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\newcommand{\bd}{\displaystyle}
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{\bf \large Problem Set 7 Solution}\\
17.872 TA Jiyoon Kim\\
Nov. 26, 2003\\
\end{center}
{\bf Bulmer Exercise 10.1}\\
You need to first calculate the probability of throwing six on a white die.  This is 3932/20000.\\
\[
\begin{array}{lllll}
E(\mbox{throwing a 6 with the white die}) &=& \frac{\bd 3932}{\bd 20000} = 0.1966\\
\\
Var(\mbox{throwing a 6 with the white die}) &=& (.1966)(1-.1966) = .1579\\
\end{array}
\]
\[
\begin{array}{llll}
\mbox{95\% of confidence interval} &=& 0.1966 \pm 1.96\sqrt{\frac{.1579}{20000}}\\
\\
    &=& .1966 \pm .0055\\
    \end{array}
    \]
    \\
{\bf Bulmer Exercise 10.2}\\
(a)  Since we would like to use proportion, I will use Table 2b.\\
\[
\begin{array}{lllll}
\Pr(\mbox{stillborn in male}) &=& \frac{.012}{.5141} = .0233\\
\\
\mbox{95 \% confidence interval} &=& .0233 \pm 1.96\sqrt{\frac{.0233(1-.0233)}{368490}}\\
\\
    &=& .0233 \pm .00049\\
    \end{array}
    \]
\\
(b) In female case, \\
\[
\begin{array}{lllll}
\Pr(\mbox{stillborn in female}) &=& \frac{.0109}{.4859} = .0224\\
\\
\mbox{95 \% confidence interval} &=& .0224 \pm 1.96\sqrt{\frac{.0224(1-.0224)}{348250}}\\
\\
    &=& .0224 \pm .00049\\
    \end{array}
    \]
\\
(c)  What about gender difference?\\
\[
\begin{array}{lllll}
\Pr(\mbox{difference}) &=& \Pr_{\mbox{male}} - \Pr_{\mbox{female}} = .01\\
\\
Var(\mbox{difference}) &=& \frac{.0233(1-.0233)}{368490} + \frac{.0224(1-.0224)}{348250} \\
\\
    &=& .0000001249\\
    \\
\mbox{95 \% confidence interval} &=& .001 \pm 1.96\sqrt{.0000001249}\\
\\
    &=& .001 \pm .00069\\
    \end{array}
    \]
    \\
{\bf Bulmer Exercise 10.5}\\
\begin{center}
\begin{tabular}{|c|c|c|c|} \hline
  % after \\: \hline or \cline{col1-col2} \cline{col3-col4} ...
   & Hyoscyamine & Hyoscine & Difference \\ \hline
  $\bar{x}$ & .75 & 2.33 & 1.58 \\ \hline
  $s$ & 1.79 & 2 & 1.23 \\ \hline
  $t = \frac{\bar{x}}{s/\sqrt{10}}$ & 1.32 & 3.68 & 4.06 \\ \hline
\end{tabular}
\end{center}
(a) Hyoscyamine:\\
\[
\begin{array}{llll}
\left|\frac{\bar{x} - \mu}{s/\sqrt{10}}\right| \leq 2.262 , \left| \bar{x} - \mu \right| \leq 2.262 \times
\frac{s}{\sqrt{10}}\\
\\
\bar{x} - 2.262 \times \frac{s}{\sqrt{10}} \leq \mu \leq \bar{x} + 2.262 \times \frac{s}{\sqrt{10}}\\
\\
.75 - 2.262 \times \frac{1.79}{\sqrt{10}} \leq \mu \leq .75 + 2.262 \times \frac{1.79}{\sqrt{10}}\\
\\
.75 - 1.28  \leq \mu \leq .75 + 1.28\\
\end{array}
\]
note : This is $t$ - distribution since it is using $s$ instead $\sigma$.  You have to refer to $t$-table.  ($d.f.
= 10 - 1 = 9$)\\
\\
(b) In case of hyoscine, \\
\[
\begin{array}{llll}
2.33 - 2.262 \times \frac{2}{\sqrt{10}} \leq \mu \leq 2.33 + 2.262 \times \frac{2}{\sqrt{10}}\\
\\
2.33 - 1.43  \leq \mu \leq 2.33 + 1.43\\
\end{array}
\]
\\
(c) It is basically asking you the difference.\\
\[
\begin{array}{lllll}
1.58 - 2.262 \times \frac{1.23}{\sqrt{10}} \leq \delta \leq 1.58 + 2.262 \times \frac{1.23}{\sqrt{10}}\\
\\
1.58 - .88 \leq \delta \leq 1.58 + .88\\
\\
.7  \leq \delta \leq 2.46\\
\end{array}
\]
\\
{\bf Bulmer Exercise 10.6}\\
Now, you have to obtain confidence interval for standard deviation.  In the table, we are only given $s$ and do
not know of real or true $\sigma$.  We were hinted that $\frac{S^2}{\sigma^2} \sim \chi^2_{9 d.f.}$, which means
that $\sigma^2$ will fall into the region of $\frac{S^2}{\chi^2_{.025}}$ and $\frac{S^2}{\chi^2_{.975}}$ with 9
degrees of freedom.  ($S^2$ is sum of square of mean deviation.  that is $\sum (x_i - \bar{x})^2$.)\\
\\
\[
\begin{array}{lllll}
\chi^2_{.025} = 19.02 \& \ \chi^2_{.975} = 2.70\\
\\
S^2 = s^2(n-1) = 4 \times 9 = 36\\
\\
\frac{36}{19.02} \leq \sigma^2 \leq \frac{36}{2.7}\\
\\
1.89 \leq \sigma^2 \leq 13.33\\
\end{array}
\]
\begin{enumerate}
  \item $\sigma^2 \geq 1.89, \ \sigma \geq 1.37 \mbox{and} \sigma \leq -1.37$\\
  \\
  \item $\sigma^2 \leq 13.33, \ -3.65 \leq \sigma \leq 3.65$\\
  \\
The range that satisfies both of (1) and (2), and also standard deviation should not be negative,\\
\[
1.37 \leq \sigma \leq 3.65\\
\]
\end{enumerate}
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