\documentclass[12pt]{article} \usepackage{fullpage} \usepackage{graphicx} \usepackage{float} \usepackage{amsmath} \usepackage{amssymb} % normal margin % \textwidth=6.40in % \textheight=9.00in % \hoffset-0.07in % \voffset+0.16in % full-size (documents with many figures) \textwidth=7.0in \textheight=9.0in \hoffset-0.25in \voffset-0.25in \footskip=1in \renewcommand{\baselinestretch}{1} \pagestyle{empty} \begin{document} \begin{flushright} \underline{ID (E.g. ED7843):~~~~~~~~~~~~~~~~~~~~~~~~~~~~~} \underline{Full Name (Print):~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~} \end{flushright} \begin{center} \textbf{MA 2210/6150 Midterm II (11/16, 5:30-7:20p)} \end{center} \noindent - Print your full name on \textbf{the first sheet} and at least first or last name on all the other sheets you will submit. \noindent - Write down your answers on the same page as the questions. You may use the backside of the sheet if the space is not enough. \noindent - Write down the process to the answer EXCEPT FOR Q1 and Q4. You may not get full credits if you only write the final answer. \noindent \hrulefill \vspace{1zh} \noindent \textbf{1.} Are the following statements correct? Circle your answers (44 points; a correct answer = 4 pts each, ``Skip'' = 2 pts each, a wrong answer = 0 pts each). \begin{description} \item[(1)] Let $X_1, X_2 \sim Bin(1,p)$, and $X_1$ and $X_2$ are independent. Then $X_1 + X_2 \sim Bin(2,2p)$. \item[(2)] Assume $0 < p <1$. A binomial distribution is symmetric if and only if $p=0.5$. \item[(3)] A Poisson distribution with $\mu=3$ is maximized at $k=2$ and $3$. \item[(4)] Let $X \sim N(0,1)$ then $P(X = 0) = 1/2$. \item[(5)] Let $X \sim N(0,1)$ and $a >0$, then $P(-a \le X \le 0) + P(a < X) = 1/2$. \item[(6)] Let $X_1 \sim N(\mu_1, \sigma_1^2)$ and $X_2 \sim N(\mu_2, \sigma_2^2)$ then $X_1 - X_2 \sim N(\mu_1 - \mu_2, \sigma_1^2 - \sigma_2^2)$. \item[(7)] Let $X_1$ and $X_2$ be independent random variables, and the distribution of $X_1$ and $X_2$ are the same. Then the shape of the distribution of $X_1$ and the shape of the distribution of $\frac{X_1+X_2}{2}$ are the same, even though the variances are different. \item[(8)] Let $X_i$ ($i=1,2,\cdots,n$) be IID random variables with $Var(X_i) = 1$, then the variance of $\bar{X} = \frac{X_1 + \cdots + X_n}{n}$ is $1/n$. \item[(9)] Let $X \sim Bin(n,p)$, then the variance of $\frac{X}{n}$ goes to zero when $n$ gets larger. \item[(10)] If a point estimator $\hat{\mu}$ is biased, it is never consistent. \item[(11)] A confidence interval $[A,B]$ for a parameter $\mu$ is (a pair of) random variables (before observing a sample). \end{description} \begin{table}[htbp] \begin{center} \begin{tabular}{|c|c|c|c|c|c|c|c|c|c|c|c|} \hline (1) & (2) & (3) & (4) & (5) & (6) & (7) & (8) & (9) & (10) & (11) \\ \hline Yes & Yes & Yes & Yes & Yes & Yes & Yes & Yes & Yes & Yes & Yes \\ No & No & No & No & No & No & No & No & No & No & No \\ Skip & Skip & Skip & Skip & Skip & Skip & Skip & Skip & Skip & Skip & Skip \\ \hline \end{tabular} \end{center} \end{table} %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% \newpage \begin{flushright} \underline{Full Name (Print):~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~} \end{flushright} \noindent \textbf{2.} (12 points) \\ Suppose that the probability that a college student becomes a victim of a violent crime during a given year is 0.002. Let us randomly choose 1,500 college students, and let $X$ be the number of students who will be victims in a year. \begin{description} \item[(a)] Write down $P(X=k)$ as a function of $k$ (note: we want the exact distribution, not an approximation). (2) \item[(b)] Calculate $P(X \le 1)$ to four places of decimals (e.g., 0.1235). (4) \item[(c)] The distribution in (a) can be approximated by a Poisson distribution. Write down the probability function $P(X=k)$ of THIS Poisson distribution. (2) \item[(d)] Approximate $P(X \le 1)$ by using (c) to four places of decimals. (4) \end{description} %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% \newpage \begin{flushright} \underline{Full Name (Print):~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~} \end{flushright} \noindent \textbf{3.