Question

An airline's public relations department claims that when luggage is lost, 88% is recovered and delivered to its owner within 24 hours. A consumer group who surveyed a large number of air travelers found that 138 out of 160 people who lost luggage on that airline were reunited with the missing items by the next day. At the 0.10 level of significance, do the data provide sufficient evidence to conclude that the proportion of times that luggage is returned within 24 hours is less than 0. 88? Identify the hypothesis statements you would use to test this. H0: p = 0.88 versus HA : p < 0.88 H0: p < 0.88 versus HA : p = 0.88 H0: p = 0.863 versus HA : p < 0.863

Answer 1

Answer:

The hypothesis statements are: H0: p = 0.88 versus HA : p < 0.88

The p-value of the test is 0.2483 > 0.1, which means that the data does not provide sufficient evidence to conclude that the proportion of times that luggage is returned within 24 hours is less than 0.88.

Step-by-step explanation:

Test if the proportion of times that luggage is returned within 24 hours is less than 0. 88

At the null hypothesis, we test if the proportion is of 0.88, that is:

$H_0: p = 0.88$

At the alternate hypothesis, we test if this proportion is less than 0.88, that is:

$H_a: p < 0.88$

The hypothesis statements are: H0: p = 0.88 versus HA : p < 0.88

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.

0.88 is tested at the null hypothesis:

This means that $\mu = 0.88, \sigma = \sqrt{0.88*0.12}$

A consumer group who surveyed a large number of air travelers found that 138 out of 160 people who lost luggage on that airline were reunited with the missing items by the next day.

This means that $n = 160, X = \frac{138}{160} = 0.8625$

Test statistic:

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

$z = \frac{0.8625 - 0.88}{\frac{\sqrt{0.88*0.12}}{\sqrt{160}}}$

$z = -0.68$

P-value of the test:

The p-value of the test is the probability of finding a sample proportion below 0.8625, which is the p-value of z = -0.68.

Looking at the z-table, the p-value of z = -0.68 is of 0.2483.

The p-value of the test is 0.2483 > 0.1, which means that the data does not provide sufficient evidence to conclude that the proportion of times that luggage is returned within 24 hours is less than 0.88.