Question

In Exercises 1–4, use the following listed arrival delay times (minutes) for American Airline flights from New York to Los Angeles. Negative values correspond to flights that arrived early. Also shown are the SPSS results for analysis of variance. Assume that we plan to use a 0.05 significance level to test the claim that the different flights have the same mean arrival delay time.

Flight 1	-32	-25	-26	-6	5	-15	-17	-36
Flight 19	-5	-32	-13	-9	-19	49	-30	-23
Flight 21	-23	28	103	-19	-5	-46	13	-3

Test Statistic What is the value of the test statistic? What distribution is used with the test statistic?

Short answer

The value of the test statistic is 1.334.

The test statistic follows F-distribution with (2,21) degrees of freedom.

Step-by-step solution

Given information

The output for ANOVA is given from SPSS.

Identify the value of test statistic

The value of the test statistic can be obtained from the SPSS output from the column header F.

Thus, the test statistic is 1.334, which is computed as follows:

\(\begin{array}{c}F = \frac{{{\rm{Mean}}\;{\rm{square}}\;{\rm{between}}\;{\rm{groups}}}}{{{\rm{Mean}}\;{\rm{square}}\;{\rm{within}}\;{\rm{groups}}}}\\ = \frac{{1287.500}}{{965.310}}\\ = 1.334\end{array}\)

Thus, the F-statistic is 1.334.

Identify the distribution of the test statistic

F-distribution describes the ratio of two variance measures.

The test statistic uses F-distribution, which has two degrees of freedom:

• The numerator degrees of freedom, which is 2
• The denominator degrees of freedom, which is 21.

Thus, the test statistic uses F-distribution with (2,21) degrees of freedom.