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

Question

Quizplus · Accepted Answer

First, we need to calculate the sample proportions for both samples: 
$$\hat{p}_1 = \frac{40}{150} = 0.267$$
$$\hat{p}_2 = \frac{30}{250} = 0.12$$
Next, we need to calculate the pooled sample proportion and the standard error: 
$$\hat{p} = \frac{40+30}{150+250} = 0.1667$$
$$SE = \sqrt{\hat{p}(1-\hat{p})\left(\frac{1}{150}+\frac{1}{250}\right)} = 0.0406$$
Now we can calculate the test statistic: 
$$Z = \frac{(\hat{p}_1 - \hat{p}_2 - 0.05)}{SE} = \frac{(0.267 - 0.12 - 0.05)}{0.0406} = 2.326$$
Since we are testing the null hypothesis that the proportion in population 1 exceeds that in population 2, we are interested in a one-tailed test with a significance level of 0.05. Consulting the standard normal distribution table or calculator, we find that the critical value for a one-tailed test at the 0.05 level is 1.645. Since our test statistic (2.326) exceeds this critical value, we reject the null hypothesis and conclude that the proportion of successes in population 1 does exceed the proportion of successes in population 2 by 0.05.

A Sample of Size 150 from Population 1 Has 40