% Use Beamer to produce an outline-style document instead of slides.
% \documentclass[serif]{beamer} % Serif means Computer Modern math font.
% \hypersetup{colorlinks,linkcolor=,urlcolor=red}

%  To create handout using article mode: Comment above and uncomment below (2 places)
\documentclass[12pt]{article}
\usepackage{beamerarticle}
\usepackage{comment}
\usepackage[colorlinks=true, pdfstartview=FitV, linkcolor=blue, citecolor=blue, urlcolor=red]{hyperref} % For live Web links with href in article mode
\usepackage{fullpage}
 
\usefonttheme{serif} % Looks like Computer Modern for non-math text -- nice!
\setbeamertemplate{navigation symbols}{} % Supress navigation symbols
% \usetheme{Berlin} % Displays sections on top
\usepackage[english]{babel}
% \definecolor{links}{HTML}{2A1B81}
% \definecolor{links}{red}
\setbeamertemplate{footline}[frame number] 

\mode<presentation>
% \mode<handout>{\setbeamercolor{background canvas}{bg=black!5}}


\title{Categorical Data Analysis\footnote{See last slide for copyright information.}}
\subtitle{STA 312: Fall 2022}
\date{} % To suppress date


\begin{document}

% More material required for handout in article mode
\title{Categorical Data Analysis\footnote{See last page for copyright information.}}
\subtitle{STA 312: Fall 2022}
\date{} % To suppress date
\maketitle


\begin{frame}
  \titlepage
\end{frame}


\begin{frame}
\frametitle{Variables and Cases}
%\framesubtitle{Optional sub-title}

  \begin{itemize}
    \item There are $n$ \textbf{cases} (people, rats, factories, wolf packs) in a data set.
    \item A \textbf{variable} is a characteristic or piece of information that can be recorded for each case in the data set. 
    \item For example cases could be patients in a hospital, and variables could be Age, Sex, Diagnosis, Have family doctor (Yes-No), Family history of heart disease (Yes-No), etc.
  \end{itemize}
\end{frame}

\begin{frame}{Variables can be Categorical, or Continuous}
  \begin{itemize}
    \item Categorical: Gender, Diagnosis, Job category, Have family doctor, Family history of heart disease, 5-year survival (Y-N)
    \item Some categories are ordered (birth order, health status)
    \item Continuous: Height, Weight, Blood pressure
    \item Some questions:
        \begin{itemize} 
            \item Are all normally distributed variables continuous?
            \item Are all continuous variables quantitative?
            \item Are all quantitative variables continuous?
            \item Are there really any data sets with continuous variables?   
        \end{itemize} 
  \end{itemize}
\end{frame}

\begin{frame}{Variables can be Explanatory, or Response}
  \begin{itemize}
    \item Explanatory variables are sometimes called ``independent variables." 
    \item The $x$ variables in regression are explanatory variables.
    \item Response variables are sometimes called ``dependent variables."
    \item The $Y$ variable in regression is the response variable.
    \item Sometimes the distinction is not useful: Does each twin get cancer, Yes or No?
  \end{itemize}
\end{frame}

\begin{frame}{Our main interest is in categorical variables}
  \begin{itemize}
    \item Especially categorical response variables
    \item In ordinary regression, outcomes are normally distributed, and so continuous.
    \item But often, outcomes of interest are categorical
        \begin{itemize} 
            \item Buy the product, or not
            \item Marital status 5 years after graduation
            \item Survive the operation, or not.
        \end{itemize} 
    \item Ordered categorical response variables, too: for example highest level of hockey ever played.
  \end{itemize}
\end{frame}

\begin{frame}{Distributions}{We will mostly use}

  \begin{itemize}
    \item Bernoulli
    \item Binomial
    \item Multinomial 
    \item Poisson
  \end{itemize}
\end{frame}


\begin{frame}{The Poisson process}{Why the Poisson distribution is such a useful model for count data}

  \begin{itemize}
    \item Events happening randomly in space or time
    \item Independent increments
    \item  For a small region or interval,
        \begin{itemize}
            \item Chance of 2 or more events is negligible
            \item Chance of an event roughly proportional to the size of the region or interval
        \end{itemize}
    \item Then (solve a system of differential equations), the probability of observing $x$ events in a region of size $t$ is 
\begin{displaymath}
    \frac{e^{-\lambda t}(\lambda t)^x }{x!} \mbox{ for } x=0,1,\ldots
\end{displaymath}
  \end{itemize}
\end{frame}

