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Among 5,924 patients treated with Eliquis, 153 developed the adverse reaction of nausea.

The null hypothesis is written as follows:

The proportion of patients who suffered from the adverse reaction is equal to 3%.

$H_0:p=0.03$

The alternative hypothesis is written as follows:

The proportion of patients who suffered from the adverse reaction is not equal to 3%.

$H_1:p鈮�0.03$

The test is two-tailed.

The sample proportion of patients who suffered from the adverse reaction is as follows:

$$\begin{gathered} \hat{p}=\frac{\text{Number}\text{of}patients\text{who}\text{developedadverse}\text{reaction}}{\text{Total}\text{number}\text{of}\text{patients}} \\ =\frac{153}{5924} \\ =0.0258 \end{gathered}$$

The population proportion of patients who developed an adverse reaction is equal to p=0.03.

The sample size (n) is equal to 5924.

The value of the test statistic is computed below:

$$\begin{gathered} z=\frac{\hat{p}-p}{\sqrt{\frac{pq}{n}}} \\ =\frac{0.0258-0.03}{\sqrt{\frac{0.03\left(1-0.03\right)}{5924}}} \\ =-1.895 \end{gathered}$$

Thus, z=-1.895.

Referring to the standard normal distribution table, the critical value of z at $伪=0.05$for a two-tailed test is equal to 1.96.

Referring to the standard normal distribution table, the p-value for the two-tailed test using absolute test statistic (1.895) is equal to 0.0581.

Since the p-value is greater than 0.05, the null hypothesis is failed to reject.

There is not enough evidence to reject the claim that the proportion of patients who developed an adverse reaction is equal to 0.03.

Only 2.6% of the subjects developed nausea as an adverse reaction of even less than 3%. Thus, it can be said that nausea does not appear to be a serious problem.