%                               Stats intro for Applied Stat I
% Notes and comments at the end

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\title{Introduction\footnote{See last slide for copyright information.}}
\subtitle{STA442/2101 Fall 2014}
\date{} % To suppress date


\begin{document}

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

\begin{frame}
\frametitle{Background Reading}
\framesubtitle{Optional} 
  \begin{itemize}
    \item Chapter 1 of \emph{Linear models with R}
    \item Chapter 1 of Davison's \emph{Statistical models}: Data, and probability models for data.
  \end{itemize}
\end{frame}

\begin{frame}
\frametitle{Goal of statistical analysis}

The goal of statistical analysis is to draw reasonable conclusions from noisy numerical data.

\end{frame}


\begin{frame}{Steps in the process of statistical analysis}
{One 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 model parameters, for example 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 estimation and inference, draw conclusions about the example or problem.
  \end{itemize}
\end{frame}

\begin{frame}
\frametitle{What is a statistical model?}
\framesubtitle{You should always be able to state the model.}

\pause

A \emph{statistical model} is a set of assertions that partly specify the probability distribution of the observable data. The specification may be direct or indirect.
 
\pause

  \begin{itemize}
    \item Let $X_1, \ldots, X_n$ be a random sample from a normal distribution with expected value $\mu$ and variance $\sigma^2$.
    
\pause

    \item For $i=1, \ldots, n$, let $Y_i = \beta_0 + \beta_1 x_{i1} + \cdots + \beta_k x_{ik} + \epsilon_i$, where
    \begin{itemize}
        \item[] $\beta_0, \ldots, \beta_k$ are unknown constants.
        \item[] $x_{ij}$ are known constants.
        \item[] $\epsilon_1, \ldots, \epsilon_n$ are independent $N(0,\sigma^2)$ random variables.
        \item[] $\sigma^2$ is an unknown constant.
        \item[] $Y_1, \ldots, Y_n$ are observable random variables.
    \end{itemize}
  \end{itemize}

\pause

Is the model the same thing as the \emph{truth}?
\end{frame}

\begin{frame}
\frametitle{Parameter Space}

The \emph{parameter space} is the set of values that can be taken on by the parameter. 

\pause
 
  \begin{itemize}
    \item Let $X_1, \ldots, X_n$ be a random sample from a normal distribution with expected value $\mu$ and variance $\sigma^2$. 
\pause    

The parameter space is $\{(\mu,\sigma^2): -\infty < \mu < \infty, \sigma^2 > 0\}$.

\pause

    \item For $i=1, \ldots, n$, let $Y_i = \beta_0 + \beta_1 x_{i1} + \cdots + \beta_k x_{ik} + \epsilon_i$, where
    \begin{itemize}
        \item[] $\beta_0, \ldots, \beta_k$ are unknown constants.
        \item[] $x_{ij}$ are known constants.
        \item[] $\epsilon_1, \ldots, \epsilon_n$ are independent $N(0,\sigma^2)$ random variables.
        \item[] $\sigma^2$ is an unknown constant.
        \item[] $Y_1, \ldots, Y_n$ are observable random variables.
    \end{itemize}

\pause

The parameter space is $\{(\beta_0, \ldots, \beta_k, \sigma^2): -\infty < \beta_j < \infty, \sigma^2 > 0\}$.
  \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 $\theta$ denote the probability that a consumer will choose the new blend, treat the data $X_1, \ldots, X_n$ as a random sample from a Bernoulli distribution. That is, independently for $i=1, \ldots, n$,
\begin{displaymath}
    P(x_i|\theta) = \theta^{x_i} (1-\theta)^{1-x_i}
\end{displaymath}
for $x_i=0$ or $x_i=1$, and zero otherwise.
% The conditional probability notation is not in the book (I believe).

