A car dealer is interested in comparing the average gas mileages of four different car models. The dealer believes that the average gas mileage of a particular car will vary depending on the person who is driving the car due to different driving styles. Because of this, he decides to use a randomized block design. He randomly selects five drivers and asks them to drive each of the cars. He then determines the average gas mileage for each car and each driver. Can the dealer conclude that there is a significant difference in average gas mileages of the four car models? The results of the study are as follows.

Find the value of the F-test statistic for testing whether the average gas mileage is the same for the four car models. Round your answer to two decimal places, if necessary.

Car A Car B Car C Car D
Driver 1 23 39 22 25
Driver 2 37 39 28 39
Driver 3 39 40 21 25
Driver 4 34 36 27 33
Driver 5 27 35 26 37

Required:
Make the decision to reject or fail to reject the null hypothesis that there is no difference in the average gas mileages of the four car models. State the conclusion in terms of the original problem. Use alpha = 0.01.

a. We reject the null hypothesis. At the 0.01 level of significance, there is sufficient evidence of a difference in average gas mileages of the four car models.
b. We fail to reject the null hypothesis. At the 0.01 level of significance, there is not sufficient evidence of a difference in average gas mileages of the four car models.
c. We fail to reject the null hypothesis. At the 0.01 level of significance, there is sufficient evidence of a difference in average gas mileages of the four car models.
d. We reject the null hypothesis. At the 0.01 level of significance, there is not sufficient evidence of a difference in average gas mileages of the four car models.

Answer:

We reject the null hypothesis at the 0.01 level of significance, there is sufficient evidence of a difference in average gas mileages of the four car models  ( A )

Step-by-step explanation:

Using the Given data we will apply the two way ANOVA without

source of variation            SS             df        MS        F

Block                                  148.80       4        37.20    1.68

Treatment                         424.40       3        141.47     6.39

Error                                  265.60       12       22.13

Total                                  838.80       19

From the Two way ANOVA table the F-statistic = 6.39

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