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Out of 227 subjects, the number of subjects who developed nausea is equal to 52.

The null hypothesis is written as follows:

The proportion of subjects who developed nausea is equal to 20%.

$H_0:p=0.20$

The alternative hypothesis is written as follows:

The proportion of subjects who developed nausea is more than 20%.

$H_1:p>0.20$

The test is right-tailed.

The sample size is equal to n=227.

The sample proportion of subjects who developed nausea isas follows:

$$\begin{gathered} \hat{p}=\frac{52}{227} \\ =0.229 \end{gathered}$$

The population proportion of subjects who developed nausea is equal to 0.20.

The value of the test statistic is computed below:

$$\begin{gathered} z=\frac{\hat{p}-p}{\sqrt{\frac{pq}{n}}} \\ =\frac{0.229-0.20}{\sqrt{\frac{0.20\left(1-0.20\right)}{227}}} \\ =1.092 \end{gathered}$$

Thus, z=1.092.

Referring to the standard normal distribution table, the critical value of z at $伪=0.05$ for a right-tailed test is equal to 1.645.

Referring to the standard normal distribution table, the p-value for the test statistic value of 1.092 is equal to 0.1374.

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

There is not enough evidence to support the claim that the proportion of subjects who developed nausea is more than 20%.

Although the null hypothesis is failed to reject, the sample proportion of subjects who developed nausea equal to 0.229 appears to be quite high.