A trucking company conducted an ANOVA analysis comparing the mean mileage associ

Question

A trucking company conducted an ANOVA analysis comparing the mean mileage associated with several different brands of tires. The ANOVA table is shown below.

d) Using a 0.05 level of significance, what is the critical point compared to the F statistic in order to make a conclusion? Give your answer to four decimal places.  
e) Using a 0.05 level of significance, what conclusion should the company reach?

Reject the hypothesis that all brands have the same mean mileage.

Reject the hypothesis that one brand has a different mean mileage than the other brands.     

Fail to reject the hypothesis that one brand has a different mean mileage than the other brands.

Fail to reject the hypothesis that all brands have the same mean

Source DF SS MS F Treatment 2 3.894 1.947 4.198835 Error 24 11.129 C Total 26 B

Explanation / Answer

As TSS = SS due to treatment + SS due Error

B = TSS = 3.894 + 11.129 = 14.023

F = MS due treatment / MSE = 1.947/C

4.198835 = 1.947/C Implies C = 1.947 / 4.198835 = 0.4637

(d) The tabulated value of F at 0.05 level of significance for (2,24) degrees of freedom is = 3.40.

(e) As the calculated value of F is more than the tabulated value of F, the hypothsis that all brands have the same mean is rejected.

Source DF SS MS F Treatment 2 3.894 1.947 4.198835 Error 24 11.129 C Total 26 B

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