbf{Step 4} \textit{(p--value)} \vsp Since both population variances are unknown, we use $t$--tables to obtain our critical value. \pause \vsp The degrees of freedom, $\nu=n_{1}+n_{2}-2$, i.e. $\nu=12+15-2=25$. \pause \vsp Under a two--tailed test, and using $t$--tables (page 158), we get the following critical values: \pause \begin{center} \begin{tabular}{|c| c c c|} \hline Significance level &10\% &5\% &1\% \\ \hline Critical value &1.708 &2.060 &2.787 \\ \hline \end{tabular} \end{center} \pause \noindent Our test statistic $t=0.939$ lies to the left of the first critical value, and so our $p$--value is bigger than 10\%. } \frame{ \frametitle{Example 6.4: Solution} \textbf{Step 5} \textit{(Conclusion)} \vsp Using Table 6.2, \pause \vspl \begin{itemize} \item We have \textcolor{purple}{\bf no} evidence against $H_{0}$ \pause \vspl \item Therefore, we \textcolor{purple}{\bf retain} $H_{0}$ \pause \vspl \item There is no significant difference between the average level of transactions at the two shops. \end{itemize} } \announce \frame{ \frametitle{Welcome back!} Two more assignments: \vspl \begin{itemize} \item Clubs case study -- next week's computer session, deadline \textbf{4pm, Friday 12th May} \pause \vspl \item CBA -- practice mode next week, assessed mode w/b 8th May, deadline \textbf{11.45pm, Sunday 14th May} \end{itemize} \pause } \frame{ \frametitle{Welcome back!} Case study: \pause \vspl \begin{itemize} \item Won't take too much work \pause \vspl \item Three/four pages max, including graphs and tables \pause \vspl \item Need to do simple stuff: graphs, means, standard deviations etc., PLUS `harder' stuff: confidence intervals and a hypothesis test \pause \vspl \item Must do analysis on Minitab - we will do all of the necessary computing in the computer sessions next week \pause \vspl \item Individual submissions! \end{itemize} \pause \vspl Next week: \pause \vspl \begin{itemize} \item Bank holiday \pause \vspl \item Monday practicals cancelled - replacement Wednesday, 1-3pm, Side cluster BS.TC \end{itemize} } \normal \normal \frame{ \vspace{1cm} \huge \textbf{Chapter 7} \vspace{1cm} \Large \textbf{Techniques of regression} } \frame{ \frametitle{What we'll cover...} \begin{itemize} \item Visualising \textcolor{purple}{\bf bivariate data} \pause \vspl \item Assessing the \textcolor{purple}{\bf association} between two continuous variables: \pause \vspl \begin{itemize} \item [--] Informally \pause \vspl \item [--] More objectively -- the sample \textcolor{purple}{\bf correlation coefficient} \pause \end{itemize} \vspl \item Assessing the \textcolor{purple}{\bf signficance} of the correlation coefficient \pause \vspl \item Modelling the relationship between two continuous variables: \textcolor{purple}{\bf Simple linear regression} \pause \vspl \item Extensions to the \textcolor{purple}{\bf multiple linear regression} model, and possibly beyond... \end{itemize} } \frame{ \frametitle{7.1 Introduction} \noindent In this chapter we will investigate relationships between \textcolor{purple}{\bf continuous variables}. \pause \vsp Initially, we will assume our data consists of $n$ pairs of observations on two variables, say $X$ and $Y$: \begin{eqnarray*} (x_{1},y_{1}), (x_{2},y_{2}), \ldots, (x_{n},y_{n}). \end{eqnarray*} \pause \begin{itemize} \item Could be a random sample of $n$ individuals from a population, on which two measurements/observations $(x_{i},y_{i})$, $i=1, \ldots, n$ were made \pause \vspl \item Could be from an experiment in which one variable ($X$) is held fixed or controlled at certain chosen levels and independent measurements of the \textcolor{purple}{\bf response} variable ($Y$), are taken at each of these levels \end{itemize} } \frame{ \frametitle{7.1 Introduction} The first step is \textit{always} to plot the data on a \textcolor{purple}{\bf scatter diagram}. \pause \vsp We considered scatter diagrams, often called \textcolor{purple}{\bf scatter plots}, in Chapter 4 (Section 4.5.3, page 92). \pause \vsp From this diagram we can get an initial impression about the relationship between $X$ and $Y$ and form some subjective assessments. } \frame{ \frametitle{7.1 Introduction} The main aim of this Chapter is to supplement such descriptive analyses with more formal techniquesb --- in terms of \pause \vspl \begin{itemize} \item \textcolor{purple}{\bf quantifying} the association between $X$ and $Y$, and \pause \vspl \item \textcolor{purple}{\bf modelling} any relationship between $X$ and $Y$ \end{itemize} } \frame{ \frametitle{7.1 Introduction} Towards the end of the course, we will also consider extending these techniques to investigate the relationship between the response variable $Y$ and more than one \textcolor{purple}{\bf predictor} variable -- perhaps to