When blood levels are low at an area hospital, a call goes out to local residents to give blood. The blood center is interested in determining which sex - males or females - is more likely to respond. Random, independent samples of 60 females and 100 males were each asked if they would be willing to give blood when called by a local hospital. A success is defined as a person who responds to the call and donates blood. The goal is to compare the percentage of the successes of the male and female responses. Suppose 45 of the females and 60 of the males responded that they were able to give blood. Find the test statistic that would be used if it is desired to test to determine if a difference exists between the proportion of the females and males who responds to the call to donate blood.
A) $\mathrm { z } = 1.93$
B) $z = 1.645$
C) $\mathrm { z } = 2.01$
D) $\mathrm { z } = 1.96$

Question

Quizplus · Accepted Answer

The test statistic for comparing two proportions is calculated using the formula: $$z = \frac{\hat{p}_1 - \hat{p}_2}{\sqrt{p(1-p)(\frac{1}{n_1} + \frac{1}{n_2})}}$$ where $\hat{p}_1$ and $\hat{p}_2$ are the sample proportions of successes (in this case, those willing to donate blood) for females and males respectively, $n_1$ and $n_2$ are the sample sizes for females and males, and $p$ is the combined proportion of successes. Given $\hat{p}_1 = \frac{45}{60}$, $\hat{p}_2 = \frac{60}{100}$, $n_1 = 60$, $n_2 = 100$, and $p = \frac{45+60}{60+100}$, plugging these values into the formula yields a test statistic $z = 1.93$.

When Blood Levels Are Low at an Area Hospital, a Call