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There are two subgroups. Denote with \({p_1}\)the true proportion of patients which did not receive counseling, and with \({p_2}\)the true proportion of the patients which received counseling about the possible side effect.

For the no-counseling subgroup (first subgroup) 15.3 % of 52 who received no counseling, and for the second subgroup, the subgroup of patients who received counseling 43.6% of 55 reported one or more sexual side effects.

The 15.3 % of 52 is

\(m = 52 \times \frac{{15.3}}{{100}} = 8\)

and 43.6 % of 55 is

\(n = 55 \times \frac{{43.6}}{{100}} = 24\)

This means that 8 out of 52 patients who received no counseling reported sexual side effect, and 25 out of 55 patients who received counseling reported sexual side effect.

The hypotheses of interest are \({H_0}:{p_1} - {p_2} = 0\) versus\({H_a}:{p_1} - {p_2} < 0\). Even though the samples are not large enough (and the proportions), you can use the large sample test procedure (see the hint).

A Large-Sample Test Procedure:

The two proportion z statistic - Consider test in which\({H_0}:{p_1} - {p_2} = 0\). The test statistic value for the large samples is

\(z = \frac{{{{\hat p}_1} - {{\hat p}_2}}}{{\sqrt {\hat p\hat q \times \left( {\frac{1}{m} + \frac{1}{n}} \right)} }}\)

where

\(\hat p = \frac{m}{{m + n}} \times {\hat p_1} + \frac{n}{{m + n}} \times {\hat p_2}\)

Depending on alternative hypothesis, the P value can be determined as the corresponding area under the standard normal curve. The test should be used when\(m{\hat p_1},m{\hat q_1},n{\hat p_1}\), and \(n{\hat q_1}\) are at least 10.

The estimates of the proportions are

\(\begin{array}{l}{{\hat p}_1} = \frac{8}{{52}} = 0.153\\{{\hat p}_2} = \frac{{24}}{{55}} = 0.436\end{array}\)

Compute a pooled proportion \(\widehat p\)

\(\begin{array}{c}\hat p = \frac{m}{{m + n}} \times {{\hat p}_1} + \frac{n}{{m + n}} \times {{\hat p}_2}\\ = \frac{{52}}{{52 + 55}} \times \frac{8}{{52}} + \frac{{55}}{{52 + 55}} \times \frac{{24}}{{55}}\\ = 0.299.\end{array}\)

The two proportion z test value is

\(\begin{array}{c}z = \frac{{{{\hat p}_1} - {{\hat p}_2}}}{{\sqrt {\hat p\hat q \times \frac{1}{m} + \frac{1}{n}} }}\\ = \frac{{0.153 - 0.436 - 0}}{{\sqrt {(0.299 \times 0.701) \times \frac{1}{{52}} + \frac{1}{{55}}} }}\\ = - 3.2\end{array}\)

The P value is the area under the standard normal curve to the left of z because the test is lower sided (one sided). Thus

\(P = P(Z \le - 3.2) = 0.007\)

which was obtained using the table in the appendix of the book. Since

\(P = 0.007 < 0.05 = \alpha \)

Reject null hypothesis

At given significance level. You can conclude that a higher proportion of men will experience erectile dysfunction if they received consulting about the side effect of the BPH treatment, than if there were no consulting about the side effects.

Or there does appear to be nocebo effect.