Exam 9: One-Way Analysis of Variance

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a. First, .68/25=.027. Then square root that to get .16. So that's the standard error.
b. Iowa-Arizona: (3.35-2.10)/.16=7.81
Arizona-California: (3.35-2.8)/.16 = 3.44
Iowa-CA: (2.80-2.10)/.16=4.12
The critical value for the Tukey test, from Appendix D, is 3.40 (with 3 groups and 60 df error).
c. All three of the observed Tukey values exceed the critical Tukey value, so all three population means differ from each other. Students in AZ have higher average GPAs than students in CA and Iowa, and students in CA have higher average GPAs than students in Iowa.
NOTE: It is possible to get significant results from the Tukey post-hoc tests even if the overall ANOVA did not produce significant differences between the group means. This happens for two reasons. First, the overall F value from the ANOVA tests for AVERAGE differences between all of the means. Even if two group means differ from each other, on average the three group means may not. Second, the Tukey test is known as a LIBERAL test. It is more likely to produce significant results than the overall ANOVA does, or than other post-hoc tests do.
d. Every time you do a t test, you have a certain alpha level which determines your Type I error rate. If you do three t tests to compare three groups, your Type I error rate increases from .05 to .15 (if you set alpha at .05). To account for the number of comparisons being made, and control the Type I error rate, you need to do an ANOVA. The df between groups takes into consideration how many groups are being compared.

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All of them divide the difference between observed means by a measure of the standard error, i.e., the expected difference between means due to random sampling error.

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a. df between = 3 - 1 = 2; df within = 75 - 3 = 72.
b. MS_B = 4.20/2 $\rightarrow$ 2.10
MS_E = 49.25/72 $\rightarrow$ .68
c. F_{(2, 72)} = 2.10/.68 $\rightarrow$ F_{(2, 72)} = 3.09
d. Well, for an alpha level of .05 with 2 and 70 degrees of freedom (this is the closest we can get to df + 72 in Appendix C), my critical F value from Appendix C will be 3.13. Because my observed F value < than my critical F value, I conclude that my population means are NOT different, and the differences in my sample means was TOO LIKELY due to chance (or random sampling error) for me to rule chance out.
e. I know the population means are NOT significantly different from each other, so I conclude that, on average, children in CA, Iowa, and Arizona have the same GPA in high school. Any observed differences between the sample means is considered to have been caused by random sampling error, or chance.