\documentclass[11pt]{article} \usepackage{amsmath} \usepackage{graphicx} \usepackage{verbatim} \allowdisplaybreaks \pagestyle{empty} \setlength{\topmargin}{-0.75in} \setlength{\textheight}{9.5in} \setlength{\oddsidemargin}{-0.1in} \setlength{\evensidemargin}{-.2in} \setlength{\textwidth}{6.7in} \font\heada=cmbx10 scaled\magstep3 \font\headb=cmsl10 scaled\magstep1 \font\headc=cmr8 \pretolerance=10000 \def\ds{\displaystyle} \begin{document} \section*{Chapter 9 - Estimation} \bigskip \subsection*{Point Estimation} \begin{enumerate} \item {\bf Point Estimator} - A formula applied to a data set which results in a single value or point. In other words, a statistic. Usually, estimators are used to give plausible values of some population parameter. \begin{itemize} \item $\overline X$, $\tilde X$, $S^2$, and $p$ are point estimators of the parameters $\mu$, $\tilde \mu$, $\sigma^2$ and $\pi$ respectively. \end{itemize} \vspace{0.05in} \item {\bf Point Estimate} - The resulting value of a point estimator, when applied to a data set. \begin{itemize} \item $\overline x$ = 27.6, $\tilde x$ = 85, $s^2 = 2.45$, and $p$ = 0.30 are point estimates of the values of $\mu$, $\tilde \mu$, $\sigma^2$ and $\pi$ respectively. \end{itemize} \end{enumerate} \vspace{0.15in} \noindent {\bf \underline{Desirable Properties of a Point Estimator}:} \begin{enumerate} \item {\bf Unbiased} - A point estimator is \underline{unbiased} for a parameter if the mean of the estimator's sampling distribution equals the value of the parameter. Otherwise, the estimator is \underline{biased}. \begin{itemize} \item $\overline X$ is an unbiased estimator of $\mu$ because $\mu_{_{\overline X}} = \mu$. \item $S^2$ is an unbiased estimator of $\sigma^2$ because $\mu_{S^2}=\sigma^2$. $S$ is NOT an unbiased estimate of $\sigma$! \item $p$ is an unbiased estimator of $\pi$ because $\mu_p = \pi$. \end{itemize} \vspace{0.05in} \item {\bf Smallest Variability} - When choosing among unbiased estimators, the one with the smallest sampling variability (i.e. smallest standard deviation) is the best, because the point estimates will be most closely concentrated around the parameter value. \end{enumerate} \begin{itemize} \item A Point Estimator that has properties 1 and 2 above is called a {\bf Minimum Variance Unbiased Estimator (MVUE)}. When the data is normal, then {$\overline X$ is an MVUE for $\mu$}. That is, $\sigma^2_{\overline X}=\frac{\sigma^2}{n}$ is smaller than the variance for any other point estimator of $\mu$. \item When the data is not normal, then $\overline X$ is not an MVUE for $\mu$ (see page 365 of your textbook). \vspace{0.25in} \end{itemize} \subsection*{Interval Estimation} \begin{enumerate} \item {\bf Interval Estimator}: A formula applied to a data set which results in an interval of plausible values for some parameter. It has the form \begin{center} {\bf point estimator $\pm$ margin of error}. \end{center} \item {\bf Confidence Interval (C.I.)