include several predictor variables $X_{1}, X_{2}, \ldots$. } \frame{ \frametitle{7.2 Example: The \textit{Saint Clair Estate Winery}} \noindent The \textit{Saint Clair Estate Winery} is a vineyard in the Marlborough region of the South Island of New Zealand. \pause \vspl \begin{itemize} \item \textcolor{purple}{\bf Multi--million} dollar industry \pause \vspl \item \textit{Saint Clair} one of the country's leading producers/exporters, of wine \end{itemize} } \frame{ \frametitle{7.2 Example: The \textit{Saint Clair Estate Winery}} \noindent The price of a bottle of wine is thought to depend on many factors, such as its age, the quality of the grapes used to produce it, the amount of rainfall during the growing season, where the wine was produced, etc. \pause \vspace{0.3cm} The table below shows the price of 10 randomly selected bottles of wine from \textcolor{blue}{\texttt{www.tanners-wines.co.uk}}, an online wine merchant. Also shown is the age of each wine selected. \pause {\scriptsize \begin{center} \begin{tabular}{|c|c c c c c c c c c c|} \hline Bottle &1 &2 &3 &4 &5 &6 &7 &8 &9 &10 \\ \hline Age ($X$)&$3\frac{1}{2}$ &5 &3 &$2\frac{1}{2}$ &3 &2 &$2\frac{1}{2}$ &1 &10 &4 \\ Price ($Y$) &4.50 &12.95 &6.50 &4.99 &7.50 &14.95 &8.25 &3.95 &18.99 &10.00 \\ \hline \end{tabular} \end{center}} } \frame{ \frametitle{7.2 Example: The \textit{Saint Clair Estate Winery}} %\centering \includegraphics[width=10cm,height=7cm]{wine1.ps} %\caption{Scatter plot of price of wine (\pounds) against age (years)} %\end{figure} } \sol \frame{ \frametitle{7.2 Example: The \textit{Saint Clair Estate Winery}} \noindent Looking at the scatter plot (and maybe just the raw data themselves!), what can you say about the relationship between age of wine and price? \pause \begin{itemize} \item \textit{Generally}, as the age of wine increases, the price also increases \pause \item There is a \textcolor{purple}{\bf linear} relationship \pause \item There is a \textcolor{purple}{\bf strong} linear relationship? \pause Or maybe \textcolor{purple}{\bf moderate}? \pause \item There is a \textcolor{purple}{\bf positive correlation} \end{itemize} } \normal \frame{ \frametitle{7.3 Quantifying the relationship: Correlation} \noindent There is clearly a relationship between the age and price of wine; the relationship is \textcolor{purple}{\bf strong}, \textcolor{purple}{\bf positive} and \textcolor{purple}{\bf linear}. \pause \vspace{0.3cm} How would you describe, in words, the relationship between $X$ and $Y$ in the following scatter plots? %\begin{figure}[!h] \begin{center} \includegraphics[width=7cm,height=4cm]{scatter1.ps} %\caption{Scatter plots showing different relationships} %\end{figure} \end{center} } \frame{ \frametitle{7.3 Quantifying the relationship: Correlation} \noindent Scatterplots such as the one in the bottom left--hand corner can be difficult to interpret using words alone, since different people might say different things. \pause \vspace{0.3cm} Some might think there is a moderate/fairly strong relationship between $X$ and $Y$ here, whilst others might conclude that there is a relatively weak relationship between these two variables. \pause \vspace{0.3cm} Interpreting such relationships with words alone can be subjective; \textcolor{purple}{\bf quantifying} such relationships \textcolor{purple}{\bf numerically} can circumvent this problem of subjectivity. } \frame{ \frametitle{7.3 Quantifying the relationship: Correlation} One way of doing this is to calculate the \textcolor{purple}{\bf product moment correlation coefficient}, often denoted by the letter $r$. \pause \vspace{0.3cm} The formula for $r$ is \pause \begin{eqnarray*} r &=& \frac{S_{xy}}{\sqrt{S_{xx}\times S_{yy}}}, \end{eqnarray*} \pause where \pause \begin{eqnarray*} S_{xy} &=& \left(\sum xy\right)-n\bar{x}\bar{y}, \\ \pause S_{xx} &=& \left(\sum x^{2}\right) - n\bar{x}^{2} \qquad \textrm{\sf and} \\ \pause S_{yy} &=& \left(\sum y^{2}\right) - n\bar{y}^{2}, \end{eqnarray*} \pause $n$ is the number of pairs and $\bar{x}$ and $\bar{y}$ correspond to the mean of $X$ and the mean of $Y$ (respectively). } \frame{ \frametitle{7.3 Quantifying the relationship: Correlation} \noindent The correlation coefficient $r$ always lies between $-1$ and $+1$. \pause \begin{itemize} \item If $r$ is close to $+1$, there is a strong \textit{positive} linear relationship \pause \item If $r$ is close to $-1$ there is a strong \textit{negative} relationship \pause \item If $r$ is close to zero, there is \textit{no} linear relationship between the variables. \end{itemize} \pause Note that $r \approx 0$ does not imply no relationship at all, simply no \textit{linear} relationship. \pause \vspace{0.3cm} \noindent Can you estimate the value of $r$ for the wine age/price