Question

Answer the within-subjects ANOVA questions using the data below. Use α = 0.05.

1	2	3	4
53 26 47 42 51 34 44	44 39 35 34 39 30 12	42 42 39 37 36 34 47	25 35 30 27 24 28 31

a) Compute the preliminary statistics below.
SS_BG =  ; df_BG =  
SS_BS =  ; df_BS =  
SS_E =  ; df_E =  
SS_T =  ; df_T =  

b) Compute the appropriate test statistic(s) to make a decision about H₀.
critical value =  ; test statistic =  
Decision:  ---Select--- Reject H0 Fail to reject H0

c) Compute the corresponding effect size(s) and indicate magnitude(s).
η² =  ;  ---Select--- na trivial effect small effect medium effect large effect

d) Make an interpretation based on the results.

At least one week differs on the GRE verbal score.No week is different on GRE verbal score.    

e) Conduct Tukey's Post Hoc Test for the following comparisons:
2 vs. 4: difference =  ; significant:  ---Select--- Yes No
1 vs. 4: difference =  ; significant:  ---Select--- Yes No

f) Conduct Scheffe's Post Hoc Test for the following comparisons:
2 vs. 3: test statistic =  ; significant:  ---Select--- Yes No
3 vs. 4: test statistic =  ; significant:  ---Select--- Yes No

Answer 1

For getting the results from the above data, all computation is done in excel based on in built command.

a) The output from the excel is

ANOVA
Source of Variation	SS	df	MS	F	P-value	F crit
Rows	232.2143	6	38.70238	0.596258	0.729545	2.661305
Columns	816.3929	3	272.131	4.192517	0.020481	3.159908
Error	1168.357	18	64.90873
Total	2216.964	27

the preliminary statistics is

SS_BG	258.3571	df_BG	6
SS_BS	895.5714	df_BS	3
SS_E	1067.929	df_E	18
SS_T	2221.857	df_T	27

B) The null hypothesis for all weeks equal on the GRE verbal score.

H₁ : al least one week is differed on the GRE verbal score

Form the table above

Critical value is 2.66

Test statistic is F test : 0.596

Decision. Since test statistic value is less than critical value, so, we fail to reject the null hypothesis and concluded that all weeks equal on the GRE verbal score.

(c)

small effect

d) No week is different on the GRE verbal score.

e) Conduct Tukey's Post Hoc Test for the following comparisons

where M_i – M_j is the difference between the pair of means. to calculate this, M_i should be larger than M_j, MS_w is the Mean Square Within, and n is the number in the group or treatment.

	Mean difference	size	df	MSE	HSD	Q-critical	significant
2 vs. 4:	4.714286	7	18	64.9	2.482732	4	Yes
1 vs. 4:	13.85714	7	18	64.9	7.297727	4	No