12-1 One-Way ANOVA 615 StatCrunch TI-83, 84 Plus SPSS JMP YOUR TURN. Do Exercise 5 “Car Size and Left Femur in Crash Tests.” Step 2: In addition to the test statistic of F = 7.6853, the displays all show that the P-value is 0.000 when rounded. Step 3: Because the P-value of 0.000 is less than the significance level of a = 0.05, we reject the null hypothesis of equal means. (If the P is low, the null must go.) INTERPRETATION There is sufficient evidence to warrant rejection of the claim that the four samples come from populations with means that are all equal. Using the samples of measurements listed in Table 12-1, we conclude that those values come from populations having means that are not all the same. On the basis of this ANOVA test, we cannot conclude that any particular mean is different from the others, but we can informally note that the sample mean for the small cars is higher than the means for the midsize, large, and SUV vehicles. It appears that in crash tests, small cars are associated with higher head injury measurements. How is the P-Value Related to the Test Statistic? Larger values of the test statistic result in smaller P-values, so the ANOVA test is right-tailed. Figure 12-2 on the next page shows the relationship between the F test statistic and the P-value. Assuming that the populations have the same variance s 2 (as required for the test), the F test statistic is the ratio of these two estimates of s 2: (1) variation between samples (based on variation among sample means); and (2) variation within samples (based on the sample variances). CAUTION When we conclude that there is sufficient evidence to reject the claim of equal population means, we cannot conclude from ANOVA that any particular mean is different from the others. (There are several other methods that can be used to identify the specific means that are different, and some of them are discussed in Part 2 of this section.)
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