Airlines compute the weight of outbound fights using either standard average wei

Question

Airlines compute the weight of outbound fights using either standard average weights provided by the Federal Aviation Admnistration (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 Econamic 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 neight for ther carry-on items of 16 pounds, the FAA standard average weight, before implementation of the checked-bag fee. The aifine 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 82 passengers and weighs their carry-on items. The sample mean is 17.8 pounds, and the sample standard deviation i pounds. the airline uses a significance level of aos to conduct its hypothesis test The hypothesis test ist test a right-tailed a left-tailed a two-tailed The test statistic follow distribution. The value of the test statist is Select a Distribution Distributions use the Distributions tooi to develop the critical region. According to the oritical region method, when do you reject the null hypothesis? O Reject Ho if z 1.645 O Reject No dts -1.664 O Reject Ho if ts1.99D or t z 1 990 0 Reject Hof t2 1.664 The p-value is using the citical region method, the null hypothesis is p-value method, the null hypothiess is beosuse Using the because Therefore, you mean weight of the ailine's passengers carry-on nems has increased after the

Explanation / Answer

a)
right tailed {we need to test if mean is more than 16 }

b)
t distribution as population sd is not known

c)
TS = (Xbar - mu)/(s/sqrt(n))
= (17.8 - 16)/(8.8/sqrt(82))
= 1.85223786916
= 1.85

d)
df = n-1 =81
t.inv(0.95,81)
=1.663883913
=1.64

p-value =
=T.DIST.RT(1.85,81)
=0.034

using critical method the null is rejected ,

1.85 > 1.664

using p-value ,the null is rejected ,

0.0340 < 0.05

therefore you can conclude

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