Question

Suppose the Acme Drug Company develops a new drug, designed to prevent colds. The company states that the drug is equally effective for men and women. 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 reject the company's claim that the drug is equally effective for men and women? Use a 0.05 level of significance.
a) Hypothesis: (In both symbols and words)
b) Decision rule:
c) Calculation:
d) Conclusion:

Answer 1

Answer:

a) Null hypothesis: the drug is equally effective for men and women (company's claim)

Alternative hypothesis: the drug effectiveness significantly differs for men and women.

$H_0: \pi_1-\pi_2=0\\\\H_a:\pi_1-\pi_2\neq 0$

b) If the P-value is smaller than the significance level, the null hypothesis is rejected. If not, the null hypothesis failed to be rejected.

c) In the explanation.

d) As the P-value (0.0342) is smaller than the significance level (0.05), the effect is significant.

The null hypothesis is rejected.

There is enough evidence to support the claim that the drug effectiveness significantly differs for men and women.

The company's claim is rejected.

Step-by-step explanation:

This is a hypothesis test for the difference between proportions.

The claim of the company, will be stated in the null hypothesis. We will test if there is evidence against that claim to reject it or not.

Then, the test claim is that the drug effectiveness significantly differs for men and women.

The null and alternative hypothesis are:

$H_0: \pi_1-\pi_2=0\\\\H_a:\pi_1-\pi_2\neq 0$

The significance level is 0.05.

The sample 1 (women), of size n1=100 has a proportion of p1=0.38.

The sample 2 (men), of size n2=200 has a proportion of p2=0.51.

The difference between proportions is (p1-p2)=-0.13.

$p_d=p_1-p_2=0.38-0.51=-0.13$

The pooled proportion, needed to calculate the standard error, is:

$p=\dfrac{X_1+X_2}{n_1+n_2}=\dfrac{38+102}{100+200}=\dfrac{140}{300}=0.467$

The estimated standard error of the difference between means is computed using the formula:

$s_{p1-p2}=\sqrt{\dfrac{p(1-p)}{n_1}+\dfrac{p(1-p)}{n_2}}=\sqrt{\dfrac{0.467*0.533}{100}+\dfrac{0.467*0.533}{200}}\\\\\\s_{p1-p2}=\sqrt{0.002489+0.001244}=\sqrt{0.003733}=0.061$

Then, we can calculate the z-statistic as:

$z=\dfrac{p_d-(\pi_1-\pi_2)}{s_{p1-p2}}=\dfrac{-0.13-0}{0.061}=\dfrac{-0.13}{0.061}=-2.1276$

This test is a two-tailed test, so the P-value for this test is calculated as (using a z-table):

$P-value=2\cdot P(z<-2.1276)=0.0342$

As the P-value (0.0342) is smaller than the significance level (0.05), the effect is significant.

The null hypothesis is rejected.

There is enough evidence to support the claim that the drug effectiveness significantly differs for men and women.