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

To find which locations differ in the means we use TUKEYS HSD

It is given by

where Mi and Mj are the 2 means being compared and their positive difference is taken. n = number of replicates in each sample. Here n = 15, MSerror = MSwithin = 330.2338

The Rule is that if Tukeys Observed is > Tukeys Critical, then there is a significant difference between the groups.

Tukeys critical is found from the critical values table for = 0.05, for k (no. of columns) = 4 on the horizontal and df error = 56 on the vertical. The Value = 3.74

M1 = 12 M2 = 13, M3 = 8 and M4 = 9

Group 1 and Group 2 : M1 - M2 = Absolute (12 - 13) = 1

Tukeys Observed = 1 / 0.76 = 1.32. Since Tukeys observed (1.32) is < Tukeys Critical (3.74), there is not a significant difference between the 2 groups.

We do the above mentioned process taking 2 groups at a time. The following table gives a clear picture.

We do the abovementioned process taking 2 groups at a time. The following table gives a clear picture.

M1	12
M2	13
M3	8
M4	9
Mserror	2.8
n	5
Sqrt(Msbetween/n)	0.76

	Absolute Difference	Observed	Critical	Obs > Crit
M1 - M2	1	1.316	3.74	No
M1 - M3	4	5.263	3.74	Yes
M1 - M4	3	3.947	3.74	Yes
M2 - M3	5	6.579	3.74	Yes
M2 - M4	4	5.263	3.74	Yes
M3 - M4	1	1.316	3.74	No

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SCHEFFES TEST

We use the sceffes comparison value to compare means and see if they are significant or not.

= SQRT (k - 1) * F * MSE * 2 / n)

Here k = number of columns = 4, F stat = 9.84, MSE = 2.88 and n = 5

Therefore Scheffes Value = SQRT (3 * 9.84 * 2.8 * 2 / 5) = 5.83

So if any of the mean differences is > 5.75, then there is a significant difference between means

	Absolute Difference	Scheffes	Obs > Crit
M1 - M2	1	5.83	No
M1 - M3	4	5.83	No
M1 - M4	3	5.83	No
M2 - M3	5	5.83	No
M2 - M4	4	5.83	No
M3 - M4	1	5.83	No

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