Answer to Question #349330 in Statistics and Probability for Yun

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 110 women and 190 men from a population of 100,000 volunteers.

At the end of the study, 28% of the women caught a cold; and 61% 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.

Answer 1

The value of the pooled proportion is computed as

$$\bar{x}=\dfrac{X_1+X_2}{n_1+n_2}=\dfrac{\hat{p}_1n_1+\hat{p}_2n_2}{n_1+n_2}$$

$$=\dfrac{0.28(110)+0.61(190)}{110+190}=0.489$$

The following null and alternative hypotheses need to be tested:

$H_0:p_1=p_2$

$H_1:p_1\not=p_1$

This corresponds to a two-tailed test, and a z-test for two population proportions will be used.

Based on the information provided, the significance level is $\alpha = 0.05,$ and the critical value for a two-tailed test is $z_c = 1.96.$

The rejection region for this two-tailed test is $R = \{z:|z|>1.96\}.$

The z-statistic is computed as follows:

$$z=\dfrac{\hat{p}_1-\hat{p}_2}{\sqrt{\bar{p}(1-\bar{p})(1/n_1+1/n_2)}}$$

$$=\dfrac{0.28-0.61}{\sqrt{0.489(1-0.489)(1/110+1/190)}}=-5.5101$$

Since it is observed that $|z|=5.5101>1.96=z_c,$ it is then concluded that the null hypothesis is rejected.

Using the P-value approach:

The p-value is $p=2P(z<-5.5101)=0,$ and since $p=0<0.05=\alpha,$ it is concluded that the null hypothesis is rejected.

Therefore, there is enough evidence to claim that the population proportion $p_1$ is different than $p_2,$ at the $\alpha = 0.05$ significance level.