\vspace{5mm}

\pause

\begin{itemize}
     \item Parameter space is the interval from zero to one.
     \item $\theta$ could be estimated by maximum likelihood.
     \item Large-sample tests and confidence intervals are available.
\end{itemize}

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

\end{frame}

% Had a slide showing calculation of MLE -- deleted.

\begin{frame}
\frametitle{Tests of statistical hypotheses}
%\framesubtitle{} 

  \begin{itemize}
    \item Model: $X \sim F_\theta $
    \item $X$ is the data vector, and $\mathcal{X}$ is the sample space: $X \in \mathcal{X}$
    \item $\theta$ is the parameter, and $\Theta$ is the parameter space: $\theta \in \Theta$
    \pause
    \item Null hypothesis is $H_0: \theta \in \Theta_0 \mbox{ v.s. } H_A: \theta \in \Theta \cap \Theta_0^c$
    \item Meaning of the \emph{null} hypothesis is that \emph{nothing} interesting is happening.
    \pause
    \item $\mathcal{C} \subset \mathcal{X}$ is the \emph{critical region}. Reject $H_0$ in favour of $H_A$ when $X \in \mathcal{C}$.
    \pause
    \item Significance level $\alpha$ (\emph{size} of the test) is the maximum probability of rejecting $H_0$ when $H_0$ is true.
    \pause
    \item $p$-value is the smallest value of $\alpha$ for which $H_0$ can be rejected.
    \item Small $p$-values are interpreted as providing stronger evidence against the null hypothesis.
  \end{itemize}
\end{frame}

\begin{frame}
\frametitle{Type I and Type II error}
\framesubtitle{A Neyman-Pearson idea rather than Fisher} 
  \begin{itemize}
    \item Type I error is to reject $H_0$ when $H_0$ is true.
    \item Type II error is to \emph{not} reject $H_0$ when $H_0$ is false.
    \pause
%    \item Can't minimize the probability of both types of error at once.
%    \item So hold the maximum probability of a Type I error (the significance level $\alpha$) to a ``small" value like  $\alpha=0.05$, and then seek tests that minimize the probability of Type II error.
    \item $1-Pr\{$Type II Error$\}$ is called \emph{power}.
    \item Power may be used to select sample size.
  \end{itemize}
\end{frame}

\begin{frame}
\frametitle{Carry out a test to determine which brand of coffee is preferred}
\framesubtitle{Recall the model is 
$X_1, \ldots, X_n \stackrel{i.i.d.}{\sim} B(1,\theta)$} 

Start by stating the null hypothesis.

\pause

  \begin{itemize}
    \item $H_0: \theta=0.50$
    \item $H_1: \theta \neq 0.50$

\pause

    \item Could you make a case for a one-sided test?

\pause

    \item $\alpha=0.05$ as usual.

\pause

    \item Central Limit Theorem says $\widehat{\theta}=\overline{X}$ is approximately normal with mean $\theta$ and variance $\frac{\theta(1-\theta)}{n}$.
  \end{itemize}
\end{frame}

\begin{frame}[fragile]
\frametitle{Several valid test statistics for $H_0: \theta=\theta_0$ are available}
{Two of them are} % Which one do you like more? Why?

\begin{displaymath}
    Z_1 = \frac{\sqrt{n}(\overline{X}-\theta_0)}{\sqrt{\theta_0(1-\theta_0)}}
\end{displaymath}

and

\begin{displaymath}
    Z_2 = \frac{\sqrt{n}(\overline{X}-\theta_0)}{\sqrt{\overline{X}(1-\overline{Y})}}
\end{displaymath}

\vspace{10mm}

\pause

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 and the $p$-value for each test}
\framesubtitle{Suppose 60 out of 100 preferred the new blend}