data? And for the four datasets shown on page 169? } \frame{ \frametitle{7.3 Quantifying the relationship: Correlation} \noindent Using the wine price/age data, we can calculate the value of $r$. \pause The easiest way to do this is to draw up a table: \pause %{\small \begin{center} \begin{tabular}{|c| c| c| c| c| } \hline $x$ &$y$ &$x^{2}$ &$y^{2}$ &$xy$ \\ \hline 3.5 &4.50 &12.25 &20.25 &15.75 \\ \pause 5 &12.95 &25 &167.7025 &64.75 \\ \pause 3 &6.50 &9 &42.25 &19.5 \\ \pause 2.5 &4.99 &6.25 &24.9001 &12.475 \\ 3 &7.50 &9 &56.25 &22.5 \\ 2 &14.95 &4 &223.5025 &29.9 \\ 2.5 &8.25 &6.25 &68.0625 &20.625 \\ 1 &3.95 &1 &15.6025 &3.95 \\ 10 &18.99 &100 &360.6201 &189.900\\ 4 &10.00 &16 &100 &40 \\ \hline \pause \textbf{36.5} &\textbf{92.58} &\textbf{188.75}&\textbf{1079.14}&\textbf{419.35} \\ \hline \end{tabular} \end{center} } \sol \frame{ \frametitle{7.3 Quantifying the relationship: Correlation} \noindent Then we have: \pause \begin{eqnarray*} \bar{x} &=& \frac{36.5}{10} \\ \pause & & \\ &=& 3.65 \qquad \textrm{\sf and} \\ \pause & & \\ \bar{y} &=& \frac{92.58}{10} \\ \pause & & \\ &=& 9.258. \end{eqnarray*} } \frame{ \frametitle{7.3 Quantifying the relationship: Correlation} \noindent We can now calculate $S_{xy}$, $S_{xx}$ and $S_{yy}$: \pause \begin{eqnarray*} S_{xy} &=& \left(\sum xy\right) - n \bar{x}\bar{y} \\ \pause &=& 419.35 - 10\times 3.65 \times 9.258 \\ \pause &=&81.433, \pause \end{eqnarray*} \begin{eqnarray*} S_{xx} &=&\left(\sum x^{2}\right) - n \bar{x}^{2} \\ \pause &=& 188.75 - 10 \times 3.65 \times 3.65 \\ \pause &=& 55.525 \qquad \textrm{\sf and} \end{eqnarray*} } \frame{ \frametitle{7.3 Quantifying the relationship: Correlation} \begin{eqnarray*} S_{yy} &=& \left(\sum y^{2}\right) - n \bar{y}^{2} \\ \pause &=& 1079.14 - 10 \times 9.258 \times 9.258 \\ \pause &=& 222.0344. \end{eqnarray*} } \frame{ \frametitle{7.3 Quantifying the relationship: Correlation} Thus, \begin{eqnarray*} r &=& \frac{S_{xy}}{\sqrt{S_{xx}\times S_{yy}}} \\ \pause & & \\ &=& \frac{81.433}{\sqrt{55.525 \times 222.0344}} \\ \pause & & \\ &=& \frac{81.433}{111.0336} \\ \pause & & \\ &=& 0.7334. \end{eqnarray*} } \normal \frame{ \frametitle{7.3 Quantifying the relationship: Correlation} Since this is fairly close to +1, we have a moderate/strong positive linear association between the age and price of wine. \pause \vspace{0.3cm} Remember that this correlation coefficient can only be used to detect \textcolor{purple}{\bf linear} associations. \pause \vspace{0.3cm} For information, the value of $r$ for the plots on page 169, from top--left and moving clockwise, is:\pause $r=1$, \pause$-0.899$, \pause $0.699$ and \pause $0.064$. \vspace{0.3cm} \pause Note there is clearly a relationship between $X$ and $Y$ in the bottom--right plot, but here $r=0.064$ which is very close to zero: this is because the relationship here is plainly non--linear. } \frame{ \frametitle{7.4 Modelling the relationship: linear regression} A correlation analysis may establish a linear relationship but it does not allow us to \textit{use} it to, say, predict the value of one variable given the value of another. \pause \vspace{0.3cm} \textcolor{purple}{\bf Regression analysis} allows us to do this and more. \pause \vsp Look at the scatter plot of the price of wine against the corresponding age of each bottle. } \frame{ \frametitle{7.4 Modelling the relationship: linear regression} %\begin{figure}[!h] \begin{center} \includegraphics[width=10cm,height=7cm]{wine1.ps} %\caption{Scatter plots showing different relationships} %\end{figure} \end{center} } \frame{ \frametitle{7.4 Modelling the relationship: linear regression} A ``line of best fit'' can be drawn through the data, and from this line we can make predictions of price based on age. \vsp \pause The problem is, everyone's line of best fit is bound to be slightly different, and so everyone's predictions will be slightly different! \vsp \pause The aim of regression analysis is to find the very best line which goes through the data in a completely objective way. \vsp \pause We do this through the \textcolor{purple}{\bf regression equation}. } \frame{ \frametitle{7.4 Modelling the relationship: linear regression} Recall from Chapter 1 that the equation of a straight line takes the general form \pause \[ y=mx +c, \] where $m$ is the \textcolor{purple}{\bf gradient} and $c$ is the \textcolor{purple}{\bf intercept}. \pause \pause \vsp Statisticians tend to use different notation for their \textcolor{purple}{\bf regression equation}: \pause \begin{eqnarray*} Y &=&\beta_{0} + \beta_{1} X + \epsilon, \end{eqnarray*} \pause where $Y$ is the \textcolor{purple}{\bf response} variable and $X$ the \textcolor{purple}{\bf predictor} variable, and $\epsilon$ (``epsilon'') is a ``random