The data in the table were collected from randomly selected fights at airports i

Question

The data in the table were collected from randomly selected fights at airports in three cities and indicate the number of minutes that each plane was behind schedule at its departure Perform a one-way ANOVA using alpha = 0.05 to determine if there is a difference in the average lateness of fights k from these three airports. a. What are the correct hypotheses for the one-way ANOVA test? A. H_0 Not all the means are equal. H_1: mu_1 = mu_2 = mu_3 B. H_0: mu_1 notequalto mu_2 notequalto mu_3 H_1: mu_1 = mu_2 = mu_3 H C. H_0: mu_1 = mu_2 = H_1 Not all the means are equal D. H_0: mu_1 = mu_2 = mu_3 H_1: mu_1 notequalto mu_2 notequalto mu_3 Complete the ANOVA summary table below. Determine the p-value for this test. p-value = State the conclusion for alpha = 0 05. Since the p-value is than the level of significance, the null hypothesis and conclude that the average lateness of flight different for these airports.

Explanation / Answer

a) option A

b) Corrected mean = (all observations)2/N = 3602/15 = 8640

SS(Total) = sum of squares of all observations - CM = 10202 - 8640 = 1562

SS(Between) = Sum of each treatment2/n - CM = 1202/5 + 722/5 + 1682/5 - 8640 = 921.6

SS(Within) = SS(Total) - SS(Between) = 1562 - 921.6 = 640.4

df(Total) = N-1 = 14

df(Between) = k-1 = 2

df(Within) = 15 - 3 = 12

MS(Between) = SS(Between)/df(Between) = 921.6/2 = 460.8

MS(Within) = SS(Error)/df(Error) = 640.4/12 = 53.3667

F = MS(Between)/MS(Within) = 460.8/53.3667 = 8.63

p = 0.005

Since the p-value is less than the level of significance. Reject the null hypothesis and conclude that the average lateness of flights is different for these airports

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