## Question 874:

1

1. Q874_1_anova874.xls

We have a two part question. For both parts we want to compare more than 2 means and so use an ANOVA. For part 1, the Null Hypothesis is that there is no difference between the average distance cars drive based on the type of gas.

For part 2, the Null Hypothesis is that there is no difference between the average distance each car drives based on the type of car.

We will reject the Null if our test statistic (F) is greater than 4.07 for Part 1 and 4.26 for Part 2. These values are found from an F table using 3 and 8 degrees of freedom and 2 and 8 degrees of freedom or using the Excel function =FINV(.05, 3,8).

The calculations are rather tedious to do by hand so see the attached excel file for the calculations. We get a test statistic of 2.20 for Part 1 with a p-value of .1658. Since this is less than the critical value we fail to reject the Null Hypothesis and conclude that there is no difference between the gas types.

For part 2 we get a test statistic of .95 and p-value of .423 which is also above the critical value. So we again fail to reject the Null. We conclude there is no difference between the cars' average distance.

Note: You could also perform a 2-Way ANOVA (2x2) which would generate very similar p-values and lead to the same conclusion.