}: An interval estimator which has a {\it level of confidence} attached to it. \item {\bf Confidence Level:} A quantity (typically stated as a percentage) describing how often a confidence interval (over all samples of a given size) captures the parameter value. \begin{itemize} \item Commonly-used confidence levels are 90\%, 95\%, and 99\%. \item A 95\% confidence level means that: \begin{itemize} \item With probability .95, the formula for the confidence interval captures the value of the parameter. \item If confidence intervals were calculated from all possible samples, then 95\% of the intervals would contain the parameter value. \end{itemize} \end{itemize} \end{enumerate} \newpage \noindent {\bf Confidence Interval for $\mu$ (when $\sigma$ is known)}\ \ \ $\longrightarrow$ \ \ \ \fbox{$\overline X~\pm~z_{_{1-\frac{\alpha}{2}}}\left(\ds \frac{\sigma}{\sqrt{n}}\right)$} \begin{verse} \noindent {\bf Assumptions:} \vspace{-0.05in} \begin{itemize} \item The data must be a SRS (so if you sample a finite population without replacement, then $.05N\ge n$). \item The sampling distribution of $\overline X$ must be at least approximately normal (so the data is normal OR $n>15$ for symmetric non-normal data OR $n>30$ for skewed data). \end{itemize} \end{verse} \vspace{0.15in} \noindent {\bf Notation:} \begin{itemize} \item the margin of error is $m = z_{_{1-\frac{\alpha}{2}}}\left(\ds \frac{\sigma}{\sqrt{n}}\right)$. \item $z_{_{1-\frac{\alpha}{2}}}$ is the {\em critical value}, the $100\left(1-\frac{\alpha}{2}\right)$ percentile of the Z distribution, Z $\sim$ N(0,~1). \item $\alpha$ is a {\em significance level} \item $C=1-\alpha$ is the confidence level \end{itemize} \begin{tabular}{||c|c|c|c||} \hline C & $\alpha$ = 1 - C & 1 - $\frac{\alpha}{2}$ & $z_{_{1-\frac{\alpha}{2}}}$\\ \hline 0.90 & 0.10 & 0.95 & \\ 0.95 & 0.05 & 0.975 & \\ 0.99 & 0.01 & 0.995 & \\ \hline \end{tabular} \vspace{0.25in} \noindent Find $z_{_{1-\frac{\alpha}{2}}}$ for 90\%, 95\% and 99\% confidence intervals using the normal table. \vspace{2.5in} \noindent \underline{EXAMPLE \#1}: (Problem 9.37 on page 392) Seventy seven students at the University of Virginia were asked to keep a diary of conversations with their mothers, recording any lies they told during the conversations. Suppose that $\overline x=.5$ and $\sigma=.4$. \begin{itemize} \item Is the sample size large enough to calculate a C.I.? \item Construct a 90\% C.I. for $\mu$, the true number of lies told to mothers during each conversation. \end{itemize} \newpage \noindent {\bf \underline{Confidence Interval Behavior}:} \\ \vspace{0.1in} \noindent The margin of error $m$, and hence the width of the interval depends, on: \begin{enumerate} \item The confidence level $C$: Increasing $C$ increases $m$ and the interval width. \item The variability of the response $\sigma$: Increasing $\sigma$ increases $m$ and the interval width. \item The sample size $n$:Increasing $n$ decreases $m$ and the interval width. \end{enumerate} \vspace{3in} \begin{itemize} \item The C.I.'s with $\overline x$'s that are less than one margin of error away from $\mu$ will be the intervals that \underline{capture $\mu$}. This happens $100C\%$ of the time. \item All $\overline x$'s that are more than one margin of error away from $\mu$ will be the intervals that \underline{do not capture $\mu$}. This happens $100\alpha\%$ of the time. \item Typically, only one sample is selected from the population and therefore only one confidence interval will be calculated. The {\bf CORRECT INTERPRETATION} of this {\it one} confidence interval is: \vspace{-0.15in} \begin{center} \noindent We are \underline{\hspace{0.4in}}\% confident that $\mu$ lies between \underline{\hspace{0.6in}} and \underline{\hspace{0.6in}}. \end{center} Usually, one interprets $\mu$ in terms of the problem. \item An {\bf INCORRECT INTERPRETATION} of this {\it one} confidence interval is: \vspace{-0.1in} \begin{center} \noindent There is a \underline{\hspace{0.4in}}\% chance (i.e. probability) that $\mu$ lies between \underline{\hspace{0.6in}} and \underline{\hspace{0.6in}}. \end{center} Probability statements can only be made about confidence intervals BEFORE you calculate the specific confidence interval estimate for the data. A given interval estimate captures the parameter with probability {\bf 0} or {\bf 1}, we simply do not know which it is. \end{itemize} \vspace{0.05in} \noindent \underline{EXAMPLE \#1} revisited: From our previous example, we'd interpret the C.I. like this: \begin{center} We are \underline{\hspace{0.4in}}\% confident that, on average, University of Virginia students lie to their mothers between \underline{\hspace{0.6in}} and \underline{\hspace{0.6in}} times per conversation. \end{center} \noindent {\bf \underline{QUESTION}: True or False?} \\ \vspace{0.05in} \noindent Suppose $\mu$ is the average weight (in pounds) of dogs in Bozeman. A 95\% C.I. for $\mu$ is (42,~48). \begin{itemize} \item[1.] T \ / \ F: \ Ninety-five percent of the weights in the population are between 42 and 48 pounds. \vspace{-0.05in} \item[2.] T \ / \ F: \ Ninety-five percent of the weights in the sample are between 42 and 48 pounds. \vspace{-0.05in} \item[3.] T \ / \ F: \ The probability that the confidence interval (42,~48) includes $\mu$ is 0.95. \vspace{-0.05in} \item[4.] T \ / \ F: \ The sample mean $\overline{x}$ is in the confidence interval with probability 0.95. \vspace{-0.05in} \item[5.] T \ / \ F: \ If 100 confidence intervals were generated using this same process, approximately 5 of the confidence intervals would not include $\mu$. \end{itemize} \noindent \underline{EXAMPLE \#2}: \\ \noindent Gas mileages (in MPG) of a SRS of 20 cars of a certain make and model are recorded and the sample mean is found to be 18.48 MPG. Suppose that the data is not normal but is fairly symmetric. Also suppose that the population standard deviation is known to be $\sigma$ = 2.9 MPG. \begin{itemize} \item Do we have a large enough sample size? \vspace{1in} \item Find and interpret a 95\% confidence interval for $\mu$, the mean MPG for this vehicle. \end{itemize} \vspace{2.25in} \newpage \noindent {\bf Confidence Interval for $\mu$ ($\sigma$ unknown)} \ \ \ $\longrightarrow$ \ \ \ \fbox{$\overline X~\pm~t_{_{1-\frac{\alpha}{2},~n-1}}\left(\ds \frac{s}{\sqrt{n}}\right)$} \begin{verse} \noindent {\bf Assumptions:} \vspace{-0.05in} \begin{itemize} \item The data must be a SRS (so if you sample a finite population without replacement, then $.05N\ge n$). \item The sampling distribution of $\overline X$ must be at least approximately normal (so the data is normal OR $n>15$ for symmetric non-normal data OR $n>30$ for skewed data). \end{itemize} \end{verse} \vspace{0.15in} \noindent {\bf \underline{t Distribution}} \ \ $\longrightarrow$ \ \ $T \sim t(df)$ \vspace{0.1in} \begin{itemize} \item The t distribution is symmetric, unimodal, bell-shaped, and centered at zero. \vspace{-0.05in} \item The t distribution