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In: Statistics and Probability

Airlines compute the weight of outbound flights using either standard average weights provided by the Federal...

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 64 passengers and weighs their carry-on items. The sample mean is x̄

= 16.9 pounds, and the sample standard deviation is s = 4.5 pounds. The airline uses a significance level of α = 0.05 to conduct its hypothesis test.

The hypothesis test is (upper tail, lower tail or two tailed) test.

The test statistic follows a (binomial, t, or standard normal) distribution. The value of the test statistic is (1.04, 0.20, 1.60, 1.10).

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

Reject H₀ if t ≤ –1.998 or t ≥ 1.998

Reject H₀ if t ≤ –1.669

Reject H₀ if t ≥ 1.669

Reject H₀ if z ≥ 1.645

The p-value is (1.6000, 0.0548, 0.0340, 0.0573)

Using the critical value approach, the null hypothesis is (rejected/not rejected) , because (1.60 < 1.998, 0.20 < 1.669, 0.0573 > 0.05, 1.60 < 1.669). Using the p-value approach, the null hypothesis is (rejected/not rejected), because (0.0548 > 0.5, 0.0573 > 0.5, 1.60 < 1.669, 0.0340 < 0.5). 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.

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