Answer: Using the Wilcoxon rank-sum test with women as the first sample, the sum of the women’s ranks is \(95\).

Under \(H_0\), \(E(R)=132\) and \(\operatorname{Var}(R)=264\).

So, using a continuity correction,

\(z=\frac{95.5-132}{\sqrt{264}}\approx -2.246\).

For a two-tailed test at the \(5\%\) level, the critical values are \(\pm 1.96\).

Since \(-2.246\lt -1.96\), reject \(H_0\).

There is sufficient evidence at the \(5\%\) level that the cholesterol levels of women and men differ.

Let the women’s sample be sample 1, with \(n_1=11\), and the men’s sample be sample 2, with \(n_2=12\).

We test:

\(H_0\): the two populations have the same median cholesterol level.

\(H_1\): the two populations have different median cholesterol levels.

This is a two-sided Wilcoxon rank-sum test.

First combine all 23 observations and rank them from smallest to largest.

There are no tied values.

Sum of ranks for women:

\(R=1+2+4+6+7+8+9+13+14+15+16=95\).

Since both sample sizes are greater than 10, we use the normal approximation.

For the rank sum \(R\), under \(H_0\),

\(\mu_R=\frac{n_1(n_1+n_2+1)}{2}=\frac{11(24)}{2}=132\).

Also,

\(\sigma_R^2=\frac{n_1n_2(n_1+n_2+1)}{12}=\frac{11\times 12\times 24}{12}=264\).

So \(\sigma_R=\sqrt{264}\).

The observed rank sum \(95\) is below the mean \(132\), so with a continuity correction we use \(95.5\):

\(z=\frac{95.5-132}{\sqrt{264}}\approx \frac{-36.5}{16.248}\approx -2.246\).

For a two-tailed test at the \(5\%\) significance level, the critical values are \(\pm 1.96\).

Since \(-2.246\lt -1.96\), the test statistic lies in the critical region, so we reject \(H_0\).

Therefore there is sufficient evidence to conclude that the population median cholesterol levels differ between women and men.