error'' term. } \frame{ \frametitle{7.4 Modelling the relationship: linear regression} The unknown parameters $\beta_{0}$ (``beta nought'') and $\beta_{1}$ (``beta one'') represent the intercept and slope of the population regression line $\beta_{0} + \beta_{1}X$. \pause \vsp Obviously, we need to find $\beta_{0}$ and $\beta_{1}$; the best values will minimise the vertical `gaps' between the regression line and the data. These `gaps' are known as the \textcolor{purple}{\bf residuals} -- see diagram. } \frame{ \frametitle{7.4 Modelling the relationship: linear regression} Each of the points $i$, $i=1, \ldots, n$, in our scatter plot has a $y$ co--ordinate $y_{i}$. \pause \vsp Recall from above that the corresponding $y$ co--ordinates of points on the line, say $\hat{y}_{i}$, are given by \[ \hat{y}_{i} = \beta_{0}+\beta_{1}x_{i}. \] \pause Thus, the vertical ``gaps'' between the points and the line are given by \begin{eqnarray*} y_{1} - \hat{y}_{1} &=&y_{1}-\beta_{0}-\beta_{1}x_{1} \\ y_{2} - \hat{y}_{2} &=&y_{2}-\beta_{0}-\beta_{1}x_{2} \\ &\vdots& \\ y_{n} - \hat{y}_{n} &=&y_{n}-\beta_{0}-\beta_{1}x_{n} \end{eqnarray*} } \frame{ \frametitle{7.4 Modelling the relationship: linear regression} Some of these ``gaps'' will be negative, as defined above, as some points will lie below the line. \pause \vsp To get rid of any ``negative gaps'', we square all of these quantities: \begin{eqnarray*} && (y_{1}-\beta_{0}-\beta_{1}x_{1})^{2} \\ \pause && (y_{2}-\beta_{0}-\beta_{1}x_{2})^{2} \\ \pause && \hspace{1cm} \vdots \\ && (y_{n}-\beta_{0}-\beta_{1}x_{n})^{2} \\ \end{eqnarray*} } \frame{ \frametitle{7.4 Modelling the relationship: linear regression} The very best line of best fit -- that is, the line which minimises the sum of these ``squared gaps'', is what we call the \textcolor{purple}{\bf regression line}. \pause \vsp So we want the regression line to minimise the quantity \[ \Delta(\beta_{0},\beta_{1}) = \sum_{i=1}^{n}(y_{i}-\beta_{0}-\beta_{1}x_{i})^{2}. \] And how do we minimise a function? We use calculus! So we solve \begin{eqnarray*} \frac{d}{d\beta_{0}}\Delta(\beta_{0},\beta_{1}) &=&0 \\ & &\\ \frac{d}{d\beta_{1}}\Delta(\beta_{0},\beta_{1}) &=&0 \end{eqnarray*} for $\beta_{0}$ and $\beta_{1}$. } \frame{ \frametitle{7.4 Modelling the relationship: linear regression} Doing so gives some very nice formulae: \begin{eqnarray*} \hat{\beta}_{1} &=& \frac{S_{xy}}{S_{xx}} \qquad \textrm{and} \\ && \\ \hat{\beta}_{0} &=& \bar{y}-\hat{\beta}_{1}\bar{x}, \end{eqnarray*} where the ``hat'' notation $\widehat{\textcolor{white}{L}}$ implies that we obtain \textcolor{purple}{\bf estimates} of the gradient and intercept from our sample data, and not the \textcolor{purple}{\bf true} values of these parameters. \pause \vsp Estimate the regression equation for the wine data, and superimpose this on the original scatter plot.} \sol \frame{ \frametitle{7.4 Modelling the relationship: linear regression} \noindent For the wine data, we have \pause \begin{eqnarray*} \hat{\beta}_{1} &=& \frac{S_{xy}}{S_{xx}} \\ \pause & & \\ &=& \frac{81.433}{55.525} \\ \pause & & \\ &=&1.467 \qquad \textrm{\sf and} \\ \pause & & \\ \hat{\beta}_{0} &=& \bar{y}-\hat{\beta}_{1}\bar{x} \\ \pause &=& 9.258 - 1.467 \times 3.65 \\ \pause &=& 3.903. \end{eqnarray*} } \frame{ \frametitle{7.4 Modelling the relationship: linear regression} \noindent Thus, the regression equation is \pause \begin{eqnarray*} Y &=& 3.903 + 1.467 X + \epsilon. \end{eqnarray*} \pause To plot this line (page 172), we just need to co--ordinates of two points. We know one of them: $(0,3.9)$. For another, choose any $x$--value -- say $x=8$: \[ Y = 3.903+1.467\times 8 = 15.639 \longrightarrow (8, 15.6). \] } \frame{ \frametitle{Modelling the relationship: simple linear regression} %\begin{figure}[!h] \begin{center} \includegraphics[width=10cm,height=7cm]{wine2.ps} %\caption{Scatter plot of price of wine (\pounds) against age (years), with regression line superimposed} %\end{figure} \end{center} } \normal \frame{ \frametitle{Modelling the relationship: simple linear regression} We can use the estimated regression equation to make predictions of wine price given a certain age. \vsp \pause For example, suppose we produce a bottle of wine that has been ageing for $4\frac{1}{2}$ years. How much should we sell it for? \vsp \pause Based on our regression equation, we could estimate a selling price per bottle as: \pause \begin{eqnarray*} Y &=& 3.903 + 1.467 \times 4.5 \\ \pause &=& 10.505, \end{eqnarray*} i.e. about \pounds 10.50. } \frame{ \frametitle{A cautionary note (I)} As far as possible, avoid \textcolor{purple}{\bf extrapolation}. \pause \vspl \begin{itemize} \item Only use the regression equation to make predictions using $X$--values that lie within the range