\pause
$ Z_1 = \frac{\sqrt{n}(\overline{X}-\theta_0)}{\sqrt{\theta_0(1-\theta_0)}}$
\pause
\begin{verbatim}
> theta0 = .5; ybar = .6; n = 100
> Z1 = sqrt(n)*(ybar-theta0)/sqrt(theta0*(1-theta0)); Z1
[1] 2
> pval1 = 2 * (1-pnorm(Z1)); pval1
[1] 0.04550026
\end{verbatim}
\pause
$Z_2 = \frac{\sqrt{n}(\overline{X}-\theta_0)}{\sqrt{\overline{X}(1-\overline{Y})}}$
\pause
\begin{verbatim}
> Z2 = sqrt(n)*(ybar-theta0)/sqrt(ybar*(1-ybar)); 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 $H_0$? \pause \emph{Yes, just barely.}

\pause

    \item Isn't the $\alpha=0.05$ significance level pretty arbitrary?  \pause \emph{Yes, but if people insist on a Yes or No answer, this is what you give them.}

\pause

    \item What do you conclude, in symbols?  \pause $\theta \neq 0.50$. \emph{Specifically,} $\theta > 0.50$.

\pause

    \item What do you conclude, in plain language? Your answer is a statement about coffee.  \pause \emph{More consumers prefer the new blend of coffee beans.}

\pause

    \item Can you really draw directional conclusions when all you did was reject a non-directional null hypothesis? \pause
\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}

\pause
  
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 $Z=1.84$?}

\pause

Here are two possibilities, in plain language.

  \begin{itemize}
    \item ``This study does not provide clear evidence that consumers prefer one blend of coffee beans over the other."

\pause

    \item ``The results are consistent with no difference in preference for the two coffee bean blends."
  \end{itemize}
  
\vspace{5mm}

\pause
  
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{Usually for individual parameters} 
  \begin{itemize}
    \item Point estimates may give a false sense of precision.
    \item We should provide a margin of probable error as well.
  \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.}

\pause

\begin{eqnarray*}
      & & \left(\overline{x} - z_{\alpha/2}\sqrt{\frac{\overline{x}(1-\overline{x})}{n}} ~~,~~
                \overline{x} + z_{\alpha/2}\sqrt{\frac{\overline{x}(1-\overline{x})}{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*}

\pause

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 $\theta$.
    \item Does this mean $Pr\{ 0.504 < \theta < 0.696 \}=0.95$?

\pause

    \item No! The quantities $0.504$, $0.696$ and  $\theta$ are all constants, so $Pr\{ 0.504 < \theta < 0.696 \}$ is either zero or one. 

\pause
 
    \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.

\pause

    \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. 

\pause

    \item The confidence interval is a guess, and the guess is either right or wrong. But the guess is the constructed by a method that is right 95\% of the time. % Take it or leave it. 
  \end{itemize}
\end{frame}




\begin{frame}
\frametitle{More on confidence intervals}
%\framesubtitle{Usually, confidence \emph{intervals} for single parameters} 

  \begin{itemize}
    \item Can have confidence \emph{regions} for the entire parameter vector or multi-dimensional functions of the parameter vector.
    \item Confidence regions correspond to tests.
  \end{itemize}

\end{frame}

























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


\begin{frame}
\frametitle{Copyright Information}

This slide show 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/appliedf14} {\footnotesize \texttt{http://www.utstat.toronto.edu/$^\sim$brunner/oldclass/appliedf14}}

\end{frame}


\end{document}



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

\begin{frame}
\frametitle{}
%\framesubtitle{} 
  \begin{itemize}
    \item 
    \item 
    \item 
  \end{itemize}
\end{frame}









 \begin{itemize}
    \item $R=R(X)$ is a function of the sample data, and \emph{not} of any unknown parameters. Otherwise you could not compute it.
    \item Probability calculation applies \emph{before} data are collected
    \item Once a particular data vector $x$ is obtained, can calculate a particular region $r=R(x)$.
    \item The region $r$ 
    \item 
    \item 
\pause

    \item 
  \end{itemize}


\begin{frame}
\frametitle{Meaning of the confidence interval}
  \begin{itemize}
    \item We calculated a 95\% confidence interval of $(0.504,0.696)$ for $\theta$.
    \item Does this mean $Pr\{ 0.504 < \theta < 0.696 \}=0.95$?