\begin{frame}{Poisson process examples}{Some variables that have a Poisson distribution}

  \begin{itemize}
    \item Calls coming in to an emergency number
    \item Customers arriving in a given time period
    \item Number of raisins in a loaf of raisin bread
    \item Number of bomb craters in a region after a bombing raid, London WWII
    \item In a jar of peanut butter \ldots
  \end{itemize}
\end{frame}


\begin{frame}{Steps in the process of statistical analysis}
{One possible approach}

  \begin{itemize}
    \item Consider a fairly realistic example or problem
    \item Decide on a statistical model
    \item Perhaps decide sample size
    \item Acquire data
    \item Examine and clean the data; generate displays and descriptive statistics
    \item Estimate parameters, perhaps by maximum likelihood
    \item Carry out tests, compute confidence intervals, or both
    \item Perhaps re-consider the model and go back to estimation
    \item Based on the results of inference, draw conclusions about the example or problem
  \end{itemize}
\end{frame}

\begin{frame}{Coffee taste test}
A fast food chain is considering a change in the blend of coffee beans they use to make their coffee. To determine whether their customers prefer the new blend, the company plans to select a random sample of $n=100$ coffee-drinking customers and ask them to taste coffee made with the new blend and with the old blend, in cups marked ``$A$" and ``$B$." Half the time the new blend will be in cup $A$, and half the time it will be in cup $B$. Management wants to know if there is a difference in preference for the two blends.
\end{frame}

\begin{frame}{Statistical model}
Letting $\pi$ denote the probability that a consumer will choose the new blend, treat the data $Y_1, \ldots, Y_n$ as a random sample from a Bernoulli distribution. That is, independently for $i=1, \ldots, n$,
\begin{displaymath}
    P(y_i|\pi) = \pi^{y_i} (1-\pi)^{1-y_i}
\end{displaymath}
for $y_i=0$ or $y_i=1$, and zero otherwise.
% The conditional probability notation is not in the book (I believe).

\vspace{10mm}

Note that $Y = \sum_{i=1}^n Y_i$ is the number of consumers who choose the new blend. Because $Y \sim B(n,\pi)$, the whole experiment could also be treated as a single observation from a Binomial.

\end{frame}

\begin{frame}{Find the MLE of $\pi$}{Show your work}

Maximize the log likelihood.
\begin{eqnarray*}
    \frac{\partial}{\partial\pi} \log \ell 
    & = & \frac{\partial}{\partial\pi} \log\left(\prod_{i=1}^n P(y_i|\pi) \right) \\
    & = & \frac{\partial}{\partial\pi} \log\left(\prod_{i=1}^n \pi^{y_i} (1-\pi)^{1-y_i} \right) \\
    & = & \frac{\partial}{\partial\pi} \log\left(\pi^{\sum_{i=1}^n y_i} 
                                       (1-\pi)^{n-\sum_{i=1}^n y_i}\right) \\
    & = & \frac{\partial}{\partial\pi}\left((\sum_{i=1}^n y_i)\log\pi + 
          (n-\sum_{i=1}^n y_i)\log (1-\pi) \right) \\
    & = & \frac{\sum_{i=1}^n y_i}{\pi} - \frac{n-\sum_{i=1}^n y_i}{1-\pi}
\end{eqnarray*}
\end{frame}

\begin{frame}{Setting the derivative to zero,}
\begin{eqnarray*}
\frac{\sum_{i=1}^n y_i}{\pi} = \frac{n-\sum_{i=1}^n y_i}{1-\pi}  
    & \Rightarrow & (1-\pi)\sum_{i=1}^n y_i = \pi (n-\sum_{i=1}^n y_i) \\
    & \Rightarrow & \sum_{i=1}^n y_i-\pi\sum_{i=1}^n y_i = n\pi - \pi\sum_{i=1}^n y_i \\
    & \Rightarrow & \sum_{i=1}^n y_i = n\pi  \\
    & \Rightarrow & \pi = \frac{\sum_{i=1}^n y_i}{n} = \overline{y} = p
\end{eqnarray*}
So it looks like the MLE is the sample proportion. Carrying out the second derivative test to be sure,
\end{frame}

