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\begin{document}

\newcommand{\smean}{\overline X}
\newcommand{\Ceil}[1]{\left\lceil{#1}\right\rceil}
\newcommand{\Parens}[1]{\left({#1}\right)}
\newcommand{\df}{\mathrm{df}}

\theoremstyle{remark}
\newtheorem*{problem*}{Problem}

\newenvironment{sol}{\begin{proof}[Solution]}{\end{proof}}
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\title{HW Solutions}
\author{Melikamp}
\date{\today}
\maketitle

\section{Problems due by July 23}

All drawings are omitted. All equalities in computations should be
assumed to be approximate, even though $=$ is used.

\begin{problem*}[4.5.4]
\end{problem*}\begin{sol}
  It is given that $\sigma = 6.3$.
  \begin{enumerate}
  \item Given $n=40$ we can write
    \begin{eqnarray*}
      P(\smean < \mu+1) &=& P\Parens{Z < \frac{\mu+1-\mu}{\sigma/\sqrt n}}\\
      &=& P\Parens{Z < \frac1{6.3/\sqrt{40}}}\\
      &=& P(Z < 1)\\
      &=& 0.8413.
    \end{eqnarray*}
  \item With $n=100$ and the same $\sigma$ we can write
    \begin{eqnarray*}
      P(\smean < \mu+1) &=& P\Parens{Z < \frac{\mu+1-\mu}{\sigma/\sqrt n}}\\
      &=& P\Parens{Z < \frac1{6.3/\sqrt{100}}}\\
      &=& P(Z < 1.59)\\
      &=& 0.9441.
    \end{eqnarray*}
  \end{enumerate}
\end{sol}

\begin{problem*}[4.5.8]
\end{problem*}\begin{sol}
  It is given that $\mu=182$ and $\sigma = 14.7$.
  \begin{enumerate}
  \item Here $n=20$, and
    \begin{eqnarray*}
      P(180 < \smean < 185) &=& P(\smean<185) - P(\smean<180)\\
      &=& P\Parens{Z < \frac{185-182}{14.7/\sqrt{20}}}
      - P\Parens{Z < \frac{180-182}{14.7/\sqrt{20}}}\\
      &=& P(Z < 0.91) - P(Z < -0.61)\\
      &=& 0.8186 - 0.2709 = 0.5477.
    \end{eqnarray*}
  \item As $\mu = 170$, $\sigma=26.8$, and $n=40$, we have
    \begin{eqnarray*}
      P(180 < \smean < 185) &=& P(\smean<185) - P(\smean<180)\\
      &=& P\Parens{Z < \frac{185-170}{26.8/\sqrt{40}}}
      - P\Parens{Z < \frac{180-170}{26.8/\sqrt{40}}}\\
      &=& P(Z < 3.54) - P(Z < 2.36)\\
      &=& 1 - 0.9909 = 0.0091.
    \end{eqnarray*}
  \end{enumerate}
\end{sol}

\begin{problem*}[5.6.4]
\end{problem*}\begin{sol}
  Here $\sigma=3.4$, the desired margin of error is $E=1$, and
  $\alpha=0.1$, so
  \[n=\Ceil{\Parens{\frac{Z_{1-\alpha/2}\sigma}{E}}^2}
  =\Ceil{\Parens{\frac{1.645\cdot3.4}{1}}^2}
  =\Ceil{31.3}=32.\]
\end{sol}

\begin{problem*}[5.6.5]
\end{problem*}\begin{sol}
  Here $n=15$, $\smean=27$, $s=4.2$, $\alpha=0.1$. We use $t$ with
  $\df=14$ rather than $Z$ because the sample size is small. The
  confidence interval is given by
  \[\smean\pm t_{1-\alpha/2}\frac{s}{\sqrt n}
  = 27\pm 1.761\frac{4.2}{\sqrt{15}}
  = 27\pm 1.91,\]
  or
  \[(25.09, 28.91).\]
\end{sol}

\begin{problem*}[5.6.6]
\end{problem*}\begin{sol}
  Here $n=40$ (so we use $Z$), $\smean=14.6$, $s=2.8$, $\alpha=0.05$.
  The confidence interval is
  \[\smean\pm Z_{1-\alpha/2}\frac{s}{\sqrt n}
  = 14.6\pm1.96\frac{2.8}{\sqrt{40}}
  = 14.6\pm0.87,\]
  or
  \[(13.73, 15.47).\]
\end{sol}

\begin{problem*}[5.6.17]
\end{problem*}\begin{sol}
  $n=10$.
  \begin{enumerate}
  \item $\smean = 7.2$.
  \item $s=3.9$.
  \item Use $t$ with $\df=9$.
    \[\smean\pm t_{1-\alpha/2}\frac{s}{\sqrt n}
    = 7.2\pm2.262\frac{3.9}{\sqrt{10}}
    = 7.2\pm2.8,\]
    or \[(4.4,10).\]
  \item \[n=\Ceil{\Parens{\frac{Z_{1-\alpha/2}s}{E}}^2}
    = \Ceil{\Parens{\frac{1.96\cdot3.9}{1}}^2}
    = \Ceil{58.4} = 59.\]
  \end{enumerate}
\end{sol}

\begin{problem*}[5.6.22]
\end{problem*}\begin{sol}
  It is given that $\mu=35.2$, $\sigma=8.8$.
  \begin{enumerate}
  \item \[P(\smean>38)
    = P\Parens{Z>\frac{38-35.2}{8.8/\sqrt{50}}}
    = P(Z>2.25) = P(Z<-2.25) = 0.0122.\]
  \item
    \begin{enumerate}
    \item $H_0 : \mu=35.2$, $H_1:\mu>35.2$, $\alpha=0.05$.
    \item $Z=\frac{\smean-\mu_0}{\sigma/\sqrt n}$.
    \item Reject $H_0$ iff $Z \geq 1.645$.
    \item $Z=\frac{38-35.2}{8.8/\sqrt{50}} = 2.25$.
    \item The data provides sufficient evidence to conclude
      that the mean ratio of students to professors is higher
      than $35.2$, $\alpha=0.05$.
    \end{enumerate}
  \end{enumerate}
\end{sol}

% For testing procedures:
\renewcommand{\labelenumi}{(\roman{enumi})}

\begin{problem*}[5.6.23]
\end{problem*}\begin{sol}
  $n=48$.
  \begin{enumerate}
  \item $H_0 : \mu=18$, $H_1:\mu<18$, $\alpha=0.05$.
  \item $Z=\frac{\smean-\mu_0}{s/\sqrt n}$.
  \item Reject $H_0$ iff $Z \leq -1.645$.
  \item $Z=\frac{16.4-18}{4.1/\sqrt{48}} = -2.7$.
  \item The data provides sufficient evidence to conclude that the
    average daily iron intake for females aged under 51 is less than
    18 mg.
  \end{enumerate}
\end{sol}

\begin{problem*}[5.6.27]
  \textbf{Run a two-sided test.}
\end{problem*}\begin{sol}
  $n=15$, so we will use $t$ with $\df=14$.
  \begin{enumerate}
  \item $H_0 : \mu=3$, $H_1:\mu\neq3$, $\alpha=0.05$.
  \item $t=\frac{\smean-\mu_0}{s/\sqrt n}$, $\df=14$.
  \item Reject $H_0$ iff $t > 2.145$ or $t < -2.145$.
  \item $t = \frac{3.9-3}{0.4/\sqrt{15}} = 8.71$.
  \item The data provides sufficient evidence to conclude that the
    average time between CD4 test is significantly different from
    3 months.
  \end{enumerate}
\end{sol}

\end{document}