\pause

    \item No! The quantities $0.504$, $0.696$ and  $\theta$ are all constants, so $Pr\{ 0.504 < \theta < 0.696 \}$ is either zero or one. 

\pause
 
    \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.

\pause

    \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. 

\pause

    \item So we sometimes say that we are ``$95\%$ confident" that $0.504 < \theta < 0.696$.
  \end{itemize}
\end{frame}

\begin{frame}
\frametitle{Confidence intervals (regions) correspond to tests}
\framesubtitle{Recall $Z_1 = \frac{\sqrt{n}(\overline{Y}-\theta_0)} {\sqrt{\theta_0(1-\theta_0)}}$ and $Z_2 = \frac{\sqrt{n} (\overline{Y}-\theta_0)}{\sqrt{\overline{Y}(1-\overline{Y})}}$.}

\pause

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}
\overline{Y} - z_{\alpha/2}\sqrt{\frac{\overline{Y}(1-\overline{Y})}{n}} < \theta_0 < \overline{Y} +  z_{\alpha/2}\sqrt{\frac{\overline{Y}(1-\overline{Y})}{n}}
\end{displaymath}

\pause


  \begin{itemize}
    \item So the confidence interval consists of those parameter values $\theta_0$ for which $H_0: \theta=\theta_0$ is \emph{not} rejected. \pause
    \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. \pause
    \item There is a confidence interval corresponding to $Z_1$ too. 
    \item In general, any test can be inverted to obtain a confidence region.
  \end{itemize}

\end{frame}


\begin{displaymath}
    \begin{array}{l}
     Y_1, \ldots, Y_n  \stackrel{i.i.d.}{\sim} F_\theta, \, \theta \in \Theta \\
     \Theta_0 = \{\theta \in \Theta: \theta_1=h_1, \ldots, \theta_r=h_r \} \\
     H_0: \theta \in \Theta_0 \mbox{ v.s. } 
     H_A: \theta \in \Theta \cap \Theta_0^c, 
    \end{array}
\end{displaymath}

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

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

\vspace{10mm}

\begin{verbatim}
> p = 60/100; p
[1] 0.6
\end{verbatim}
\end{frame}


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

\pause

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

\pause

  \begin{itemize}
    \item Could express this as $\overline{Y} \pm  z_{\alpha/2}\sqrt{\frac{\overline{Y}(1-\overline{Y})}{n}}$ \pause 
    \item $z_{\alpha/2}\sqrt{\frac{\overline{Y}(1-\overline{Y})}{n}}$ 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.}

\pause

\begin{eqnarray*}
      & & \left(\overline{x} - z_{\alpha/2}\sqrt{\frac{\overline{x}(1-\overline{x})}{n}} ~~,~~
                \overline{x} + z_{\alpha/2}\sqrt{\frac{\overline{x}(1-\overline{x})}{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*}

\pause

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 $\theta$.
    \item Does this mean $Pr\{ 0.504 < \theta < 0.696 \}=0.95$?

\pause

    \item No! The quantities $0.504$, $0.696$ and  $\theta$ are all constants, so $Pr\{ 0.504 < \theta < 0.696 \}$ is either zero or one. 

\pause
 
    \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.

\pause

    \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. 

\pause

    \item So we sometimes say that we are ``$95\%$ confident" that $0.504 < \theta < 0.696$.
  \end{itemize}
\end{frame}




%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
Notes and comments

In 2013, Cut out non-central chisq part, added nice what's a model, parameter space
Non-central chisq is in 2012 version.
In 2014, based on a one-vs. 2-sample t-test rather than binomial. Version 2014a wandered back into the taste test example.