Question

3. Hypothesis tests about a population mean, population standard deviation unknown Aa Aa 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 x = 17.2 pounds, and the sample standard deviation is s = 6.2 pounds. The airline uses a significance level of a = .05 to conduct its hypothesis test The hypothesis test is test The test statistic follows a distribution. The value of the test statistic is Use the Distributions tool to develop the critical value rejection rule. According to the critical value approach, the rejection rule is OReject Ho if z 2 1.645 O Reject Ho if t s -1.669 O Reject Ho if t 2 1.669 O Reject Ho if t s -1.998 or t 2 1.998 The p-value is Using the critical value approach, the null hypothesis is because Using the Therefore, you p-value approach, the null hypothesis is , because 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

The hypothesis test is a one tailed test.

The test statistic follows a normal distribution. The value of the test statistic is:

$\small z=\frac{17.2-16}{\frac{6.2}{\sqrt{64}}}=1.5483$

We search the z table to find the z value for which Pr(z)>x=0.95. According to the z table that value is 1.645.

Therefore, we will reject $\small H_0$ if $\small H_0\geq 1.645$

The p value is 0.0618

Since this is not the case, we fail to reject $\small H_0$

Using the critical value approach, the null hypothesis is not rejected, because
1.5483 1.645
. Using the p value approach, the null hypothesis is accepted, p value is greater than the significance level. Therefore, you fail to conclude that the mean weight has increased.