Data Set 15 in textbook Appendix B lists 48 different departure delay times minu
ID: 3252529 • Letter: D
Question
Data Set 15 in textbook Appendix B lists 48 different departure delay times minutes) for American Airlines flights from New York (JFK) to Los Angeles. Negative departure delay times correspond to flights that departed early. The mean of the 48 times is 10.5 min and the standard deviation is 30.8 min. The following STATDISK display results from a test of the claim that the mean departure delay time for all such flights is less than 12.0 min. Hypothesis Test: Mean-One Sample Alternative Hypothesis: 3) Pop. Mean Claimed Mean Significance: 0.01 Alternative Hypothesis: Claimed Mean: 12.0 t Test Population St. Dev: Test Statistic t: 0.3374 (if known) Critical t: 2.4083 P-value: 0.3687 sample size, n: 48 986 Confidence interval: Sample Mean: 10.5 -0.2065424 21.20654 Sample St. Dev, s 30.8 Evaluate Copy (a) What is the a used in the test? What does a mean? (b) Is the test two-tailed, left-tailed, or right-tailed? (c) What is the test statistic? Is it a z test or t test? (d) What is the P-value? (e) What is Ho and Hi? Do you reject Ho or fail to reject? f) Justify your decision above using the P-value method. g) Justify your decision above using the critical value method. (h) What is the final conclusion? Is a flight operations manager justified in reporting that the mean departure time is less than 12.0 min?Explanation / Answer
Solution:-
State the hypotheses. The first step is to state the null hypothesis and an alternative hypothesis.
Null hypothesis: > 12
Alternative hypothesis: < 12
b) Note that these hypotheses constitute a one-tailed test.
a) Formulate an analysis plan. For this analysis, the significance level is 0.01.
The test method is a one-sample t-test.
Analyze sample data. Using sample data, we compute the standard error (SE), degrees of freedom (DF), and the t statistic test statistic (t).
SE = s / sqrt(n)
S.E = 4.45
DF = n - 1 = 48 - 1
D.F = 47
c) t = (x - ) / SE
t = - 0.3371
tcritical = - 2.41
where s is the standard deviation of the sample, x is the sample mean, is the hypothesized population mean, and n is the sample size.
The observed sample mean produced a t statistic test statistic of - 0.337. We use the t Distribution Calculator to find P(t < - 0.337) = 0.369
d) Thus the P-value in this analysis is 0.369
f) Interpret results. Since the P-value (0.369) is greater than the significance level (0.01), we cannot reject the null hypothesis.
g) Since the t test is greater than the t critical, hence we have to accept the null hypothesis.
h) Do not reject H0, We do not have sufficient evidence in the favor of the claim that the mean departure is less than 12 minutes.
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