of the data observed \pause \vspl \item So we should not use our regression equation to estimate the selling price of a bottle of wine that has been ageing for 12 years \pause \vspl \item We cannot be sure that this linear association will continue beyond the range of our data! \end{itemize} } \frame{ \frametitle{A cautionary note (II)} Also care should be taken not to read too much into the regression equation. \pause \vsp For example, consider sales of ice cream and sales of sun tan lotion. \pause \vspl \begin{itemize} \item In hot weather sales of ice cream increase and sales of sun tan lotion also increase \pause \vspl \item So ice cream sales may be a useful predictor of sun tan lotion sales \pause \vspl \item However, the act of buying an ice cream does not \textit{cause} someone to by some sun tan lotion... \pause \vspl \item ...What is happening is that both ice cream sales and sun tan lotion sales are directly influenced by a third factor: in this case, the weather. \end{itemize} } \frame{ \frametitle{7.5 Testing the strength of a correlation} In Section 7.3 we thought about how we can quantify the strength of (linear) association between a pair of variables $X$ and $Y$. \pause \vsp We then moved on, in Section 7.4, to think about how we can \textcolor{purple}{\bf model} this relationship through the simple linear regression model. \pause \vsp Surely, though, there is no point in estimating the regression equation if there is little, or no, linear association between $X$ and $Y$? \vsp That is, if the correlation coefficient is close zero, we have shown that the strength of linear association is negligible and so why would we then be interested in modelling this negligible/non--existent relationship? } \frame{ \frametitle{Example 7.1} \noindent The following table shows the total market value of 14 companies (in \pounds million) and the number of stock exchange transactions in that company's shares occurring on a particular day. Underneath, you are given some numerical summaries; the graph below shows a scatter plot for these data, as produced by \texttt{Minitab}. {\tiny \begin{center} \begin{tabular}{|c|c c c c c c c c c c c c c c|} \hline Market value &6.5 &5.2 &0.4 &1.7 &1.9 &2.4 &3.2 &4.7 &10.1 &12.5 &13.1 &5.5 &2.5 &1.5 \\ \hline Transactions &380 &200 &42 &50 &40 &78 &350 &18 &295 &190 &200 &55 &38 &20 \\ \hline \end{tabular} \end{center}} \normalsize \[ \sum_{i=1}^{14}x_{i} = 71.2 \qquad \sum_{i=1}^{14}y_{i} = 1956 \] \[ \sum_{i=1}^{n}x_{i}^{2} = 582.66 \qquad \sum_{i=1}^{14}y_{i}^{2} = 487166 \qquad \sum_{i=1}^{14}x_{i}y_{i} = 13481.6 \] } \frame{ \frametitle{Example 7.1} \begin{center} \includegraphics[width=11cm,height=6cm]{shares.ps} %\caption{Scatter plot showing market value versus number of stock exchange transactions for 14 companies} %\end{figure} \end{center} } \frame{ \frametitle{Example 7.1} \begin{itemize} \item [(a)] Find the sample correlation coefficient $r$, and comment. \vspl \item [(b)] Formally test the strength of correlation as suggested by your answer to part (a). \end{itemize} } \sol \frame{ \frametitle{Example 7.1(a): Solution} We have \begin{eqnarray*} S_{xx} &=& 582.66 - 14 \times \left(\frac{71.2}{14}\right)^{2} = \pause 220.55714 \\ \pause S_{yy} &=& 487166 -14 \times \left(\frac{1956}{14}\right)^{2} = \pause 213884.8571 \\ \pause S_{xy} &=&13481.6 - 14 \times \left(\frac{71.2}{14}\right)\times \left(\frac{1956}{14}\right) = \pause 3533.942857 \end{eqnarray*} \pause So \[ r = \frac{3533.942857}{\sqrt{220.55714 \times 213884.8571}} = 0.515. \]\pause \textbf{Comment:} Moderate, positive linear correlation. } \frame{ \frametitle{Example 7.1(b): Solution} We have \begin{eqnarray*} H_{0} &:& \rho = 0 \qquad \textsf{(i.e. no correlation in the population)} \pause \\ H_{1} &:& \rho \neq 0 \quad \textsf{ \hspace{0.3cm} (i.e. there {\it is} a correlation!)} \pause \end{eqnarray*} The test statistic is \[ t = r\sqrt{\frac{n-2}{1-r^{2}}} \pause = 0.515\sqrt{\frac{12}{1-0.515^{2}}}=\pause 2.08. \] \pause \vspl To get the $p$--value, we use $t$ tables (see \textcolor{purple}{\bf Page 156}, Chapter 6) with $\nu=n-2 = 14-2 = 12$, giving: \pause \vspl \begin{center} \begin{tabular}{|c|c c c|} \hline $p$--value &10\% &5\% &1\% \\ \hline Critical value &1.782 &2.179 &3.055 \\ \hline \end{tabular} \end{center} \vspl So here, $p$ lies between 5\% and 10\%. } \frame{ \frametitle{Example 7.1(b): Solution} Thus \pause \vsp \begin{itemize} \item There is \textcolor{purple}{\bf slight} evidence against $H_{0}$ \vspl \item We \textcolor{purple}{\bf retain } $H_{0}$ \vspl \item The correlation observed in the sample is not strong enough to suggest a true correlation between $X$ and $Y$ in the population \end{itemize} } \normal \frame{ \frametitle{7.6 Multiple