Question

uestion 3 (15 points) The table shows the costs per mile (in cents) for a sample of automobiles. At a 0.01, can you conclude that at least one mean cost per mile is different from the others? Please perform the following steps: (a) identify the claim and state Ho and Ha, (b) find the critical value and identify the rejection region, (c) find the test statistic F, (d) decide whether to reject or fail to reject the null hypothesis, and (e) interpret the decision in the context of the original claim. Assume the samples are random and independent, the populations are normally distributed, and th e population variances are equal Small sedan Medium sedan Large sedan SUV 4WD Minivan 65 47 61 57 62 50 79 58 67 70 60 39 47 52 79 71 76 68 64 74 57 49 68

Answer 1

The following table is obtained Group1 Group 2 65 47 61 57 62 50 Group 3 60 69 79 71 76 68 Group 4 79 58 67 70 Group 5 64 74 57 49 68 39 47 52 Sum- 223 44.6 10051 5.128 105.2 342 57 19748 7.127 254 6 423 70.5 30043 6.656 221.5 6 274 68.5 18994 8.66 225 4 312 62.4 19846 Average- St. Dev. 377.2

The total sample size is N 26. Therefore, the total degrees of freedom are dftotal 26-1-2i Also, the between-groups degrees of freedom are dfbetwen -5-1 degrees of freedom are 4, and the within-groups dfvithin - dftotal - dfbetween - 25-4-21 First, we need to compute the total sum of values and the grand mean. The following is obtained X223+342 + 423+274+312- 1574 i.j Also, the sum of squared values is X 10051 19748 +30043 18994 19846 98682 i.j Based on the above calculations, the total sum of squares is computed as follows ss-X152 x, I = i.j total = X2 X986823394.462 i.j

The within sum of squares is computed as shown in the calculation below: SSvithinSih1 105.2 254 +221.5 + 225 + 377.2 1182.9 The between sum of squares is computed directly as shown in the calculation below:

Now that sum of squares are computed, we can proceed with computing the mean sum of squares SSbetween2211.562 M Sbetueendfbetwr MSuiin dfwithin SSuithin 1182.9 21 Finally, with having already calculated the mean sum of squares, the F-statistic is computed as follows betueen552.89 within56.329 MS (1) Null and Alternative Hypotheses The following null and alternative hypotheses need to be tested Ha: Not all means are equal The above hypotheses will be tested using an F-ratio for a One-Way ANOVA.

(2) Rejection Region Based on the information provided, the significance level is a 0.01, and the degrees of freedom are dfi 4 and df 4, therefore, the rejection region for this F-test is R F:F>Fe 4.369 (3)_Test Statistics F= MS,tween M Suithin 552.89 56.329 = Vl Dwithin Since it is observed that F- 9.815 >F 4.369, it is then concluded that the null hypothesis is rejected Using the P-value approach: The p-value is p = 0.0001, and since p = 0.0001 < 0.01, it is concluded that the null hypothesis is rejected. (5) Conclusion It is concluded that the null hypothesis Ho is rejected. Therefore, there is enough evidence to claim that not all 5 population means are equal, at the a 0.01 significance level.