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\begin{center}\huge MATH 1113 - EXAM 3 SPRING 2017\\
SOLUTION\\
\end{center}

\begin{flushleft}\vspace{0.5in}
Friday 21 April 2017 \\
Instructor: Tom Cuchta
\end{flushleft}

\textbf{Instructions:}
\begin{itemize}
	\item  Show all work, clearly and in order, if you want to get full
	credit. If you claim something is true \textbf{you must show work backing up your claim}. I reserve the right to take off points if I cannot see how you arrived at your answer (even if your final answer is correct).
	\item Justify your answers algebraically whenever possible to ensure full credit. 	
	\item  Circle or otherwise indicate your final answers.
	\item  Please keep your written answers brief; be clear and to the point.
	\item  Good luck!
\end{itemize}
\vspace*{10pt}

\begin{minipage}{0.5\textwidth}
\begin{tabular}{|l|}
\hline
$E=z_c \sigma_{\bar{x}}; \sigma_{\bar{x}}=\dfrac{\sigma}{\sqrt{n}}; \bar{x}-E < \mu < \bar{x}+E$\\
\hline
$E=t_c \dfrac{s}{\sqrt{n}}; \bar{x}-E < \mu < \bar{x}+E$\\
\hline 
$E=z_c\sqrt{\dfrac{\hat{p}\hat{q}}{n}}; \hat{p}-E < p < \hat{p}+E$ \\
\hline
$\dfrac{(n-1)s^2}{\chi_R^2} < \sigma^2 < \dfrac{(n-1)s^2}{\chi_L^2}$ \\
\hline
$\sqrt{\dfrac{(n-1)s^2}{\chi_R^2}} < \sigma < \sqrt{\dfrac{(n-1)s^2}{\chi_L^2}}$ \\
\hline
\end{tabular}
\end{minipage}
\begin{minipage}{0.5\textwidth}
\begin{tabular}{|l|}
\hline
$z=\dfrac{\bar{x}-\mu}{\sigma_{\bar{x}}}; \sigma_{\bar{x}}=\dfrac{\sigma}{\sqrt{n}}$ \\
\hline
$t=\dfrac{\bar{x}-\mu}{s_{\bar{x}}}; s_{\bar{x}}=\dfrac{s}{\sqrt{n}}$\\
\hline
$z=\dfrac{\hat{p}-p}{\sigma_{\hat{p}}}; \sigma_{\hat{p}}=\sqrt{\dfrac{pq}{n}}$ \\
\hline
$\chi^2 = \dfrac{(n-1)s^2}{\sigma^2}$ \\
\hline
$z=\dfrac{(\bar{x}_1-\bar{x}_2)-(\mu_1-\mu_2)}{\sigma_{\bar{x}_1,\bar{x}_2}}; \sigma_{\bar{x}_1,\bar{x}_2}=\sqrt{\dfrac{\sigma_1^2}{n_1} + \dfrac{\sigma_2^2}{n_2}} $ \\
\hline
$t=\dfrac{(\bar{x}_1-\bar{x}_2)-(\mu_1-\mu_2)}{s_{\bar{x}_1,\bar{x}_2}}$: \\
\underline{Case I (variances equal)}: \\
$s_{\bar{x}_1,\bar{x}_2}=\hat{\sigma}\sqrt{\dfrac{1}{n_1}+\dfrac{1}{n_2}}; \hat{\sigma}=\sqrt{\dfrac{(n_1-1)s_1^2+(n_2-1)s_2^2}{n_1+n_2-2}}$; \\
d.f. $= n_1+n_2-2$ \\
\underline{Case II (variances not equal)}: \\
$s_{\bar{x}_1,\bar{x}_2}=\sqrt{\dfrac{s_1^2}{n_1} + \dfrac{s_2^2}{n_2}}$\\
d.f. $=$ smaller of $n_1-1$ and $n_2-1$\\
\hline
\end{tabular}
\end{minipage}

