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A sample of 11,037 male physicians was treated with aspirin. Among them, 139 experienced myocardial infarctions. Another sample of 11,034 male physicians was given placebos. Among them, 239 experienced myocardial infarctions.

It is claimed that aspirin has no effect on myocardial infarctions.

Null hypothesis: Theproportion of physicians who experienced myocardial infarctions after taking aspirin is equal to the proportion of physicians who experienced myocardial infarctions after taking the placebo.

\({H_0}:{p_1} = {p_2}\)

Alternative hypothesis: Theproportion of physicians who experienced myocardial infarctions after taking aspirin is not equal to the proportion of physicians who experienced myocardial infarctions after taking the placebo.

\({H_1}:{p_1} \ne {p_2}\)

The test is two-tailed.

Let \({\hat p_1}\) denote the sampleproportionof physicians who experienced myocardial infarctions after taking aspirin.

\(\begin{array}{c}{{\hat p}_1} = \frac{{139}}{{11037}}\\ = 0.0126\end{array}\)

Let \({\hat p_2}\) denote the sampleproportionof physicians who experienced myocardial infarctions after taking the placebo.

\(\begin{array}{c}{{\hat p}_2} = \frac{{239}}{{11034}}\\ = 0.0217\end{array}\)

The sample size of physicians who weretreated with aspirin\(\left( {{n_1}} \right)\)is equal to 11037.

The sample size of physicians who weretreated with a placebo\(\left( {{n_2}} \right)\)is equal to 11034.

The value of the pooled sample proportion is computed as follows.

\(\begin{array}{c}\bar p = \frac{{\left( {{x_1} + {x_2}} \right)}}{{\left( {{n_1} + {n_2}} \right)}}\,\\ = \frac{{\left( {139 + 239} \right)}}{{\left( {11037 + 11034} \right)}}\\ = 0.0171\end{array}\)

\(\begin{array}{c}\bar q = 1 - \bar p\\ = 1 - 0.0171\\ = 0.9829\end{array}\).

The test statistic is computed as follows.

\(\begin{array}{c}z = \frac{{\left( {{{\hat p}_1} - {{\hat p}_2}} \right) - \left( {{p_1} - {p_2}} \right)}}{{\sqrt {\left( {\frac{{\bar p\bar q}}{{{n_1}}} + \frac{{\bar p\bar q}}{{{n_2}}}} \right)} }}\\ = \frac{{\left( {0.0126 - 0.0217} \right) - 0}}{{\sqrt {\left( {\frac{{0.0171 \times 0.9829}}{{11037}} + \frac{{0.0171 \times 0.9829}}{{11034}}} \right)} }}\\ = - 5.1907\end{array}\).

The value of the test statistic is -5.1907.

Referring to the standard normal distribution table, the critical value of z corresponding to\(\alpha = 0.05\)for a two-tailed test is equal to 1.96.

Referring to the standard normal distribution table, the corresponding p-value for z equal to -5.1907 is equal to 0.000.

As the p-value is less than 0.05, the null hypothesis is rejected.

a.

There is sufficient evidence to reject the claim thataspirin has no effect on myocardial infarctions.

b.

The general formula for the confidence interval of the difference of proportions is written below.

\({\rm{Confidence}}\,\,{\rm{Interval}} = \left( {\left( {{{\hat p}_1} - {{\hat p}_2}} \right) - E\,\,,\,\,\left( {{{\hat p}_1} - {{\hat p}_2}} \right) + E} \right)\,\,\,\,\,\,\,\,...\left( 1 \right)\)\(\)

Here, E is the margin of error and has the following formula:

\(E = {z_{\frac{\alpha }{2}}} \times \sqrt {\left( {\frac{{{{\hat p}_1} \times {{\hat q}_1}}}{{{n_1}}} + \frac{{{{\hat p}_2} \times {{\hat q}_2}}}{{{n_2}}}} \right)} \).

For computing the confidence interval, first, find the critical value\({z_{\frac{\alpha }{2}}}\).

Here,\(\alpha = 0.05\).

Hence,

\(\begin{array}{c}\frac{\alpha }{2} = \frac{{0.05}}{2}\\ = 0.025\end{array}\).

The value of\({z_{\frac{\alpha }{2}}}\)from the standard normal table is equal to 1.96.

Now, the margin of error (E) is equal to

\(\begin{array}{c}E = {z_{\frac{\alpha }{2}}} \times \sqrt {\left( {\frac{{{{\hat p}_1} \times {{\hat q}_1}}}{{{n_1}}} + \frac{{{{\hat p}_2} \times {{\hat q}_2}}}{{{n_2}}}} \right)} \\ = 1.96 \times \sqrt {\left( {\frac{{0.0126 \times 0.9874}}{{11037}} + \frac{{0.0217 \times 0.9783}}{{11034}}} \right)} \\ = 0.00342\end{array}\).

Substitute the value of E in equation (1), as follows.

\(\begin{array}{c}{\rm{Confidence}}\,\,{\rm{Interval}} = \left( {\left( {{{\hat p}_1} - {{\hat p}_2}} \right) - E\,\,,\,\,\left( {{{\hat p}_1} - {{\hat p}_2}} \right) + E} \right)\\ = \left( {\left( {0.0126 - 0.0217} \right) - 0.00342\,\,,\,\,\left( {0.0126 - 0.0217} \right) + 0.00342} \right)\\ = \left( { - 0.012\,,\, - 0.006} \right)\end{array}\).

Thus, the 95% confidence interval is equal to (-0.12, -0.006).

The interval does not contain the value of 0 and contains all negative values.

This implies that the values of the two proportions cannot be equal.

There is sufficient evidence to reject the claim thataspirin has no effect on myocardial infarctions.

c.

As the confidence interval consists of negative values, the proportion of physicians who experienced myocardial infarctions after taking aspirin is less as compared to the physicians who took the placebo.

Therefore, it can be concluded that aspirin is effective in reducing the case of myocardial infarction.