Question

Airlines compute the weight of outbound flights using either standard average weights provided by the Federal Aviation Administration (FAA) or weights obtained from their own sample surveys. The FAA standard average weight for a passenger’s carry-on items (personal items plus carry-on bags) is 16 pounds.

Many airline companies have begun implementing fees for checked bags. Economic theory predicts that passengers will respond to the increase in the price of a checked bag by substituting carry-on bags for checked bags. As a result, the mean weight of a passenger’s carry-on items is expected to increase after the implementation of the checked-bag fee.

Suppose that a particular airline’s passengers had a mean weight for their carry-on items of 16 pounds, the FAA standard average weight, before implementation of the checked-bag fee. The airline conducts a hypothesis test to determine whether the current mean weight of its passengers’ carry-on items is more than 16 pounds. It selects a random sample of 63 passengers and weighs their carry-on items. The sample mean is x̄ x̄ = 17.4 pounds, and the sample standard deviation is s = 5.7 pounds. The airline uses a significance level of α = 0.01 to conduct its hypothesis test.

The hypothesis test is _____(an upper-tail, a lower-tail, a two-tailed) test

The test statistic follows a _____(binomial, standard normal, t) distribution. The value of the test statistic is _____(0.25, 1.95, 1.23, 1.85)

Use the Distributions tool to develop the critical value rejection rule. According to the critical value approach, the rejection rule is:

Reject H₀ if z ≥ 1.645

Reject H₀ if t ≤ –2.657 or t ≥ 2.657

Reject H₀ if t ≤ –2.388

Reject H₀ if t ≥ 2.388

The p-value is _____ (0.0279, 1.9495, 0.0256, 0.0514)

Using the critical value approach, the null hypothesis is _____(not rejected, rejected), because _____(1.95 < 2.388, 0.0279 > 0.01, 1.95 < 2.657, 0.25 < 2.388) . Using the p-value approach, the null hypothesis is _____(rejected, not rejected), because _____(0.0256 > 0.01, 1.95 < 2.388, 0.0279 > 0.01, 0.0514 > 0.01) . Therefore, you _____(can, cannot) conclude that the mean weight of the airline’s passengers’ carry-on items has increased after the implementation of the checked-bag fee.

Answer 1

Right tailed test

t-distribution

Test statistic,
t = (xbar - mu)/(s/sqrt(n))
t = (17.4 - 16)/(5.7/sqrt(63))
t = 1.95

This is right tailed test, for α = 0.01 and df = 62
Critical value of t is 2.388.
Hence reject H0 if t >= 2.388

P-value = 0.0279

Using the critical value approach, the null hypothesis is not rejected, because 1.95 < 2.388.

Using the p-value approach, the null hypothesis is not rejected, because 0.0256 > 0.01.

Therefore, you cannot conclude that the mean weight of the airline’s passengers’ carry-on items has increased after the implementation of the checked-bag fee.