\end{coverpages}

\begin{questions}

\question[10] Construct a $90\%$ confidence interval for the population mean: from a random sample of 48 days in a recent year, U.S. gasoline prices had a mean of $\$3.43$. Assume the population standard deviation is $\$0.19$. \\
\begin{parts}
\part[5] Summarize the data in this problem by filling out the table below: \\
\textit{Solution:}
\begin{center}\begin{tabular}{|l|l|}
\hline
$c$ & $= 0.9$ \\
(Critical value) $z_c$ & $=1.645$  \\
$n$ & $=48$ \\
$\bar{x}$ & $= 3.43$ \\
$\sigma$ & $= 0.19$ \\
$E$ & $=0.045$ \\
\hline
\end{tabular}\end{center}~\\

Notice that since $n\geq 30$ and this is a confidence interval for population mean, $$E=\dfrac{z_c\sigma}{\sqrt{n}}=\dfrac{1.645\left(0.19\right)}{\sqrt{48}}=0.04511.$$ 
\part[5] (Fill in the \textbf{three} blanks) The confidence interval is:\\
\textit{Solution:}
\begin{center} $3.38<\mu<3.48$ \end{center}
From our work above we see that the confidence interval for the population mean is given by $\bar{x}-E<\mu<\bar{x}+E$. Thus, we have $3.43-0.045<\mu<3.43+0.045$, or equivalently, $3.38<\mu<3.48.$
\end{parts}

%\question[] Construct a $95\%$ confidence interval for the population mean: in a random sample of $11$ people, the mean commute time to work was $33.9$ minutes and the sample standard deviation was $7.2$ minutes.
\newpage
\question[10] Construct a $95\%$ confidence interval for the population proportion: in a survey of $1782$ U.S. adults, $659$ think that air travel is much more reliable than taking cruises. 
\begin{parts}
\part[5] Summarize the data in this problem by filling out the table below: \\
\textit{Solution:}
\begin{center}\begin{tabular}{|l|l|}
\hline
$c$ & $= 0.95$ \\
(Critical value) $z_c$ & $=1.96$  \\
$n$ & $=1782$ \\
$\hat{p}$ & $=\dfrac{659}{1782}=0.369$ \\
$E$ & $=0.0224 $ \\
\hline
\end{tabular}\end{center}~\\

Notice that since this is a confidence interval for a proportion, $$E=z_c\sqrt{\dfrac{\hat{p}\hat{q}}{n}}=1.96\sqrt{\dfrac{0.369(1-0.369)}{1782}}=0.0224.$$

\part[5] (Fill in the \textbf{three} blanks) The confidence interval is:\\
\textit{Solution:}
\begin{center} $0.347<\hat{p}<0.391$ \end{center}
From our work above we see that the confidence interval for the population proportion is given by $\hat{p}-E<p<\hat{p}+E$. Thus, we have $0.369-0.0224<p<0.369+0.0224$, or equivalently, $0.347<p<0.391$.
\end{parts}

%\question[] The average annual precipitations (in inches) of a random sample of of 30 years in Citytown have a sample standard deviation is $6.32$. 
%\begin{parts}
%\part[] Construct a $90\%$ confidence interval for the population variance.
%\part[] Use your result in part (a) to write a $90\%$ confidence interval for the population standard deviation.
%\end{parts}
%\hrule
\vfill
\newpage
\question[16] A fast food restaurant estimates that the mean sodium content in one of its breakfast sandwiches is no more than 920 milligrams. A random sample of 44 breakfast sandwiches has a mean sodium content of 925 milligrams. Assume the population standard deviation is 18 milligrams. At $\alpha=0.10$, do you have enough evidence to reject the restaurant's claim? 
\begin{parts}
\part[4] \textbf{Identify the claim} and the hypotheses $H_0$ and $H_a$\\
\textit{Solution:}
$$\left\{ \begin{array}{ll}
H_0: & \mu\leq 920 \text{ \textit{(claim)}}\\
H_a: & \mu> 920
\end{array} \right.$$
\part[3] Summarize the data in this problem by filling out the table below: \\
\textit{Solution:}
\begin{center}\begin{tabular}{|l|l|}
\hline
$\alpha$ & $= 0.10$ \\ 
\hline
$n$ & $=44$ \\ 
\hline
$\bar{x}$ & $=925$ \\
\hline
$\sigma$ & $=$ 18 \\
\hline
\end{tabular}\end{center}
\part[2] The critical value is $z_0=1.28$. What is the rejection region?\\
\textit{Solution:}

Since this is a right-tailed test, the rejection region is $z>z_c$, that is, $z>1.28$.

