Question

A researcher had conducted an ANOVA and found that at least one of the following methods of studying for an exam:

1) office hours regularly,

2) both office hours and review sessions,

3) just review sessions, and

4) neither office hours nor review sessions.

differed significantly from the other methods in terms of stats exam scores [F(3, 16) = 9.84, p< .05]. Below is the original ANOVA source table:

Source             SS                    DF                   MS                   F

Between              85                3                      28.33 9.84    

Within             46                    16                    2.88

The researcher wanted to conduct pairwise comparisons to find which of the methods differed from the others in resulting stats exam scores. The researcher’s raw data were as follows:

Office (O)	Office & Review (O&R)	Review (R)	None (N)
12	12	8	6
11	10	7	9
14	15	10	11
11	14	9	10
12	14	6	9
X̄=12 SS=6	X̄=13 SS=16	X̄=8 SS=10	X̄=9 SS=14

Conduct pairwise comparisons via both Tukey’s HSD and Scheffe’s test. Show all work

Answer 1

SOLUTION

(1) Ho: All four groups have the same mean (μ1 = μ2 = μ3 = μ4)

Ha: At least two of the four groups have significantly different means

(2) α = 0.05, dfB = 4 - 1 = 3, dfT = 20 - 1 = 19, dfW = 19 - 3 = 16

(3 and 4)

Group 1 Group 2 Group 3 Group 4 12 12 15 14 14 10 12 Sum- 65 45 Average- 12 726 1.225 419 1.871 14 330 1.581 10 861 St. Dev.- 16

The total sample size is N 20. Therefore, the total degrees of freedom are: dfrotal 20 - 1-19 Also, the between-groups degrees of freedom are dfbetween-4-1-3, and the within-groups degrees of freedom are: d-thin=dftot al dfteeen-19 3=16 First, we need to compute the total sum of values and the grand mean. The following is obtained Xij = 60 + 65 + 40 + 45 210 2.3 Also, the sum of squared values is Σ =726 + 861 +330+419 = 2336 2.0 Based on the above calculations, the total sum of squares is computed as follows 1.J 2.0

The within sum of squares is computed as shown in the calculation below SSuithin S Suithingroup 6+16+10+14 46 The between sum of squares is computed directly as shown in the calculation below The between sum of squares is computed directly as shown in the calculation below Now that sum of squares are computed, we can proceed with computing the mean sum of squares SSitc 858333 M Sbetweendfbeteen within 46 MSithin2.875 dfuithin Now that sum of squares are computed, we can proceed with computing the mean sum of squares SSbetureen28.333 M Sbetweendfbeteen 46 rithhin42.875 Finally, with having already calculated the mean sum of squares, the F-statistic is computed as follows M Sbetureen 28.333 MS within 2.875

(1) Null and Alternative Hypotheses The following null and altermative hypotheses need to be tested Ha: Not all means are equal The above ypoiheses wto ftr a ony (2) Rejection Region Based on the information provided, the significance level is α 0.05, and the degrees of freedom are dh-3 and df-= 3, therefore, the rejection region for this F-test is R-(F : F > F 3.239 (3) Test Statistics MSeteen28.333 MS FMSictren 28.333 ithin 287 = 9.855 (4) Decision about the null hypothesis Since it is observed that F rejected 9.855 > Fe 3.239, it is then concluded that the null hypothesis is Using the P-value approach: The p-value is p 0.0006, and since p 0.0006< 0.05, it is concluded that the null hypothesis is rejected.

(5 Conclusion It is concluded that the null hypothesis Ho is rejected. Therefore, there is enough evidence to claim that not all 4 population means are equal, at the a 0.05 significance level.