Question

After once again losing a football game to the archrival, a college’s alumni association conducted a survey to see if alumni were in favor of firing the

coach. An SRS of 100 alumni from the population of all living alumni was taken, and 64 of the alumni in the sample were in favor of firing the coach.

Suppose you wish to see if a majority of living alumni are in favor of firing the coach. The appropriate test statistic is

a. $z=\frac{0.64-0.5}{\sqrt{{\displaystyle \frac{0.64(0.36)}{100}}}}$

b.$t=\frac{0.64-0.5}{\sqrt{{\displaystyle \frac{0.64(0.36)}{100}}}}$

c.$z=\frac{0.64-0.5}{\sqrt{{\displaystyle \frac{0.5(0.5)}{100}}}}$

d.$z=\frac{0.64-0.5}{\sqrt{{\displaystyle \frac{0.64(0.36)}{64}}}}$

e.$z=\frac{0.5-0.64}{\sqrt{{\displaystyle \frac{0.5(0.5)}{100}}}}$

Short answer

Option a is correct

Step-by-step solution

Step 1. Test statsitic 

The test statistic for the population proportion is as follows:

$z=\frac{p-\widehat{p}}{\sqrt{\frac{p\left(1-p\right)}{100}}}$

where $p=0.64$, role="math" localid="1654256041572" $\widehat{p}=0.5$and $n=100$.

Step 2. Substitute the value in the test statistic.

Substitute the respective values in the test statistic.

$\begin{array}{c}z=\frac{0.64-0.5}{\sqrt{\frac{0.64\left(1-0.64\right)}{100}}}\\ =\frac{0.64-0.5}{\sqrt{\frac{0.64\times 0.36}{100}}}\end{array}$

So, option a is the correct option.