\begin{frame}{Second derivative test}
\begin{eqnarray*}
    \frac{\partial^2\log \ell}{\partial\pi^2}  
    & = & \frac{\partial}{\partial\pi} \left(\frac{\sum_{i=1}^n y_i}{\pi} 
                                     - \frac{n-\sum_{i=1}^n y_i}{1-\pi} \right) \\
    & = & \frac{-\sum_{i=1}^n y_i}{\pi^2} --- \frac{n-\sum_{i=1}^n y_i}{(1-\pi)^2} \\
    & = & -n\left(\frac{1-\overline{y}}{(1-\pi)^2} + \frac{\overline{y}}{\pi^2} \right) < 0 \\
\end{eqnarray*}

\vspace{5mm}

Concave down, maximum, and $\widehat{\pi}=\overline{y}=p$.
\end{frame}



\begin{frame}[fragile]{Numerical estimate} % Note use of fragile to make verbatim work.
Suppose 60 of the 100 consumers prefer the new blend. Give a point estimate the parameter $\pi$. Your answer is a number.

\vspace{10mm}

\begin{verbatim}
> p = 60/100; p
[1] 0.6
\end{verbatim}
%\verb:f(x):
\end{frame}

\begin{frame}
\frametitle{Carry out a test to answer the question}
\framesubtitle{Is there a difference in preference for the two blends?}
{Start by stating the null hypothesis}

  \begin{itemize}
    \item $H_0: \pi=0.50$
    \item $H_1: \pi \neq 0.50$
    \item A case could be made for a one-sided test, but we'll stick with two-sided.
    \item $\alpha=0.05$ as usual.
    \item Central Limit Theorem says $\widehat{\pi}=\overline{Y}$ is approximately normal with mean $\pi$ and variance $\frac{\pi(1-\pi)}{n}$.
  \end{itemize}
\end{frame}

\begin{frame}[fragile]{Several valid test statistics for $H_0: \pi=\pi_0$ are available}
{Two of them are} % Which one do you like more? Why?

\begin{displaymath}
    Z_1 = \frac{\sqrt{n}(p-\pi_0)}{\sqrt{\pi_0(1-\pi_0)}}
\end{displaymath}

and

\begin{displaymath}
    Z_2 = \frac{\sqrt{n}(p-\pi_0)}{\sqrt{p(1-p)}}
\end{displaymath}

\vspace{10mm}

What is the critical value? Your answer is a number.

\begin{verbatim}
> alpha = 0.05
> qnorm(1-alpha/2)
[1] 1.959964
\end{verbatim}
\end{frame}

\begin{frame}[fragile]
\frametitle{Calculate the test statistic(s)}
\framesubtitle{and the $p$-value(s)}

\begin{verbatim}
> pi0 = .5; p = .6; n = 100
> Z1 = sqrt(n)*(p-pi0)/sqrt(pi0*(1-pi0)); Z1
[1] 2
> pval1 = 2 * (1-pnorm(Z1)); pval1
[1] 0.04550026
> 
> Z2 = sqrt(n)*(p-pi0)/sqrt(p*(1-p)); Z2
[1] 2.041241
> pval2 = 2 * (1-pnorm(Z2)); pval2
[1] 0.04122683
\end{verbatim}
\end{frame}

\begin{frame}
\frametitle{Conclusions}
%\framesubtitle{In symbols and words: Words are more important}

  \begin{itemize}
    \item Do you reject $H_0$? \emph{Yes, just barely.}
    \item Isn't the $\alpha=0.05$ significance level pretty arbitrary? \emph{Yes, but if people insist on a Yes or No answer, this is what you give them.}
    \item What do you conclude, in symbols? $\pi \neq 0.50$. \emph{Specifically,} $\pi > 0.50$.
    \item What do you conclude, in plain language? Your answer is a statement about coffee. \emph{More consumers prefer the new blend of coffee beans.}
    \item Can you really draw directional conclusions when all you did was reject a non-directional null hypothesis? \emph{Yes. Decompose the two-sided size $\alpha$ test into two one-sided tests of size $\alpha/2$. This approach works in general.}
  \end{itemize}
It is very important to state directional conclusions, and state them clearly in terms of the subject matter. \textbf{Say what happened!} If you are asked state the conclusion in plain language, your answer \emph{must} be free of statistical mumbo-jumbo.
\end{frame}

\begin{frame}
\frametitle{What about negative conclusions?}
\framesubtitle{What would you say if $Z=1.84$?}

