A sample of size 150 from population 1 has 40 successes. A sample of size 250 from population 2 has 30 successes. The value of the test statistic for testing the null hypothesis that the proportion of successes in population 1 exceeds the proportion of successes in population 2 by 0.05 is:
A)1.645.
B)2.327.
C)1.960.
D)1.977.

Question

Quizplus · Accepted Answer

We are testing the difference in proportions between two independent samples, so we will use the formula for the two-sample z-test:
$$\large z=\frac{(\hat{p}_1-\hat{p}_2)-\Delta}{\sqrt{\frac{\hat{p}(1-\hat{p})}{n_1}+\frac{\hat{p}(1-\hat{p})}{n_2}}}$$

where $\hat{p}_1$ and $\hat{p}_2$ are the sample proportions, $\Delta$ is the hypothesized difference between the population proportions, $\hat{p}$ is the pooled proportion, and $n_1$ and $n_2$ are the sample sizes.

We are given $\hat{p}_1=40/150=0.267$, $\hat{p}_2=30/250=0.12$, and $\Delta=0.05$. We will assume the null hypothesis and use the pooled proportion under the null hypothesis, which is:
$$\large \hat{p}=\frac{x_1+x_2}{n_1+n_2}=\frac{40+30}{150+250}=0.146$$

Plugging in these values, we get:
$$\large z=\frac{(0.267-0.12-0.05)}{\sqrt{\frac{(0.146)(1-0.146)}{150}+\frac{(0.146)(1-0.146)}{250}}}$$

Simplify this to get:
$$\large z=2.327$$

Using a significance level of 0.05 and a two-sided test, the critical z-value is 1.96. Since our calculated z-value is greater than the critical value, we reject the null hypothesis and conclude that the proportion of successes in population 1 is significantly different from the proportion of successes in population 2.

A Sample of Size 150 from Population 1 Has 40