(a):
In order to obtain a 95% large sample confidence interval you need to use facts that the standard deviations of a normally distributed random variable \(\ln \hat \theta \)
\(\begin{array}{l}is{\rm{ }}\sigma = \sqrt {\frac{{m - x}}{{mx}} + \frac{{n - y}}{{ny}}} {\rm{ and the mean value is }}\mu = \ln \theta \cdot {\rm{ }}\\{\rm{When }}\end{array}\)
Alpha=0.05, at 95 %
\(\begin{array}{l}{\rm{ confidence interval can be obtained from }}P\left( {{z_1} < \ln \hat \theta < {z_2}} \right) = 1 - \alpha {\rm{ }}\\{\rm{ and the mentioned facts above as }}\end{array}\)
\(\begin{array}{c}P\left( {{z_1} < \ln \hat \theta < {z_2}} \right) = P\;\;\left( {\;\frac{{{z_1} - \mu }}{\sigma } < \frac{{\ln \hat \theta - \mu }}{\sigma } < \frac{{{z_2} - \mu }}{\sigma }} \right)\\ = P\;\;\left( {\;\frac{{{z_1} - \mu }}{\sigma } < Z < \frac{{{z_2} - \mu }}{\sigma }} \right) = 0.95\end{array}\)
Where,
Z is standard normal distribution and the upper and lower bounds are
\[\begin{array}{l}\frac{{{z_2} - \mu }}{\sigma } = {z_{\alpha /2}}\\\frac{{{z_1} - \mu }}{\sigma } = - {z_{\alpha /2}}\end{array}\]
Thus, upper bound for a 95 % large sample confidence interval for \(\ln \theta \)is
\(\begin{array}{l}\frac{{{z_2} - \mu }}{\sigma } = {z_{\alpha /2}}\\{z_2} - \mu = {z_{\alpha /2}} \times \sigma \\{z_2} = \mu + {z_{\alpha /2}} \times \sigma \\{z_2} = \ln \hat \theta + {z_{\alpha /2}} \times \sqrt {\frac{{m - x}}{{mx}} + \frac{{n - y}}{{ny}}} \end{array}\)
and the lower bound is
\(\begin{array}{l}\frac{{{z_1} - \mu }}{\sigma } = - {z_{\alpha /2}}\\{z_1} - \mu = - {z_{\alpha /2}} \times \sigma \\{z_1} = \mu - {z_{\alpha /2}} \times \sigma \\{z_1} = \ln \hat \theta - {z_{\alpha /2}} \times \sqrt {\frac{{m - x}}{{mx}} + \frac{{n - y}}{{ny}}} \end{array}\)
Finally, the formula for a 95 % large sample confidence interval is
\(\ln \hat \theta \pm {z_{\alpha /2}} \times \sqrt {\frac{{m - x}}{{mx}} + \frac{{n - y}}{{ny}}} \)
The value needed to obtain confidence interval is the natural log
\(\begin{array}{c}\ln \hat \theta = \ln 1.818\\ \approx 0.598\end{array}\)
Remember that\({z_{\alpha /2}} = {z_{0.025}} = 1.96\)from the table in the appendix of the book. Now, the 95% confidence interval for\(\ln \theta \)is
\(\begin{array}{c}\ln \hat \theta \pm {z_{\alpha /2}} \times \sqrt {\frac{{m - x}}{{mx}} + \frac{{n - y}}{{ny}}} \\ = 0.598 \pm 1.96 \times \sqrt {\frac{{11,034 - 189}}{{11,034 \cdot 189}} + \frac{{11,037 - 104}}{{11,037 \cdot 104}}} \\ = 0.598 \pm 1.96 \times 0.1213\\ = (0.36,0.84).\end{array}\)
Taking the anti-logs of the computed\({\rm{Cl}}\), the 95 % confidence interval for \(\theta \)is\((1.43,2.32)\)
There is 95% confident that individuals who do not take aspirin treatment is about \(1.43\) to \(2.32\) times more likely to have a heart attack than from those from control group. This means that the aspirin therapy is good to reduce number of people who have heart attack.
