Using the z-distribution, as we are working with proportions, it is found that the correct option is:
b) Reject the null hypothesis concluding that there is no evidence that the proportion is higher for men.
What are the hypothesis tested?
At the null hypothesis, it is tested if the proportions are the same, that is:
[tex]H_0: p_M - p_W = 0[/tex]
At the alternative hypothesis, it is tested if it is higher for men, that is:
[tex]H_1: p_M - p_W > 0[/tex]
What is the distribution of the difference of proportions?
For each sample, we have that:
[tex]p_M = \frac{155}{254} = 0.6102, s_M = \sqrt{\frac{0.6102(0.3898)}{254}} = 0.0306[/tex]
[tex]p_W = \frac{157}{301} = 0.5216, s_W = \sqrt{\frac{0.5216(0.4784)}{301}} = 0.0288[/tex]
Then, for the distribution of differences, the mean and the standard error are given by:
[tex]\overline{p} = p_M - p_W = 0.6102 - 0.5216 = 0.0886[/tex]
[tex]s = \sqrt{s_M^2 + s_W^2} = \sqrt{0.0306^2 + 0.0288^2} = 0.042[/tex]
What is the test statistic?
It is given by:
[tex]z = \frac{\overline{p} - p}{s}[/tex]
In which p = 0 is the value tested at the null hypothesis, hence:
[tex]z = \frac{\overline{p} - p}{s}[/tex]
[tex]z = \frac{0.0886 - 0}{0.042}[/tex]
[tex]z = 2.11[/tex]
What is the decision?
Considering a right-tailed test, as we are testing if the proportion is higher than a value, with a significance level of 0.05, the critical value is of [tex]z^{\ast} = 1.645[/tex].
Since the test statistic is greater than the critical value for the right-tailed test, there is enough evidence to reject the null hypothesis that the there is no evidence that the proportion is higher for men, hence option b is correct.
More can be learned about the z-distribution at https://brainly.com/question/26454209
