Deck 9: One-Way Analysis of Variance

Question 1

Suppose I want to know whether students in Santa Clara, U.C. Berkeley, and Stanford differ how much sleep they get per night, on average. I select random samples of 25 students from each school and measure how much they sleep. I calculate an ANOVA and this produces an SSb of 20 and an SSe of 200. Do the analysis to tell me whether the difference between the groups is statistically significant. Show your work and wrap words around your results. Be thorough.

Answer 1

MSb = 10. MSe = 200/73 = 2.74. Observed F = 10/2.74 = 3.65. Critical F_(2,72) = 3.13

Question 2

Suppose I want to know whether students from Iowa, California, and Arizona differ in their average GPA in high school. I collect my data and find the following statistics from three groups of 25 students each (total N=75):
Average GPA: Iowa = 2.10, California = 2.80, Arizona = 3.35
SS_b = 4.20 SS_e = 49.25
Set your alpha level at .05. Then…
_a) Report your degrees of freedom for between groups and for within groups;
b) Calculate and report your MS_b and your MS_e
c) Calculate and report your F-ratio:
d) Decide whether you have a statistically significant result. Give your reasoning.
e) Interpret your results. What do you now know?

Answer 2

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.

Question 3

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?

Answer 3

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.

Question 4

Suppose that I perform a one-way ANOVA to examine whether students in Finland, China, and Canada differ in their average scores on an achievement test. I use random samples of 2000 students from each country and find that the F value in my ANOVA is statistically significant. What can I conclude from this? What do I know?

Question 5

I've said that the t test formulas and the F formula (from ANOVA) are all doing basically the same thing. Tell me how they are similar.