\part[4] Compute the relevant test statistic.\\
\textit{Solution:}

Since this is a test about the population mean and $n\geq 30$, we use the $z$-test. Thus,
\begin{align*}
z&=\dfrac{\bar{x}-\mu}{\sigma_{\bar{x}}}\\
&=\dfrac{925-920}{18/\sqrt{44}}\\
&=1.84.
\end{align*}

\part[2] Circle one: \quad \fbox{Reject $H_0$} \quad \quad Fail to reject $H_0$ 

\part[1] Write a sentence to interpret the decision made in part (e) for the claim.\\
\textit{Solution:}

There is not sufficient evidence that the mean sodium content in the breakfast sandwiches is less than or equal to 920 mg. 
\end{parts}

\newpage

\question[16] A company claims that the mean battery life of their cell phone is at least 30 hours. You suspect this claim is incorrect and find that a random sample of 18 cell phones has a mean battery life of 28.5 hours and a standard deviation of 1.7 hours. Is there enough evidence to reject the claim at $\alpha=0.01$? The data in this problem is summarized in the following table:

\begin{center}\begin{tabular}{|l|l|}
\hline
$\alpha$ & $= 0.01$ \\ 
\hline
$n$ & $=$ 18 \\ 
\hline
$\bar{x}$ & $=$ 28.5 \\
\hline
$s$ & $=$ 1.7 \\
\hline
\end{tabular}\end{center}

\begin{parts}

\part[4] \textbf{Identify the claim} and the hypotheses $H_0$ and $H_a$\\
\textit{Solution:}
$$\left\{ \begin{array}{ll}
H_0: & \mu\geq 30 \text{\textit{ (claim)}}\\
H_a: & \mu<30
\end{array} \right.$$
\part[4] What is the critical value for this problem? What is the rejection region?\\
\textit{Solution:}
We must use the $t$-distribution, with $d.f.=17$. With $\alpha=0.01$, $t_0=-2.567$. Since this is a left-tailed test, the rejection region is $t<-2.567$. 

\part[4] Compute the relevant test statistic.\\
\textit{Solution:}
Since this is a test about the population mean using the $t$-distribution,
\begin{align*}
	t&=\dfrac{\bar{x}-\mu}{s/\sqrt{n}}\\
	&=\dfrac{28.5-30}{1.7/\sqrt{18}}\\
	&=-3.74
\end{align*}

\part[2] Circle one: \quad \fbox{Reject $H_0$} \quad \quad Fail to reject $H_0$ 

\part[2] Write a sentence to interpret the decision made in part (d) for the claim.\\
\textit{Solution:}
There is sufficient evidence to reject the claim that the mean battery life of the cell phone is at least 30 hours.
\end{parts}

\newpage

%\question[] 7.4
%\begin{parts}
%\part[] Identify the claim and the hypotheses $H_0$ and $H_a$: \\
%\part[] Find the critical value(s) and identify the rejection region.
%\part[] Compute the relevant test statistic.
%\part[] Circle one: \quad Reject $H_0$ \quad \quad Fail to reject $H_0$ 
%\part[] Write a sentence to interpret the decision made in part (d) for the claim.
%\end{parts}

\question[16] A hospital spokesperson claims that the standard deviation of the waiting times experienced by patients in its minor emergency department is no more than $30$ seconds. A random sample of 25 waiting times has a standard deviation of $21$ seconds. At $\alpha=0.10$, is there enough evidence to reject the agent's claim?

\begin{parts}
\part[4] \textbf{Identify the claim} and the hypotheses $H_0$ and $H_a$\\
\textit{Solution:}
$$\left\{ \begin{array}{ll}
H_0: & \sigma\leq 30 \text{\textit{ (claim)}}\\
H_a: & \sigma>30
\end{array} \right.$$

\part[1] Summarize the data in this problem by filling out the table below: \\
\textit{Solution:}
\begin{center}\begin{tabular}{|l|l|}
\hline
$\alpha$ & $= 0.10$ \\ 
\hline
$n$ & $=$ 25 \\ 
\hline
$s$ & $=$ 21\\
\hline
\end{tabular}\end{center}

\part[3] What is the rejection region?\\
\textit{Solution:}
Since this is a test about the population standard deviation, we use the $\chi^2$-distribution with 24 df and significance level 0.10. Further, since this is a right-tailed test, we find that $\chi_R^2=33.196$ so that the rejection region is $\chi^2>33.196.$