has heavier tails than the Z distribution because s (an estimate of $\sigma$) is used instead of $\sigma$. \vspace{-0.05in} \item As the degrees of freedom (df) increases, the t distribution approaches the Z distribution. \end{itemize} \vspace{1.75in} \noindent {\bf Notation:} \begin{itemize} \item the margin of error is $m = t_{_{1-\frac{\alpha}{2},~n-1}}\left(\ds \frac{s}{\sqrt{n}}\right)$. \item Use the t-table or R to calculate $t_{_{1-\frac{\alpha}{2},~n-1}}$, the t {\em critical value}, the $100\left(1-\frac{\alpha}{2}\right)$ percentile of the t distribution with n - 1 degrees of freedom, $t(n-1)$. Since $t(n-1)$ has thicker tails than $N(0,1)$, then $t_{1-\frac{\alpha}{2},~n-1}>z_{1-\frac{\alpha}{2}}$. \item $\frac{s}{\sqrt{n}}$ is the {\em standard error} of $\overline X$ \end{itemize} \vspace{.05in} \noindent \underline{EXAMPLE \#3}: From a SRS of 8 shipments of corn soy blend, a highly nutritious food sent for emergency relief, the mean vitamin C content (in mg/100g) is $\overline x = 22.5$ and the sample standard deviation is $s = 7.19$. \begin{itemize} \item Do we have a large enough sample size? \item Calculate and interpret a 99\% confidence interval for $\mu$, the true mean vitamin C content of corn soy blend. \end{itemize} \vspace{2.25in} \newpage \noindent {\bf Confidence Interval for $\pi$} \ \ \ $\longrightarrow$ \ \ \ \fbox{$p~\pm~z_{1-\frac{\alpha}{2}}\left(\ds \sqrt{\frac{p(1-p)}{n}}\right)$} \begin{verse} \noindent {\bf Assumptions:} \vspace{-0.05in} \begin{itemize} \item The data must be a SRS (so if you sample a finite population without replacement, then $.05N\ge n$). \item The sampling distribution for $p$ must be approximately normal (so $np\ge 10$ and \ $n(1-p) \ge 10$) \end{itemize} \end{verse} \noindent \underline{EXAMPLE \#4}: \\ \noindent In a study of heavy drinking on college campuses, 17096 students were interviewed. Of these, 3314 admitted ti consuming more than 5 drinks at a time, three times a week. \begin{itemize} \item Give a point estimate of the proportion of college students who are ``heavy" drinkers. \vspace{1in} \item Is the sample large enough to assume that the sampling distribution of $p$ is approximately normal? \vspace{1in} \item Give a $99\%$ C.I. for the proportion of all college students who are heavy drinkers. \end{itemize} \vspace{2.25in} \newpage \subsection*{Sample Size Calculations} Before any study, a researcher often already knows the confidence level and the {\em precision} (or margin of error) of a desired confidence interval. For a fixed confidence level and margin of error, the only other factor under the researcher's control is the sample size. So a researcher must collect enough data to be able to construct the desired C.I.'s. \noindent {\bf Sample Size Calculation for Estimating $\mu$}: \ \ \ $\longrightarrow$ \ \ \ \fbox{$n = \left(\ds \frac{z_{_{1 - \frac{\alpha}{2}}}~\sigma}{m}\right)^2$} \begin{itemize} \item $z_{1 - \frac{\alpha}{2}}$ is the critical value for the desired confidence level $C=1-\alpha$. \item $m$ is the desired margin of error \item $\sigma$ is the standard deviation of the population \end{itemize} \noindent \underline{If $\sigma$ is unknown, two options for estimating $\sigma$ are}: \begin{enumerate} \item Use a sample standard deviation, $s$, from a previous