Here are two possibilities, in plain languages.

  \begin{itemize}
    \item ``This study does not provide clear evidence  that consumers prefer one blend of coffee beans over the other."
    \item ``The results are consistent with no difference in preference for the two coffee bean blends."
  \end{itemize}

\vspace{5mm}
  
In this course, we will not just casually \emph{accept} the null hypothesis. 
% We are taking the side of Fisher over Neyman and Pearson in an old and very nasty theoretical dispute. 
\end{frame}

\begin{frame}
\frametitle{Confidence Intervals}
\framesubtitle{Approximately for large $n$,}

\begin{eqnarray*}
    1-\alpha & \approx & Pr\{ -z_{\alpha/2} < Z_2 < z_{\alpha/2} \} \\
             & =  &    Pr\left\{ -z_{\alpha/2} < \frac{\sqrt{n}(p-\pi)}{\sqrt{p(1-p)}} 
                      < z_{\alpha/2} \right\} \\
             & = &    Pr\left\{ p - z_{\alpha/2}\sqrt{\frac{p(1-p)}{n}} < \pi <
                                p +  z_{\alpha/2}\sqrt{\frac{p(1-p)}{n}} \right\} 
\end{eqnarray*}

  \begin{itemize}
    \item Could express this as $p \pm  z_{\alpha/2}\sqrt{\frac{p(1-p)}{n}}$
    \item $z_{\alpha/2}\sqrt{\frac{p(1-p)}{n}}$ is sometimes called the \emph{margin of error}.
    \item If $\alpha=0.05$, it's the 95\% margin of error.
  \end{itemize}

\end{frame}

\begin{frame}
\frametitle{Give a 95\% confidence interval for the taste test data.}
\framesubtitle{The answer is a pair of numbers. Show some work.}

\begin{eqnarray*}
      & & \left(p - z_{\alpha/2}\sqrt{\frac{p(1-p)}{n}} ~~,~~
                p + z_{\alpha/2}\sqrt{\frac{p(1-p)}{n}} \right) \\
      & &                                                       \\
      &=&  \left(0.60 - 1.96\sqrt{\frac{0.6\times 0.4}{100}} ~~,~~
                 0.60 + 1.96\sqrt{\frac{0.6\times 0.4}{100}} \right) \\
      & &                                                       \\
      &=&  (0.504,0.696)
\end{eqnarray*}

In a report, you could say 
  \begin{itemize}
    \item The estimated proportion preferring the new coffee bean blend is $0.60 \pm 0.096$, or
    \item ``Sixty percent of consumers preferred the new blend. These results are expected to be accurate within 10 percentage points, 19 times out of 20."
  \end{itemize}
\end{frame}

\begin{frame}
\frametitle{Meaning of the confidence interval}
  \begin{itemize}
    \item We calculated a 95\% confidence interval of $(0.504,0.696)$ for $\pi$.
    \item Does this mean $Pr\{ 0.504 < \pi < 0.696 \}=0.95$?
    \item No! The quantities $0.504$, $0.696$ and  $\pi$ are all constants, so $Pr\{ 0.504 < \pi < 0.696 \}$ is either zero or one.  
    \item The endpoints of the confidence interval are random variables, and the numbers $0.504$ and $0.696$ are \emph{realizations} of those random variables, arising from a particular random sample.
    \item Meaning of the probability statement: If we were to calculate an interval in this manner for a large number of random samples, the interval would contain the true parameter around $95\%$ of the time. 
    \item So we sometimes say that we are ``$95\%$ confident" that $0.504 < \pi < 0.696$.
  \end{itemize}
\end{frame}

\begin{frame}
\frametitle{Confidence intervals (regions) correspond to tests}
\framesubtitle{Recall $Z_1 = \frac{\sqrt{n}(p-\pi_0)}{\sqrt{\pi_0(1-\pi_0)}}$ and $Z_2 = \frac{\sqrt{n}(p-\pi_0)}{\sqrt{p(1-p)}}$.}