\part[4] Compute the relevant test statistic.\\
\textit{Solution:}
\begin{align*}
	\chi^2&=\dfrac{(n-1)s^2}{\sigma^2}\\
		&=\dfrac{24(21^2)}{30^2}\\
		&=11.76
\end{align*}
\part[2] Circle one: \quad Reject $H_0$ \quad \quad \fbox{Fail to reject $H_0$} 


\part[2] Write a sentence to interpret the decision made in part (e) for the claim.\\
\textit{Solution:} There is not sufficient evidence to reject the claim that the standard deviation of waiting times is no more than 30 seconds.
\end{parts}

\newpage

\question[16] To compare braking distances for two types of tires, a safety engineer conducts 35 braking tests for each type. The mean breaking distance for Type A is 42 feet. Assume the population standard deviation for Type A is 4.7 feet. The mean braking distance for Type B is 45 feet. Assume the population standard deviation for Type B is 4.3 feet. At $\alpha=0.10$, can the engineer support the claim that the mean braking distances for Type A tires are larger than the mean braking distances for Type B tires? The data for this problem is summarized in the following tables:\\

\hspace*{\fill}
\begin{minipage}{0.25\textwidth}
\begin{tabular}{|l|l|}
\hline
Type A $(\mu_1)$\\
\hline
$n_1= 35$ \\ 
$\bar{x}_1=42$ \\
$\sigma_1=4.7$ \\
\hline
\end{tabular}
\end{minipage}\begin{minipage}{0.25\textwidth}
\begin{tabular}{|l|l|}
\hline
Type B $(\mu_2)$\\
\hline
$n_2= 35$ \\ 
$\bar{x}_2=45$ \\
$\sigma_2=4.3$ \\
\hline
\end{tabular}
\end{minipage}
\hspace*{\fill}

\begin{parts}

\part[4] \textbf{Identify the claim} and the hypotheses $H_0$ and $H_a$\\
\textit{Solution:}
$$\left\{ \begin{array}{ll}
H_0: & \mu_1\leq \mu_2\\
H_a: & \mu_1>\mu_2 \text{\textit{ (claim)}}
\end{array} \right.$$
where $\mu_1$ is the mean breaking distance for tire Type A and $\mu_2$ is the mean breaking distance for tire Type B.

\part[3] What is the rejection region?\\
\textit{Solution:}
Since we have the population standard deviations, we can use the $z$-distribution. With $\alpha=0.1$ and a right-tailed test, we have that $z_0=1.285$ so that the rejection region is $z>1.285$.

\part[4] Compute the relevant test statistic.\\
\textit{Solution:} First,
\begin{align*}
	\sigma_{\bar{x}_1,\bar{x}_2}&=\sqrt{\dfrac{\sigma_1^2}{n_1}+\dfrac{\sigma_2^2}{n_2}}\\
								&=\sqrt{\dfrac{4.7^2}{35}+\dfrac{4.3^2}{35}}\\
								&=1.0768
\end{align*}
Then,
\begin{align*}
	z&=\dfrac{\bar{x}_1-\bar{x}_2-(\mu_1-\mu_2)}{\sigma_{\bar{x}_1,\bar{x}_2}}\\
		&=\dfrac{42-45-0}{1.0768}\\
		&=-2.786
\end{align*}

\part[3] Circle one: \quad Reject $H_0$ \quad \quad \fbox{ Fail to reject $H_0$ }

\part[2] Write a sentence to interpret the decision made in part (d) for the claim.\\
\textit{Solution:} There is not sufficient evidence to support the claim that the mean breaking distance of the Type A tires is greater than the mean breaking distance for the Type B tires.