linear regression} \noindent In this section we will show how the linear regression model can be \textcolor{purple}{\bf extended} to include any number of predictor variables. \pause \vsp \noindent The model we have considered so far, namely \begin{eqnarray*} Y &=& \beta_{0} + \beta_{1}X + \epsilon, \end{eqnarray*} has been, and is often, referred to as the \textcolor{purple}{\bf simple} linear regression model, because it only involves a single predictor variable. \pause \vsp However, frequently two or more predictor variables may be useful together to predict $Y$. } \frame{ \frametitle{7.6 Multiple linear regression} For instance, the sales of a product may depend on \pause \vspl \begin{itemize} \item the product's unit price, \textbf{and} \pause \vspl \item the amount of advertising expenditure \textbf{and}\pause \vspl \item the price of a competing product \end{itemize} \pause \vspl The simple linear regression model can be extended to include any number of predictor $X$ variables, in which case it is called the \textcolor{purple}{\bf multiple linear regression model}. } \frame{ \frametitle{7.6.1 Back to the simple linear regression model} Recall the example relating age of wine ($X$ years) to price of wine ($Y$ pounds). \pause \vspl By hand, we found the equation of our regression line (the very best line of best fit!) to be \pause \[ Y = 3.903+1.467X+\epsilon. \] \pause Is \textcolor{purple}{\bf Age} an important predictor of \textcolor{purple}{\bf price}? } \frame{ \frametitle{7.6.1 Back to the simple linear regression model} We can attempt to answer this by testing \pause \begin{eqnarray*} H_{0} &:& \beta_{1} = 0 \qquad \textsf{versus} \\ H_{1} &:& \beta_{1} \neq 0. \end{eqnarray*} \pause We will use \texttt{Minitab} here! } \frame{ \frametitle{7.6.1 Back to the simple linear regression model} Notice that $p=1.6\%$, and so \pause \vspl \begin{itemize} \item We have \textcolor{purple}{\bf moderate} evidence against $H_{0}$ \pause \vspl \item We should \textcolor{purple}{\bf reject} $H_{0}$ in favour of $H_{1}$ \pause \vspl \item $\beta_{1}$ is significantly different from zero, and so Age \textbf{is} an important predictor of price! \end{itemize} } \frame{ \frametitle{7.6.2 Extending the simple linear regression model} \begin{center} %\begin{table}[!h] %\centering {\tiny \begin{tabular}{|c|c c c c c c c c c c|} \hline Bottle &1 &2 &3 &4 &5 &6 &7 &8 &9 &10 \\ \hline Price (\pounds$Y$) &4.50 &12.95 &6.50 &4.99 &7.50 &14.95 &8.25 &3.95 &18.99 &10.00 \\ Age ($X_{1}$ years) &$3\frac{1}{2}$ &5&3 &$2\frac{1}{2}$&3&2 &$2\frac{1}{2}$&1&10 &4 \\ Rainfall ($X_{2}$ mm) &126 &121 &125 &106 &107 &112 &124 &105 &116 &108 \\ Temp. ($X_{3}^{\textrm{o}}$C)&16&20&17 &18 &18 &22 &19 &15 &21 &20 \\ \hline \end{tabular}} %\caption{Price of bottles of wine, with corresponding Age, total rainfall and average temperature during growing season} %\end{table} \end{center}} \frame{ \frametitle{7.6.2 Extending the simple linear regression model} A multiple linear regression model that may be suitable simply extends on the simple linear regression model: \begin{eqnarray*} Y &=& \beta_{0} + \beta_{1}X_{1} \pause + \beta_{2}X_{2} + \beta_{3}X_{3} + \epsilon; \end{eqnarray*} \pause as before, $\epsilon$ is our ``random error'' term, and $\beta_{0}$, $\beta_{1}$, $\beta_{2}$ and $\beta_{3}$ are parameters in the model that we need to estimate.} \frame{ \frametitle{7.6.2 Extending the simple linear regression model} \noindent So how do we find $\hat{\beta}_{0}$, $\hat{\beta}_{1}$, $\hat{\beta}_{2}$ and $\hat{\beta}_{3}$ -- the estimated parameters of the model? \pause \vsp We \textit{can} compute these by hand, as we did in Section 7.4 for the simple linear regression model, but this requires knowledge of matrix algebra which many of you won't have (even if you did A Level maths!). \pause \vsp Anyway, \texttt{Minitab} can perform the calculations for us! } \frame{ \frametitle{7.6.2 Extending the simple linear regression model} Thus, the full (multiple) regression model is: \begin{eqnarray*} Y &=&-22.5 +0.807X_{1}-0.0004X_{2}+1.55X_{3} + \epsilon, \end{eqnarray*} where $X_{1}$, $X_{2}$ and $X_{3}$ represent the age of a bottle of wine, the total rainfall during the growing season and the corresponding average afternoon temperature (respectively). } \frame{ \frametitle{7.6.2 Extending the simple linear regression model} The estimated coefficients of the model indicate the \textcolor{purple}{\bf direction} of the relationship between the price of a bottle of wine and each of the corresponding predictors. \pause \vsp For example: \begin{itemize} \item Since $\hat{\beta}_{1}=0.807$ is positive -- positive relationship between age and price (i.e. generally, older wines are more expensive); \pause \item Since $\hat{\beta}_{2}=-0.0004$ is negative -- negative relationship between rainfall and price (i.e. generally, wines from regions with higher rainfall are cheaper); \pause \item Since $\hat{\beta}_{3}=1.55$ is positive -- positive relationship between temperature and price (i.e. generally, wines from regions with higher temperatures are more expensive). \end{itemize}} \frame{ \frametitle{7.6.3 Testing the importance of our predictor variables} \noindent Recall Section 7.6.1, where we used \texttt{Minitab} to test the significance of the parameter $\beta_{1}$ in our model. \pause \vsp The null hypothesis here was $H_{0}: \beta_{1}=0$; retention of this hypothesis would imply that the predictor variable attached to this parameter (in Section 7.6.1 this was ``Age'') is not an important predictor of the response variable (Price). \pause \vsp The output from \texttt{Minitab} for our multiple linear regression, which also uses rainfall and temperature as predictors, is shown at the top of page 182 and can be used in a similar way. } \frame{ \frametitle{7.6.3 Testing the importance of Age} \noindent For example, let us once again consider the importance of \textbf{Age} in our model: \pause \begin{eqnarray*} H_{0} &:& \beta_{1}=0 \qquad \textrm{\sf versus} \\ H_{1} &:& \beta_{1} \neq 0. \pause \end{eqnarray*} \noindent The $p$--value for this is 3\%. Thus, \textit{age appears to be important in our model}. } \frame{ \frametitle{7.6.3 Testing the importance of Rainfall} \noindent \textbf{Rainfall} is variable $X_{2}$, which has coefficient $\beta_{2}$. Our hypotheses are: \begin{eqnarray*} H_{0} &:& \beta_{2}=0 \qquad \textrm{\sf versus} \\ H_{1} &:& \beta_{2} \neq 0. \end{eqnarray*} \pause \noindent The $p$--value for this is 99.6\%. Thus, \textit{rainfall is NOT important in our model}. } \frame{ \frametitle{7.6.3 Testing the importance of Temperature} \noindent \textbf{Temperature} is variable $X_{3}$, which has coefficient $\beta_{3}$. Our hypotheses are: \pause \begin{eqnarray*} H_{0} &:& \beta_{3}=0 \qquad \textrm{\sf versus} \\ H_{1} &:& \beta_{3} \neq 0. \end{eqnarray*} \noindent The $p$--value for this is 0.2\%. Thus, \textit{temperature appears to be important in our model}. } \frame{ \frametitle{7.6.3 Testing the importance of our predictor variables} Since rainfall is not an important linear predictor in our model, we should now remove it and re--fit the model using only age and temperature. \pause \vsp In \texttt{Minitab}, we perform the regression again, but this time include only age and temperature as predictor variables. } \frame{ \frametitle{7.6.3 Testing the importance of our predictor variables} \noindent Notice that the regression equation has now changed, and now only includes age and temperature. \pause \vsp We now have: \begin{eqnarray*} Y &=& -22.6 + 0.806 X_{1} + 1.55 X_{3} + \epsilon, \end{eqnarray*} where $X_{1}$ represents the age of a bottle of wine and $X_{3}$ represents the average temperature during the growing season. \pause \vsp Notice that the $p$--values for both age and temperature are still less than 0.05, so performing a hypothesis test for both would conclude that both are important in the model. } \frame{ \frametitle{7.6.3 Testing the importance of our predictor variables} \noindent The regression equation above represents our ``\textcolor{purple}{\bf final}'' model, in that we have excluded all variables that are not important predictors of price, and the model now includes only those predictors that \textit{are} important. \pause \vsp We could now use this model to make predictions. } \frame{ \frametitle{7.6.3 Testing the importance of our predictor variables} For example, suppose you run a vineyard and have just produced a 7 year--old vintage wine. \pause \vsp During the growing season, the average afternoon temperature was $18.5^{\textsf{o}}$C and the total amount of rainfall was 117mm. How much, per bottle, might this wine sell for? } \sol \frame{ \frametitle{7.6.3 Testing the importance of our predictor variables} We have $X_{1}=7$, $X_{2}=117$ and $X_{3}=18.5$. Thus \pause \[ Y = -22.6 + 0.806\times 7 + 1.55 \times 18.5 \pause = 11.717, \] i.e. about \pounds 11.72 per bottle. } \normal \frame{ \frametitle{7.6.4 The $R^{2}$ statistic} In the \texttt{Minitab} output shown in these lecture notes so far, you may have noticed something called \texttt{R-Sq}. \pause \vsp Each time you perform a regression analysis in \texttt{Minitab}, the output includes the value of the $R^{2}$ statistic, and this is sometimes used as an overall assessment of the quality of our model. \pause \vsp Technically, the $R^{2}$ statistic tells us how much of the variation in our $Y$ data is explained by the predictors in the model. } \frame{ \frametitle{7.6.4 The $R^{2}$ statistic} For