In: Statistics and Probability
For each test below, follow the following steps.
(1) state the hypotheses,
(2) formulate an analysis plan
(3) analyze sample data
(4) interpret the results
You may use the TI-83 to calculate the test statistic and p-value,
but be sure you indicate which test you are using.
Test III – Test a New Drug (Part 2)
Suppose the Acme Drug Company develops a new drug, designed to prevent colds. The company states that the drug is more effective for women than for men. To test this claim, they choose a simple random sample of 100 women and 200 men from a population of 100,000 volunteers.
At the end of the study, 38% of the women caught a cold; and 51% of the men caught a cold. Based on these findings, can we conclude that the drug is more effective for women than for men? Use a 0.01 level of significance.
(1) state the hypotheses
H0: The proportions of women and men getting caught with cold after medications are equal. That is p1 = p2.
Ha: The proportions of women getting caught with cold after medications is less than the proportions of men getting caught with cold after medications. That is p1 < p2.
(2) formulate an analysis plan
For this analysis, the significance level is 0.01. The test method is a two-proportion z-test as,
(3) analyze sample data
Pooled proportion, p = (p1 * n1 + p2 * n2) / (n1 + n2)
= (0.38 * 100 + 0.51 * 200) / (100 + 200)
= 0.4667
Standard error of differnce in proportions, SE =
= 0.06110128
Test statistic, z = (p1 - p2)/SE = (0.38 - 0.51) / 0.06110128 = -2.13
P-value = P(z < -2.13) = 0.0166
(4) interpret the results
Since, p-value is greater than 0.01 significance level, we fail to reject null hypothesis H0 and conclude that there is no significant evidence that the proportions of women getting caught with cold after medications is less than the proportions of men getting caught with cold after medications. Thus, there is no significant evidence that the that the drug is more effective for women than for men at 0.01 level of significance.