\end{parts}

\newpage 

%\question[] A marine biologist claims that the mean length of mature pink seaperch is different in fall and winter. A sample of 26 mature pink seaperch collected in fall has a mean length of 127 millimeters and a standard deviation of 14 millimeters. A sample of 31 mature pink seaperch collected in winter has a mean length of 117 millimeters and a standard deviation of 9 millimeters. At $\alpha=0.01$, can you support the marine biologist's claim? Assume the population variances are equal. The data for this problem is summarized in the following tables: \\
%\hspace*{\fill}
%\begin{minipage}{0.25\textwidth}
%\begin{tabular}{|l|l|}
%\hline
%Fall $(\mu_1)$\\
%\hline
%$n_1= 26$ \\ 
%$\bar{x}_1=127$ \\
%$s_1=14$ \\
%\hline
%\end{tabular}
%\end{minipage}\begin{minipage}{0.25\textwidth}
%\begin{tabular}{|l|l|}
%\hline
%Winter $(\mu_2)$\\
%\hline
%$n_2= 31$ \\ 
%$\bar{x}_2=117$ \\
%$s_2=9$ \\
%\hline
%\end{tabular}
%\end{minipage}
%\hspace*{\fill}

%\begin{parts}
%\part[] \textbf{Identify the claim} and the hypotheses $H_0$ and $H_a$
%$$\left\{ \begin{array}{ll}
%H_0: & \\
%H_a: & 
%\end{array} \right.$$
%\part[] What is the rejection region?
%\part[] Compute the relevant test statistic.
%\part[] Circle one: \quad Reject $H_0$ \quad \quad Fail to reject $H_0$ 
%\part[] Write a sentence to interpret the decision made in part (d) for the claim.
%\end{parts}

%\newpage 

\question[16] A personnel director claims that the mean household income is the same in Kauai and Maui. In Kauai, a sample of 18 residents has a mean household income of $\$56,900$ and a standard deviation of $\$12,100$. In Maui, a sample of 20 residents has a mean household income of $\$57,800$ and a standard deviation of $\$8000$. At $\alpha=0.10$, can you reject the personnel director's claim? Assume the population variances are not equal. The data for this problem is summarized in the following tables: \\
\hspace*{\fill}
\begin{minipage}{0.25\textwidth}
\begin{tabular}{|l|l|}
\hline
Kauai $(\mu_1)$\\
\hline
$n_1= 18$ \\ 
$\bar{x}_1=56900$ \\
$s_1=12100$ \\
\hline
\end{tabular}
\end{minipage}\begin{minipage}{0.25\textwidth}
\begin{tabular}{|l|l|}
\hline
Maui $(\mu_2)$\\
\hline
$n_2= 20$ \\ 
$\bar{x}_2=57800$ \\
$s_2=8000$ \\
\hline
\end{tabular}
\end{minipage}
\hspace*{\fill}

\begin{parts}

\part[4] \textbf{Identify the claim} and the hypotheses $H_0$ and $H_a$\\
\textit{Solution:}
$$\left\{ \begin{array}{ll}
H_0: & \mu_1=\mu_2 \text{\textit{ (claim)}}\\
H_a: & \mu_1\neq\mu_2
\end{array} \right.$$

\part[4] What is the rejection region?\\
\textit{Solution:}
First, we note that since we do not have the population standard deviations and $n_1$ and $n_2$ are both less than 30, we must use the $t$-distribution. Further, since the population variances are assumed to be not equal, we have $18-1=17$ df. With $\alpha=0.05$ and a two-tailed test, we find that $t_0=1.740$. Thus the rejection region is $t>1.740$ or $t<-1.740$.


\part[4] Compute the relevant test statistic.
First,
\begin{align*}
	s_{\bar{x}_1,\bar{x}_2}&=\sqrt{\dfrac{s_1^2}{n_1}+\dfrac{s_2^2}{n_2}}\\
							&=\sqrt{\dfrac{(12100)^2}{18}+\dfrac{(8000)^2}{20}}\\
							&=3366.58
\end{align*}
Then,
\begin{align*}
	t&=\dfrac{\bar{x}_1-\bar{x}_2-(\mu_1-\mu_2)}{s_{\bar{x}_1,\bar{x}_2}}\\
		&=\dfrac{56900-57800-0}{3366.58}\\
		&=-0.2673
\end{align*}

\part[2] Circle one: \quad Reject $H_0$ \quad \quad \fbox{Fail to reject $H_0$ }

\part[2] Write a sentence to interpret the decision made in part (d) for the claim.\\
\textit{Solution:}
There is not sufficient evidence to reject the claim that the mean household incomes in Kauai and Maui are the same.
\end{parts}

\end{questions}

%\includepdf[pages=1-4]{tables.pdf}
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