From the derivation of the confidence interval, 

\begin{displaymath}
    -z_{\alpha/2} < Z_2 < z_{\alpha/2}
\end{displaymath}
if and only if
\begin{displaymath}
p - z_{\alpha/2}\sqrt{\frac{p(1-p)}{n}} < \pi_0 < p +  z_{\alpha/2}\sqrt{\frac{p(1-p)}{n}}
\end{displaymath}


  \begin{itemize}
    \item So the confidence interval consists of those parameter values $\pi_0$ for which $H_0: \pi=\pi_0$ is \emph{not} rejected.
    \item That is, the null hypothesis is rejected at significance level $\alpha$ if and only if the value given by the null hypothesis is outside the $(1-\alpha)\times 100\%$ confidence interval.
    \item There is a confidence interval corresponding to $Z_1$ too. Maybe it's better -- See Chapter~One. 
    \item In general, any test can be inverted to obtain a confidence region.
  \end{itemize}

\end{frame}

\begin{comment}

Cutting out power in the whole course!

\begin{frame}
\frametitle{Selecting sample size}

  \begin{itemize}
    \item Where did that $n=100$ come from?
    \item Probably off the top of someone's head.
    \item We can (and should) be more systematic.
    \item Sample size can be selected
         \begin{itemize}
             \item To achieve a desired margin of error
             \item To achieve a desired statistical power
             \item In other reasonable ways
         \end{itemize}
  \end{itemize}
\end{frame}

\begin{frame}
\frametitle{Power}

The power of a test is the probability of rejecting $H_0$ when $H_0$ is false.

  \begin{itemize}
    \item More power is good.
    \item Power is not just one number. It is a \emph{function} of the parameter. 
    \item Usually, 
         \begin{itemize}
             \item For any $n$, the more incorrect $H_0$ is, the greater the power.
             \item For any parameter value satisfying the alternative hypothesis, the larger $n$ is, the greater the power.
         \end{itemize}
  \end{itemize}
\end{frame}

\begin{frame}
\frametitle{Statistical power analysis}
\framesubtitle{To select sample size}

  \begin{itemize}
    \item Pick an effect you'd like to be able to detect -- a parameter value such that $H_0$ is false.  It should be just over the boundary of interesting and meaningful.
    \item Pick a desired power, a probability with which you'd like to be able to detect the effect by rejecting the null hypothesis.
    \item Start with a fairly small $n$ and calculate the power. Increase the sample size until the desired power is reached.
  \end{itemize}

\vspace{5mm}
There are two main issues.
\vspace{5mm}

  \begin{itemize}
    \item What is an ``interesting" or ``meaningful" parameter value?
    \item How do you calculate the probability of rejecting $H_0$?
  \end{itemize}

\end{frame}

\begin{frame}
\frametitle{Calculating power for the test of a single proportion}
\framesubtitle{True parameter value is $\pi$}

\begin{displaymath}
    Z_1 = \frac{\sqrt{n}(p-\pi_0)}{\sqrt{\pi_0(1-\pi_0)}}
\end{displaymath}

