Question

A study reported that in a sample of 90 men, 27 had elevated total cholesterol levels (more than 200 milligrams per deciliter). In a sample of 110 women, 22 had elevated cholesterol levels. At the 5% level of significance, can you conclude that the proportion of people with elevated cholesterol levels for men is larger than the proportion of people with elevated cholesterol levels for women? Compute the p-value of the test.   Use at least five decimal places for the denominator during your computations.

A)0.1010

B)0.4602

C)0.2450

D)0.0505

Answer 1

Given that in a sample of n1 = 90 men, X1 = 27 had elevated total cholesterol levels (more than 200 milligrams per deciliter), and In a sample of n2 = 110 women, X2 = 22 had elevated cholesterol levels.

Thus the sample proportions are calculated as:

Here to test the claim that the proportion of people with elevated cholesterol levels for men is larger than the proportion of people with elevated cholesterol levels for women, we need to run the hypothesis using the Z-statistic.

Thus we need pooled proportion which is calculated as:

Now based on the claim the hypotheses created are:

Now based on the hypothesis there will be a right-tailed test.

Rejection region:

Reject Ho if P-value is less than 0.05.

Test statistic:

The test statistic is calculated as:

Z = 1.64

P-value:

The P-value for the right-tailed test is calculated using the excel formula for Z-distribution which is =1-NORM.S.DIST(1.64, TRUE). Thus the p-value is computed as:

D) P-value = 0.0505.

Conclusion:

Since the P-value is greater than 0.05 hence we fail to reject the null hypothesis and conclude that there is insufficient evidence to support the claim that  the proportion of people with elevated cholesterol levels for men is larger than the proportion of people with elevated cholesterol levels for women