In the study reported in question 2, there were 25 students in each group.
a. Calculate and report the standard error of the mean.
b. Conduct the three post-hoc Tukey HSD tests.
c. Interpret the results of these tests.
d. Why would you do a One-way ANOVA to answer this question rather than 3 separate independent t tests?

Question

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The Answer is 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.

In the Study Reported in Question 2, There Were 25