Question

1. An educational psychologist does an experiment to compare three different methods of teaching math. Subjects are randomly assigned to the different treatments, then are instructed by the relevant method for one month. After completing instruction, all three groups take the same test of achievement. A summary of the results are tabled below. Test whether there is a difference among the 3 methods, summarizing your results in an ANOVA table and in English.

	Method A	Method B	Method C
Mean =	80	70	67
Variance =	25	30	20
n =	30	30	30

1. State your hypotheses.

2. Determine the characteristics of the comparison distribution.

3. Determine the cut-off value for a 5% significance level.
4. Compute the observed F ratio. Summarize results in an ANOVA table.

5. Decide whether or not to reject the null and give a verbal explanation.

6. Calculate and interpret h².

7. Compute the power of the test. How many subjects are needed to detect an effect with a power of .80 and the given effect size?

8. Compute a planned comparison between Method A and Method C. Given the number of comparisons that you plan to make, you apply the Bonferroni correction, which suggests testing at the .01 level. What are your conclusions?

Assume now that the educational psychologist had no a priori hypotheses about the nature of any differences. Instead, the researcher wants to compute all possible pairwise comparisons post-hoc. Compute the appropriate post-hoc comparison between Methods A and B (assuming that all possible pairwise comparisons are being computed as well). What are your conclusions?

Answer 1

a. Let Hi true mean score due to Method i, i = A, B, C. Null hypothesis, Ho : A = B = uc US. Alternative hypothesis, Hı: At least one li is different from others. 6. Comparison distribution is F distribution. Here the degrees of freedom (df) of F distribution is (3,87) and F distribution is positively skewed and critical region lies right side of F- distribution. c. n = total no. of observations = 30 *3 = 90 k= no. of groups = 3 Cut-off value = F0.05,3,87 = 2.7094 (R code : ”round(qf (1–0.05, 3, 87), 4)").

d. SSE = sum of squares due to error = 29 * (25 +30 + 20) = 2175 with df = 87 MSE= = mean sum of squares due to error 2175/87 = 25. T = grand mean = (80+ 70 +67)/3 = 72.3333 SSB = sum of squares between methods = 30 + (80 – 72.3333)2 + (70 – 72.3333)2 + (67 – 72.3333)2) = 2780 with df = 3-1 = 2. MSB = mean sum of squares between methods = 2780/2 = 1390 F ratio 1390/25 = 55.6 ANOV A table SST = sum of squares due to total SSE + SSB= 2175 + 2780 = 4955 Source SS df MSF Between 2780 2 1390 55.6 Error 2175 87 25 Total 4955 89 Since value of F > Cut-off value, we reject Ho at 5% level of significance. SSB e. n 2780/4955 = 0.56 and large effect size. SST