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A contingency table is constructed that shows that the number of patients that get/not get an infection when different treatments to reduce cholesterol are administered to them.

The null hypothesis is as follows:

Getting an infection is independent of the treatment.

The alternative hypothesis is as follows:

Getting an infection is not independent of the treatment.

It is considered a right-tailed test.

If the test statistic value is greater than the critical value, then the null hypothesis is rejected, otherwise not.

Observed Frequency:

The frequencies provided in the table are the observed frequencies. They are denoted by\(O\).

The following contingency table shows the observed frequency for each cell:

Expected Frequency:

The formula for computing the expected frequency for each cell is shown below:

\(E = \frac{{\left( {row\;total} \right)\left( {column\;total} \right)}}{{\left( {grand\;total} \right)}}\)

The row total for the first row is computed below:

\(\begin{aligned}{c}Row\;Tota{l_1} = 27 + 89 + 8 + 7\\ = 131\end{aligned}\)

The row total for the second row is computed below:

\(\begin{aligned}{c}Row\;Tota{l_2} = 243 + 774 + 71 + 87\\ = 1175\end{aligned}\)

The column total for the first column is computed below:

\(\begin{aligned}{c}Column\;Tota{l_1} = 27 + 243\\ = 270\end{aligned}\)

The column total for the second column is computed below:

\(\begin{aligned}{c}Column\;Tota{l_2} = 89 + 774\\ = 863\end{aligned}\)

The column total for the third column is computed below:

\(\begin{aligned}{c}Column\;Tota{l_3} = 8 + 71\\ = 79\end{aligned}\)

The column total for the fourth column is computed below:

\(\begin{aligned}{c}Column\;Tota{l_4} = 7 + 87\\ = 94\end{aligned}\)

The grand total can be computed as follows:

\(\begin{aligned}{c}Grand\;Total = \left( {131 + 1175} \right)\\ = \left( {270 + 863 + 79 + 94} \right)\\ = 1306\end{aligned}\)

Thus, the following table shows the expected frequencies for each of the corresponding observed frequencies:

The test statistic is computed as shown below:

\[\begin{aligned}{c}{\chi ^2} = \sum {\frac{{{{\left( {O - E} \right)}^2}}}{E}\;\;\; \sim } {\chi ^2}_{\left( {r - 1} \right)\left( {c - 1} \right)}\\ = \frac{{{{\left( {27 - 27.083} \right)}^2}}}{{27.083}} + \frac{{{{\left( {89 - 86.564} \right)}^2}}}{{86.564}} + ..... + \frac{{{{\left( {87 - 84.571} \right)}^2}}}{{84.571}}\\ = 0.773\end{aligned}\]

Thus, \({\chi ^2} = 0.773\).

Let r denote the number of rows in the contingency table.

Let c denote the number of columns in the contingency table.

The degrees of freedom are computed below:

\(\begin{aligned}{c}df = \left( {r - 1} \right)\left( {c - 1} \right)\\ = \left( {2 - 1} \right)\left( {4 - 1} \right)\\ = 3\end{aligned}\)

The critical value of\({\chi ^2}\)for 3 degrees of freedom at 0.01 level of significance for a right-tailed test is equal to 11.3449.

The corresponding p-value is approximately equal to 0.855914.

Since the value of the test statistic is less than the critical value and the p-value is greater than 0.05, the null hypothesis is failed to reject.

There is not enough evidence to warrant rejection of the claim thatgetting an infection is independent of the treatment.

No, atorvastatin treatment does not have an effect on infection because getting infected does not depend on the type of treatment administered.