Question

Recently, the number of airline companies that offer in-flight Wi-Fi service to passengers has increased. However, it is estimated that only 10% of the passengers who have Wi-Fi available to them are willing to pay for it. Suppose Gogo, the largest provider of airline Wi-Fi service, would like to test this hypothesis by randomly sampling 125 passengers and asking them if they would be willing to pay $4.95 for 90 minutes of onboard Internet access. Suppose that 20 passengers indicated they would use this service. Using α = 0.05, answer the following questions:
a. Based on this sample, can we conclude that the proportion of airline passengers willing to pay for onboard Wi-Fi service is different than 10%?












b. Determine the p-value for this test.

Answer 1

a.

Let p be the proportion of airline passengers willing to pay for onboard Wi-Fi service.

Null Hypothesis H0: p = 0.10

Alternative Hypothesis Ha: p 0.10

np(1-p) = 125 * 0.10 * (1 - 0.10) = 11.25

Since np(1-p) > 10, the sample size is large enough to approximate the sampling distribution of proportion as normal distribution and conduct a one sample z test.

The sample can be assumed to be a random sample, the value of np(1-p) is 11.25, which is greater than or equal to 10 and the sample size can be assumed to be less than or equal to 5% of the population size of airline passengers. Thus, all assumptions of one sample z test are satisfied.

Standard error of sample proportion, SE = = 0.02683282

Sample proportion, = 20/125 =  0.16

Test statistic, z = ( - p) / SE = (0.16 - 0.10)/0.02683282 = 2.236

For two-tail test, p-value = P(z > 2.236) = 0.0254

Since, p-value is less than 0.05 significance level, we reject null hypothesis H0 and conclude that there is significant evidence from the sample data that the true proportion of airline passengers willing to pay for onboard Wi-Fi service is different from 0.10.

b.

p-value for this test = 0.0254