91Ó°ÊÓ

(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.

b).

From the example \(1.4\) (in example 1.3 there are no aspirin treatments thus assume that there is a typing mistake), the given are

\(m = 11,034,n = 11,037,x = 189,y = 104\)

Wherem stands for control group, n stands for aspirin group,x is number of people who had heart attack from control group, and y is number of people who had heart attack from aspirin group. From

\(\hat \theta = \frac{{{{\hat p}_1}}}{{{{\hat p}_2}}}\)

Where

\({\hat p_1} = \frac{{189}}{{11,034}} = 0.017129\)

And

\({\hat p_2} = \frac{{104}}{{11,037}} = 0.009423\)

Thus the estimate is\(\hat \theta = \frac{{0.0171}}{{0.0094}} = 1.818\)

\(\begin{array}{l}{\rm{ a}}{\rm{. }}\ln \hat \theta \pm {z_{\alpha /2}} \cdot \sqrt {\frac{{m - x}}{{mx}} + \frac{{n - y}}{{ny}}} ;\exp \;\;\;\ln \hat \theta \pm {z_{\alpha /2}} \cdot \sqrt {\frac{{m - x}}{{mx}} + \frac{{n - y}}{{ny}}} ;{\rm{ }}\\{\rm{ a}}{\rm{. }}(1.43,2.32)\end{array}\)

b) \(\hat \theta = \frac{{0.0171}}{{0.0094}} = 1.818\)

Aspirin appears to be beneficial.