An airline reports that its flights are on time with an average delay of 15 minu
ID: 3261452 • Letter: A
Question
An airline reports that its flights are on time with an average delay of 15 minutes with a variance of 150 based upon a study of 25 recent flights. Doubting the consistency part of the claim, a disgruntled traveler calculates the delays for his next 25 flights. The average delay for those 25 flights was 22 minutes with a variance of 225. Based upon tha data was the disgruntled traveler correct in believing the variance of the delayed flights was greater than the reported variance of 150. Evaluate the disgruntled traveler claim by performing a hypothesis test with a significant level of .05. Assume the data is normally distributed.
What is the null and alternate hypothesis?
a Null hypothesis: The variance reported by the airline is not the same as the variance calculated by the disgruntled traveler.
Alternate hypothesis: The variance calculated by the disgruntled traveler is the same as the variance reported by the airline.
b Null hypothesis: The variance reported by the airline is the same as the variance calculated by the disgruntled traveler.
Alternate hypothesis: The variance calculated by the disgruntled traveler is not the same as the variance reported by the airline.
c Null hypothesis: The variance reported by the airline is the same as the variance calculated by the disgruntled traveler.
Alternate hypothesis: The variance calculated by the disgruntled traveler is greater than the variance reported by the airline.
d Null hypothesis: The variance reported by the airline is the same as the variance calculated by the disgruntled traveler.
Alternate hypothesis: The variance calculated by the disgruntled traveler is less than the variance reported by the airline.
Enter the correct answer by selecting the appropriate letter.
What is the appropriate distribution for performing the hypothesis test is:
a Uniform distribution
b z distribution
c t distribution
d Chi Square distribution
e F distribution
Enter the correct answer by selecting the appropriate letter.
What is the numerical value of the test statistic?
Enter answer as a 1 place decimal.
What is the pvalue for the test statistic?
Enter answer as a decimal with 2 decimal places.
What is the decision based upon the analysis of the data?
a Accept the alternate hypothesis. The data is strong enough to reject the null hypothesis. The disgruntled traveler is justified on being a disgruntled traveler.
b Do not reject null hypothesis. The data is not string enough to reject the null hypothesis. The disgruntled traveler is not justified on being a disgruntled traveler.
c There is insufficient data to make a descision as to which hypothesis is correct at the .05 significance level. Additional data needs to be obtained.
Explanation / Answer
using minitab
Two-Sample T-Test and CI
Sample N Mean StDev SE Mean
1 25 15.0 12.3 2.5
2 25 22.0 15.0 3.0
Difference = (1) - (2)
Estimate for difference: -7.00
95% lower bound for difference: -13.50
T-Test of difference = 0 (vs >): T-Value = -1.81 P-Value = 0.961 DF = 46
Null hypothesis: The variance reported by the airline is the same as the variance calculated by the disgruntled traveler.
Alternate hypothesis: The variance calculated by the disgruntled traveler is greater than the variance reported by the airline.
t distribution
T-Value = -1.8
P-Value = 0.961
p value greater than
(failed to reject null hypothesis.
accept alternative hypothesis)
a Accept the alternate hypothesis. The data is strong enough to reject the null hypothesis. The disgruntled traveler is justified on being a disgruntled traveler.
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