Question

The U.S. and many other countries have recently tightened airport security. A sociologist is interested in estimating the proportion of the travelers who feel that airports are safe. The sociologist found that 375 of the 500 travelers randomly selected and interviewed indicated that the airports were safe. With an alpha level of 0.05, conduct a test to determine whether the true proportion of interest is higher than 0.7.

Answer 1

Answer:

0.0073 < 0.05, which means that we reject the null hypothesis and conclude that the true proportion of interest is higher than 0.7.

Step-by-step explanation:

Conduct a test to determine whether the true proportion of interest is higher than 0.7.

This means that the null hypothesis is: $H_{0}: p = 0.7$

And the alternate hypothesis is: $H_{a}: p > 0.7$

The test statistic is:

$z = \frac{X - \mu}{\frac{\sigma}{\sqrt{n}}}$

In which X is the sample mean, $\mu$ is the value tested at the null hypothesis, $\sigma$ is the standard deviation and n is the size of the sample.

0.7 is tested at the null hypothesis:

This means that $\mu = 0.7, \sigma = \sqrt{0.7*0.3}$

The sociologist found that 375 of the 500 travelers randomly selected and interviewed indicated that the airports were safe.

This means that $n = 500, X = \frac{375}{500} = 0.75$

Value of the z-statistic:

$z = \frac{X - \mu}{\frac{\sigma}{\sqrt{n}}}$

$z = \frac{0.75 - 0.7}{\frac{\sqrt{0.7*0.3}}{\sqrt{350}}}$

$z = 2.44$

P-value of the test:

Probability of z being larger than 2.44, that is, a proportion larger than 0.75.

This is, looking at the z-table, 1 subtracted by the pvalue of Z = 2.44. S

Z = 2.44 has a pvalue of 0.9927

1 - 0.9927 = 0.0073

0.0073 < 0.05, which means that we reject the null hypothesis and conclude that the true proportion of interest is higher than 0.7.