study. \item Use the anticipated range divided by 4. \end{enumerate} \noindent \underline{EXAMPLE}: \\ \noindent Find the sample size necessary to estimate the mean level of phosphate in the blood of dialysis patients to within 0.05 with 90\% confidence. A previous study calculated a sample standard deviation of $s=1.6$. \vspace{1.75in} \noindent {\bf Sample Size Calculation for Estimating $\pi$}: \ \ \ $\longrightarrow$ \ \ \ \fbox{$n = \pi(1-\pi)\left(\ds \frac{z_{_{1 - \frac{\alpha}{2}}}}{m}\right)^2$} \vspace{0.1in} \noindent \underline{Two options for the value of $\pi$}: \begin{enumerate} \item Use an estimate, $p$, from a previous study. \item Use $\pi = \frac{1}{2}$. This is the more conservative choice because using it will result in a sample size $n$ even larger than needed. \end{enumerate} \noindent \underline{EXAMPLE}: \\ \noindent Your company would like to carry out a survey of customers to determine the degree of satisfaction with your customer service. You want to estimate the proportion of customers who are satisfied. What sample size is needed to attain 95\% confidence and a margin of error less than or equal to 3\%, or 0.03? \vspace{2.25in} \newpage \subsection*{R commands} \verbatiminput{Chapter9Rcode.txt} \newpage \subsection*{Estimating $\mu$ after transforming data} For large sample sizes ($n>30$), we do not need to assume that the data $\{x_i\}$ is normal in order to find a C.I. for $\mu_X$. But when we have a small sample ($n<15$) from a population which is clearly non-normal, then a ``transform to normality" $Y=f(X)$ may be appropriate (such as the Box-Cox family of transforms in Chapter 7 notes). In many cases, one can not directly interpret the point estimate $\overline y$ or a C.I. for $\mu_Y$. \begin{enumerate} \item \textbf{The General Case $Y=f(X)$} \begin{itemize} \item $f^{-1}({\rm point ~~estimator ~for ~} \mu_Y$) is a point estimator for $f^{-1}(\mu_Y)$ (invariance of MLE's). \item $f^{-1}({\rm point ~~estimator ~for ~} \mu_Y$) is a $1^{st}$ order Taylor series (a better $2^{nd}$ order estimator is given below) point estimator for $\mu_X$. \item When $f$ is monotone, the limits of the C.I. for $\mu_Y$ can ALWAYS be back-transformed to get a C.I. for $f^{-1}(\mu_Y)$. SOMETIMES, the limits of the C.I. for $\mu_Y$ can be back-transformed to get a C.I. for $\mu_X$. \item \textbf{Delta Method:} Point estimators of $\mu_Y$ and $\sigma^2_Y$ (in addition to $\overline Y$ and $S^2_Y$) can be derived by considering the Taylor series expansion $$Y=f(X)\approx f(\mu_X) + f'(\mu_X)(X-\mu_X) + \frac{1}{2}f''(\mu_X)(X-\mu_X)^2$$ and taking the mean (to $2^{nd}$ order) and variance (to $1^{st}$ order) of both sides \begin{eqnarray*} \mu_Y\approx f(\mu_X) + \frac{1}{2}f''(\mu_X)\sigma^2_X ~~~~~~~~~~~~ \sigma^2_Y \approx \left(f'(\mu_X)\right)^2\sigma^2_X. \end{eqnarray*} \item $X$ and $Y$ are random variables which can be viewed as either data or statistics. In the second case, if $\{X_i\}$ is a sequence of statistics such that $X_i~\dot\sim~ N(\mu_X,\sigma^2_X)$ (and approximation gets better as $i$ increases, with $\sigma_X\to 0$, eg when $X_i$ is a sample mean), then $$Y_i=f(X_i)~\dot\sim~ N\left(f(\mu_X),\left(f'(\mu_X)\right)^2\sigma_X^2\right)$$ (as long as $f'$ is continuous). Thus $$\sigma^2_{Y_i}\to \left(f'(\mu_X)\right)^2\sigma_X^2$$ where convergence is in probability. The technical formulation is that if $\sqrt{n}(X_i-\mu_X) \to N(0,\sigma^2)$, then $$\sqrt{n}\left(Y_i-f(\mu_X)\right)=\sqrt{n}\left(f(X_i)-f(\mu_X)\right)~\to~ N\left(0,\left(f'(\mu_X)\right)^2\sigma^2\right)$$ (as long as $f'$ is continuous), with $\sigma^2_X=\frac{\sigma^2}{n}$. \item The case where either $X$ or $Y$ is multivariate is easily dealt with using a multivariate Taylor series, so that the covariance matrix $\Sigma_Y\approx J^T\Sigma_X J$ where $J$ is a Jacobian evaluated at $\mu_X$, $J_{ij}=\frac{\partial f_j}{\partial X_i}(\mu_X)$. When the statistics $X_i$ converge to a multivariate normal distribution, then $$Y_i=f(X_i)~\dot\sim~ N\left(f(\mu_X),J^T\Sigma_X J\right)$$ where the approximation gets better as $i$ increases (see Casella and Berger p.329). \end{itemize} \item \textbf{Box-Cox transform $Y=X^\lambda$ with $\lambda \ne 0$} \begin{itemize} \item By the Delta Method, $\mu_Y^{\frac{1}{\lambda}}\approx \mu_X\left(1+\frac{\lambda(\lambda-1)}{2}\frac{\sigma_X^2}{\mu_X^2}\right)^{\frac{1}{\lambda}}.$ \item \fbox{$\overline y^{\frac{1}{\lambda}}\approx \mu_X$} if $\frac{s^2}{\overline x^2}\approx\frac{\sigma_X^2}{\mu_X^2}$ is ``small" \item If $Y=X^\lambda$ is monotone (you need to check this), then the back-transformed interval \hspace{1in}\fbox{$\left(\left(\overline y~-~t_{_{1-\frac{\alpha}{2},~n-1}}\ds \frac{s_y}{\sqrt{n}}\right)^{\frac{1}{\lambda}}, \left(\overline y~+~t_{_{1-\frac{\alpha}{2},~n-1}}\ds \frac{s_y}{\sqrt{n}}\right)^{\frac{1}{\lambda}}\right)$} \begin{itemize} \item is a $100(1-\frac{\alpha}{2})$ C.I. for $\left(\mu_Y\right)^{\frac{1}{\lambda}}$. \item is an approximate $100(1-\frac{\alpha}{2})$ C.I. for $\mu_X$ when $\frac{s^2}{\overline x^2}\approx\frac{\sigma_X^2}{\mu_X^2}$ is ``small" \end{itemize} \item When $\lambda<0$, you'll need to swap the endpoints of the C.I. \end{itemize} \item \textbf{The Box-Cox transform $Y = \ln(X)$} \begin{itemize} \item By the Delta Method, $\exp(\mu_Y)\approx \mu_X\left(\exp\left(-\frac{\sigma_X^2}{2\mu_X^2}\right)\right).$ \item \fbox{$\exp(\overline y)\approx \mu_X$} if $\frac{s^2}{\overline x^2}\approx\frac{\sigma_X^2}{\mu_X^2}$ is ``small" \item \fbox{$\exp(\overline y)=\left(\Pi_ix_i\right)^{\frac{1}{n}}$} is the geometric mean of the untransformed data %$\{x_i\}$ $\mu_Y\approx 1/n \sum_i \ln X_i=\ln\left(\Pi_iX_i\right)^{\frac{1}{n}}$ \item Since the log transform is strictly monotone, then the back-transformed interval \hspace{1in}\fbox{$\left(\exp\left(\overline y~-~t_{_{1-\frac{\alpha}{2},~n-1}}\ds \frac{s_y}{\sqrt{n}}\right), \exp\left(\overline y~+~t_{_{1-\frac{\alpha}{2},~n-1}}\ds \frac{s_y}{\sqrt{n}}\right)\right)$.} \begin{itemize} \item is a $100(1-\frac{\alpha}{2})$ C.I. for $\exp\left(\mu_Y\right)$, the population geometric mean of $X$. \item is an approximate $100(1-\frac{\alpha}{2})$ C.I. for $\mu_X$ when $\frac{s^2}{\overline x^2}\approx\frac{\sigma_X^2}{\mu_X^2}$ is ``small" \end{itemize} \end{itemize} \end{enumerate} \subsection*{Exercises} \noindent 9.1 on p366: 1-7 odd \noindent 9.2 on p379 (C.I. for $\pi$): 11-17 odd, 21-25 odd, 27 (sample size calculation). \noindent 9.3 on p391 (C.I. for $\mu$): 29-35 odd, 39-45 odd, 47 (sample size calculation) \subsection*{Reading} Sections 9.1-9.3 \end{document}