Expedia would like to test the hypothesis that the proportion of Southwest Airline flights that arrive on-time is less than 0.90. A random sample of 140 United Airline flights found that 119 arrived on-time. Expedia would like to set α = 0.05. The conclusion for this hypothesis test would be that because the test statistic is ____________________________________________.

a. more than the critical value, we can conclude that the proportion of United Airline flights that arrive on-time is more than 0.90
b. more than the critical value, we cannot conclude that the proportion of United Airline flights that arrive on-time is less than 0.90
c. less than the critical value, we can cannot conclude that the proportion of United Airline flights that arrive on-time is less than 0.90
d. less than the critical value, we can conclude that the proportion of United Airline flights that arrive on-time is less than 0.90

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Answer:

d. less than the critical value, we can conclude that the proportion of United Airline flights that arrive on-time is less than 0.90

Step-by-step explanation:

Expedia would like to test the hypothesis that the proportion of Southwest Airline flights that arrive on-time is less than 0.90.

At the null hypothesis, we test that the proportion is of 0.9, that is:

[tex]H_0: p = 0.9[/tex]

At the alternate hypothesis, we test that it is less than 0.9, that is:

[tex]H_a: p < 0.9[/tex]

The test statistic is:

[tex]z = \frac{X - \mu}{\frac{\sigma}{\sqrt{n}}}[/tex]

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

0.9 is tested at the null hypothesis:

This means that [tex]\mu = 0.9, \sigma = \sqrt{0.9*0.1}[/tex]

A random sample of 140 United Airline flights found that 119 arrived on-time.

This means that [tex]n = 140, X = \frac{119}{140} = 0.85[/tex]

Value of the test-statistic:

[tex]z = \frac{X - \mu}{\frac{\sigma}{\sqrt{n}}}[/tex]

[tex]z = \frac{0.85 - 0.9}{\frac{\sqrt{0.9*0.1}}{\sqrt{140}}}[/tex]

[tex]z = -1.97[/tex]

P-value of the test and decision:

The p-value of the test is the probability of finding a sample proportion of 0.85 or less, which is the p-value of z = -1.97.

Looking at the z-table, z = -1.97 has a p-value of 0.0244.

The p-value is 0.0244 < 0.05, which means that the test statistic is less than the critical value, and thus, we can conclude that the proportion of United Airline flights that arrive on-time is less than 0.90.

The correct answer is given by option d.