{\tiny
\begin{eqnarray*}
    \mbox{Power} &=& 1-Pr\{-z_{\alpha/2} < Z_1 < z_{\alpha/2} \} \\
                 &=& 1-Pr\left\{-z_{\alpha/2} < \frac{\sqrt{n}(p-\pi_0)}{\sqrt{\pi_0(1-\pi_0)}} 
                    < z_{\alpha/2} \right\} \\
                 &=& \ldots \\
                 &=& 1-Pr\left\{\frac{\sqrt{n}(\pi_0-\pi)}{\sqrt{\pi(1-\pi)}} -  z_{\alpha/2}\sqrt{\frac{\pi_0(1-\pi_0)}{\pi(1-\pi)}} 
< {\color{red} \frac{\sqrt{n}(p-\pi)}{\sqrt{\pi(1-\pi)}}   } \right. \\
                 & & \hspace{8mm} < \left. \frac{\sqrt{n}(\pi_0-\pi)}{\sqrt{\pi(1-\pi)}} +  z_{\alpha/2}\sqrt{\frac{\pi_0(1-\pi_0)}{\pi(1-\pi)}} \right\}\\
                 &\approx& 1-Pr\left\{\frac{\sqrt{n}(\pi_0-\pi)}{\sqrt{\pi(1-\pi)}} -  z_{\alpha/2}\sqrt{\frac{\pi_0(1-\pi_0)}{\pi(1-\pi)}} 
< {\color{red} Z } < \frac{\sqrt{n}(\pi_0-\pi)}{\sqrt{\pi(1-\pi)}} +  z_{\alpha/2}\sqrt{\frac{\pi_0(1-\pi_0)}{\pi(1-\pi)}} \right\} \\
                 &=& 1 - \Phi\left( \frac{\sqrt{n}(\pi_0-\pi)}{\sqrt{\pi(1-\pi)}} +  z_{\alpha/2}\sqrt{\frac{\pi_0(1-\pi_0)}{\pi(1-\pi)}} \right) 
                       + \Phi\left( \frac{\sqrt{n}(\pi_0-\pi)}{\sqrt{\pi(1-\pi)}} -  z_{\alpha/2}\sqrt{\frac{\pi_0(1-\pi_0)}{\pi(1-\pi)}} \right),
\end{eqnarray*}
}

where $\Phi(\cdot)$ is the cumulative distribution function of the standard normal.
\end{frame}

\begin{frame}[fragile]
\frametitle{An R function to calculate approximate power}
\framesubtitle{For the test of a single proportion}

{\tiny
\begin{displaymath}
    \mbox{Power} = 1 - \Phi\left( \frac{\sqrt{n}(\pi_0-\pi)}{\sqrt{\pi(1-\pi)}} +  z_{\alpha/2}\sqrt{\frac{\pi_0(1-\pi_0)}{\pi(1-\pi)}} \right) 
                       + \Phi\left( \frac{\sqrt{n}(\pi_0-\pi)}{\sqrt{\pi(1-\pi)}} -  z_{\alpha/2}\sqrt{\frac{\pi_0(1-\pi_0)}{\pi(1-\pi)}} \right)
\end{displaymath}
}

%{\small
\begin{verbatim}
Z1power = function(pi,n,pi0=0.50,alpha=0.05)
    {
    a = sqrt(n)*(pi0-pi)/sqrt(pi*(1-pi))
    b = qnorm(1-alpha/2) * sqrt(pi0*(1-pi0)/(pi*(1-pi)))
    Z1power = 1 - pnorm(a+b) + pnorm(a-b)
    Z1power
    } # End of function Z1power
\end{verbatim}
%}

\end{frame}

\begin{frame}[fragile]
\frametitle{Some numerical examples}

\begin{verbatim}
> Z1power(0.50,100)
[1] 0.05
>
> Z1power(0.55,100)
[1] 0.168788 
> Z1power(0.60,100)
[1] 0.5163234
> Z1power(0.65,100)
[1] 0.8621995
> Z1power(0.40,100)
[1] 0.5163234
> Z1power(0.55,500)
[1] 0.6093123
> Z1power(0.55,1000)
[1] 0.8865478
\end{verbatim}

\end{frame}

\begin{frame}[fragile]
\frametitle{Find smallest sample size needed to detect $\pi=0.55$ as different from $\pi_0=0.50$ with probability at least 0.80}

\begin{verbatim}
> samplesize = 50
> power=Z1power(pi=0.55,n=samplesize); power
[1] 0.1076602
> while(power < 0.80)
+ {
+ samplesize = samplesize+1
+ power = Z1power(pi=0.55,n=samplesize)
+ }
> samplesize; power
[1] 783
[1] 0.8002392
\end{verbatim}
\end{frame}

\begin{frame}[fragile]
\frametitle{Find smallest sample size needed to detect $\pi=0.60$ as different from $\pi_0=0.50$ with probability at least 0.80}