simplicity, suppose you have the simple linear regression model \[ Y= \beta_{0}+\beta_{1}X +\epsilon. \] \pause If we observe a perfect relationship between $X$ and $Y$, -- i.e. all of our points lie perfectly on a straight line -- then if we know $X$, we know $Y$! \pause \vsp The $Y$ value could just be ``read off'' from the regression line. \pause \vsp In this case, we would say that $X$ explains 100\% of the variation in $Y$, and the corresponding value of the $R^2$ statistic would be 100\%. } \frame{ \frametitle{7.6.4 The $R^{2}$ statistic} Thus, the closer the value of the $R^2$ statistic to 100\%, the better the model. \pause \vsp We can compare the $R^2$ statistic for the various fits we have tried out for the wine data: \begin{center} \begin{tabular}{l| c} Model &$R^2$ \\ \hline Simple -- including just ``Age'' (page 180) &53.8\% \\ Multiple -- including ``Age'', ``Rain'' and ``Temp'' (page 182) &91.2\% \\ Multiple -- including ``Age'' and ``Temp'' (page 185) &91.2\% \\\hline \end{tabular} \end{center} } \normal \frame{ \frametitle{Example 7.3} \noindent On a small island the government would like to be able to predict the number of mortgage loans issued by the state mortgage company ($Y$) from: \pause \vspl \begin{itemize} \item The amount of \textcolor{purple}{\bf personal income} in millions of local currency ($X_{1}$) \pause \vspl \item The \textcolor{purple}{\bf interest rate} ($X_{2}$)\pause \vspl \item The \textcolor{purple}{\bf year} ($X_{3}$) \end{itemize} } \sol \frame{ \frametitle{Example 7.3: Solution} \begin{itemize} \item [\textbf{1.}] Write down the ``full'' multiple linear regression model that might be suitable for this scenario. \pause \vsp \[ Y = \beta_{0}+\beta_{1}X_{1}+\beta_{2}X_{2}+\beta_{3}X_{3} + \epsilon \] \end{itemize} } \frame{ \frametitle{Example 7.3: Solution} \begin{itemize} \item [\textbf{2.}] Use \texttt{Minitab} to fit the model in (\textbf{1}). Summarise the regression output in the space below, making sure you write down the fitted model. \pause \vsp We get: \[ Y = 741022+6772X_{1}-1684X_{2}-372X_{3} + \epsilon \] \pause Notice also that $R^{2} = 94.4\%$. \end{itemize} } \frame{ \frametitle{Example 7.3: Solution} \begin{itemize} \item [\textbf{3.}] Which, if any, of the variables \textit{do not appear to be important predictors} in your model? \pause \vsp All $p$--values for the predictors are above 0.05 -- suggesting that we would \textit{retain} all null hypotheses: \pause \begin{eqnarray*} H_{0} &:& \beta_{1}=0 \\ H_{0} &:&\beta_{2}=0 \\ H_{0} &:&\beta_{3}=0 \pause \end{eqnarray*} and so all three predictors would drop out! However, we only remove one variable at a time -- here, ``Year'', as it's $p$--value is the largest. \end{itemize} } \frame{ \frametitle{Example 7.3: Solution} \begin{itemize} \item [\textbf{4.}] Remove the least important variable from your model, and re--fit using \texttt{Minitab}. Again, summarise the regression output in the space below, making sure you write down the fitted model. \pause \vsp Removing $X_{3}$ and re-fitting, gives: \pause \[ Y = 6582+3932X_{1}-1850X_{2}+\epsilon. \] Notice that now, $R^{2} = 94.2\%$. \end{itemize} } \frame{ \frametitle{Example 7.3: Solution} \begin{itemize} \item [\textbf{5.}] Repeat step 4 until you have removed all variables that do not appear to be important predictors in your model. Write down your ``final'' regression equation in the space below: \pause \vsp Our \textcolor{purple}{\bf final} model includes ``Interest'' only, giving: \pause \[ Y = 30439-3234X_{2} + \epsilon. \] \pause Notice that now, $R^{2} = 90.9\%$. \end{itemize} } \frame{ \frametitle{Example 7.3: Solution} \begin{itemize} \item [\textbf{6.}] Comment on the $R^{2}$ statistic in your regression analyses. \pause \vsp The $R^{2}$ Statistic has changed from \pause \[ 94.4\% \longrightarrow 94.2\% \longrightarrow 90.9\%. \] Still quite high, even though we've removed two variables! \end{itemize} } \frame{ \frametitle{Example 7.3: Solution} \begin{itemize} \item [\textbf{7.}] This year (2014), the average income is forecast to rise to 4.9 million units of the local currency and the interest rate is set to rise to 5\%. How many mortgages can the state mortgage company expect to issue this year? \pause \vsp So we have $X_{1}= 4.9$, $X_{2}=5$ and $X_{3}=2014$. So\pause \[ Y = 30439-3234X_{2} = \pause 30439-(3234\times 5) = 14,269\textsf{ mortgages}. \] \pause Notice we didn't need $X_{1}$ nor $X_{3}$! \end{itemize} } \end{document} \frame{ \frametitle{Example 7.3: Solution} We remove ``Year'' first, as it is least significant (it's $p$--values is the greatest). \pause \vsp Then we re--fit! } \frame{ \frametitle{Example 7.3: Solution} } \end{document} %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%