} (9 points) \\ \begin{description} \item[(a)] Let $Z \sim N(0,1)$. Calculate $P(-0.3 \le Z \le 1.3)$. (3) \item[(b)] Let $X \sim N(-3,4^2)$. Calculate $P(-10 \le X \le 5)$. (4) \item[(c)] Let $f(z)$ be the density function of $Z \sim N(0,2^2)$. Calculate $f(0)$. (2) \end{description} %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% \newpage \begin{flushright} \underline{Full Name (Print):~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~} \end{flushright} \noindent \textbf{4.} (13 points; 1 pt each for A-G, 2 pts for H, 4 pts for I) \\ There are three cards: `1', `2' and `3'. Let $X$ be the number on the card when we randomly pick up one, then $EX =$ \fbox{A}. Since $E [X^2] = 1^2 \times \frac{1}{3} + 2^2 \times \frac{1}{3} + 3^2 \times \frac{1}{3} = \frac{14}{3}$, $Var X =$ \fbox{B}. Next, randomly pick up a card three times with replacement (so we are able to pick up the same card more than once). The average number $\bar{X}$ of the three times is between 1 and 3, and its probability distribution is in the table below. The variance of $\bar{X}$ is \fbox{C}. \begin{table}[h] \begin{center} \begin{tabular}{|c|c|c|c|c|c|c|c|} \hline average & 1 & 4/3 & 5/3 & 2 & 7/3 & 8/3 & 3 \\ \hline probability & $\frac{1}{27}$ & \fbox{D} & \fbox{E} & \fbox{F} & \fbox{E} & \fbox{D} & \fbox{G} \\ \hline \end{tabular} \end{center} \end{table} Finally when we randomly pick up a card 600 times with replacement, the variance of the average number $\bar{X}$ is \fbox{H}. Since the distribution of $\bar{X}$ is approximately normal by the central limit theorem, $P(1.9 \le \bar{X} \le 2.1)$ is approximately \fbox{I}. \begin{table}[h] {\Large \centering \begin{tabular}{|l|l|l|} %\begin{tabular*}{6.4in}{@{\extracolsep{\fill}}|l|l|l|} \hline A: ~~~~~~~~~~~~~~~~~~~~~~~~ & B: ~~~~~~~~~~~~~~~~~~~~~~~~ & C: ~~~~~~~~~~~~~~~~~~~~~~~~ \\ [+30pt] \hline D: ~~~~~~~~~~~~~~~~~~~~~~~~ & E: ~~~~~~~~~~~~~~~~~~~~~~~~ & F: ~~~~~~~~~~~~~~~~~~~~~~~~ \\ [+30pt] \hline G: ~~~~~~~~~~~~~~~~~~~~~~~~ & H: ~~~~~~~~~~~~~~~~~~~~~~~~ & I: ~~~~~~~~~~~~~~~~~~~~~~~~~~ \\ [+30pt] \hline \end{tabular} } \end{table} %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% \newpage \begin{flushright} \underline{Full Name (Print):~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~} \end{flushright} \noindent \textbf{5.} (11 points) \\ A random sample of 121 human body temperatures had a mean of 97.7 degrees and a standard deviation of 0.9 degrees. \begin{description} \item[(a)] Find the value $a$ such that $P(Z > a) = 0.02$ from a normal distribution table. (2) \item[(b)] Estimate the 96\% confidence interval for the population mean of body temperatures $\mu$. (4) \item[(c)] When we construct 50 confidence intervals from 50 random samples (you may assume each sample has different 121 observations) in the same way as (b), how many of them include $\mu$ in expectation? (2) \item[(d)] When we have 3 confidence intervals in the same way as (c), what is the probability that all 3 confidence intervals include $\mu$? (Answer three places of decimals, e.g. 0.123.) (3) \end{description} %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% \newpage \begin{flushright} \underline{Full Name (Print):~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~} \end{flushright} \noindent \textbf{6.} (11 points) \\ We want to see the effectiveness of two pain relievers: A and B, and obtained the following data. \begin{table}[h] \begin{center} \begin{tabular}{|c|cc|} \hline & A & B \\ \hline sample size & 300 & 400 \\ \hline the number of effective cases & 250 & 320 \\ \hline \hline true proportion of effective cases & $p_1$ & $p_2$ \\ \hline \end{tabular} \end{center} \end{table} \begin{description} \item[(a)] Let $\hat{p}_1$ be the sample proportion for the effective cases for drug A with sample size 300, and $\hat{p}_2$ for drug B with sample size 400. Estimate the standard error of $\hat{p}_1 - \hat{p}_2$, based on the above data. (3) \item[(b)] Explain why $\hat{p}_1 - \hat{p}_2$ is approximately normal. (5) \item[(c)] Estimate the 90\% confidence interval for $p_1 - p_2$. (3) \end{description} %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% \end{document} %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%