\begin{verbatim}
> samplesize = 50
> power=Z1power(pi=0.60,n=samplesize); power
[1] 0.2890491
> while(power < 0.80)
+ {
+ samplesize = samplesize+1
+ power = Z1power(pi=0.60,n=samplesize)
+ }
> samplesize; power
[1] 194
[1] 0.8003138
\end{verbatim}
\end{frame}

\begin{frame}[fragile]
\frametitle{Conclusions from the power analysis}
%\framesubtitle{} 
  \begin{itemize}
    \item Detecting true $\pi=0.60$ as different from $0.50$ is a reasonable goal.
    \item Power with $n=100$ is barely above one half -- pathetic.
    \item As Fisher said, ``To call in the statistician after the experiment is done may be no  
more than asking him to perform a postmortem examination: he may be  
able to say what the experiment died of."
    \item $n=200$ is much better.
    \item How about $n=250$?
  \end{itemize}

\begin{verbatim}
> Z1power(pi=0.60,n=250)
[1] 0.8901088
\end{verbatim}

It depends on what you can afford, but I like $n=250$.
\end{frame}


\begin{frame}
\frametitle{What is required of the scientist}
\framesubtitle{Who wants to select sample size by power analysis}

The scientist must specify 
  \begin{itemize}
    \item Parameter values that he or she wants to be able to detect as different from $H_0$ value.
    \item Desired power (probability of detection)
  \end{itemize}

\vspace{20mm}

It's not always easy for the scientist to think in terms of the parameters of a statistical model.

\end{frame}

End commenting out power
\end{comment}


%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%


\begin{frame}
\frametitle{Copyright Information}

This document was prepared by  \href{http://www.utstat.toronto.edu/~brunner}{Jerry Brunner},
Department of Statistics, University of Toronto. It is licensed under a 
\href{http://creativecommons.org/licenses/by-sa/3.0/deed.en_US}
     {Creative Commons Attribution - ShareAlike 3.0 Unported License}. Use any part of it as you like and share the result freely. The \LaTeX~source code is available from the course website:
\href{http://www.utstat.toronto.edu/~brunner/oldclass/312f22} {\texttt{http://www.utstat.toronto.edu/$^\sim$brunner/oldclass/312f22}}

\end{frame}


\end{document}




%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
These two additional slides may be good for STA2101 downtown.


\begin{frame}[fragile]
\frametitle{Check power calculations by simulation}
\framesubtitle{First develop and illustrate the code}

{\small
\begin{verbatim}
> pi = 0.50; pi0 = 0.50; n = 100; m = 10
> p = rbinom(m,n,pi)/n; p
 [1] 0.49 0.49 0.56 0.51 0.48 0.53 0.51 0.40 0.54 0.51
> Z1 = sqrt(n)*(p-pi0)/sqrt(pi0*(1-pi0))
> rbind(p,Z1)
    [,1]  [,2] [,3] [,4]  [,5] [,6] [,7] [,8] [,9] [,10]
p   0.49  0.49 0.56 0.51  0.48 0.53 0.51  0.4 0.54  0.51
Z1 -0.20 -0.20 1.20 0.20 -0.40 0.60 0.20 -2.0 0.80  0.20
> sig = (abs(Z1)>1.96); sig
 [1] FALSE FALSE FALSE FALSE FALSE FALSE FALSE  TRUE FALSE FALSE
\end{verbatim}
}
\end{frame}



\begin{frame}[fragile]
\frametitle{Now real simulation: }
\framesubtitle{First estimated probabiliity should equal about 0.05 because $\pi=\pi_0$}

{\small
\begin{verbatim}
> pi = 0.50; pi0 = 0.50; n = 100; m = 10000
> p = rbinom(m,n,pi)/n
> Z1 = sqrt(n)*(p-pi0)/sqrt(pi0*(1-pi0))
> sig = (abs(Z1)>1.96)
> sum(sig)/m
[1] 0.055

> # Power calculation for pi=0.55 said power = 0.168788
> pi = 0.55; pi0 = 0.50; n = 100; m = 10000
> p = rbinom(m,n,pi)/n
> Z1 = sqrt(n)*(p-pi0)/sqrt(pi0*(1-pi0))
> sig = (abs(Z1)>1.96)
> sum(sig)/m
[1] 0.1837
\end{verbatim}
}
\end{frame}