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

Why Not Test Two at a Time? Refer to the sample data given in Exercise 1. If we want to test for equality of the three means, why don’t we use three separate hypothesis tests for\({\mu _1} = {\mu _2},{\mu _1} = {\mu _3}\;and\;{\mu _2} = {\mu _3}\)?

The observations for arrival time are known, along with the SPSS output.

Let there be three hypothesis tests to compare the mean arrival delay time of three flights.

\(\begin{aligned}{l}{H_0}:{\mu _1} = {\mu _2}\\{H_0}:{\mu _2} = {\mu _3}\\{H_0}:{\mu _1} = {\mu _3}\end{aligned}\)

Assume that each of the hypothesis tests is tested on a 0.05 significance level or 0.95 level of confidence.

To test three different hypotheses, the level of confidence transforms to\({0.95^3} = 0.857\).

The associated level of significance for the overall test of the three hypotheses is \(1 - 0.857 = 0.143\).

When the significance level is as high as 0.143,the type 1 error increases as a result of testing three pairs of hypotheses.

Consequently, the chances ofobtaining the result for the test accurately lower.

On the other hand, ANOVA is used to conduct the test at a 0.05 level of significance, which results in better accuracy.

Thus, conducting three different hypothesis tests is not preferred.