Question

A researcher aims to determine the difference among four proposed interventions for children with ADHD. Participants were assigned to four groups. Follow all hypothesis testing steps.
a) Using excel analysis, perform the appropriate hypothesis test at α = 0.01.

b) Measure effect size for this data.

A	B	C	D
45	23	10	43
32	22	12	42
33	10	23	39
40	40	18	44
27	23	22	39
27	25	31	29
25	30	30	29
30	40	29	42
29	23	45	33
33	19	34	38

Answer 1

Solution:

Ho: All the 4 groups have the same mean

H₁: At least one of the group have different mean

Data:

A	B	C	D
45	23	10	43
32	22	12	42
33	10	23	39
40	40	18	44
27	23	22	39
27	25	31	29
25	30	30	29
30	40	29	42
29	23	45	33
33	19	34	38

Anova: Single Factor
SUMMARY
Groups	Count	Sum	Average	Variance	Sd
A	10	321	32.1	38.54444	6.208417
B	10	255	25.5	83.83333	9.156054
C	10	254	25.4	112.4889	10.60608
D	10	378	37.8	31.28889	5.593647

ANOVA
Source of Variation	SS	df	MS	F	P-value	F crit
Between Groups	1065	3	355	5.335226	0.003814	4.377096
Within Groups	2395.4	36	66.53888889
Total	3460.4	39

Conclusion: since our p value is less than 0.01 we reject null hypothesis and conclude that at least one of the group has different mean.

Effect Size: 'Effect size' is simply a way of quantifying the size of the difference between two groups.

Effect size can be computed as:

((Mean of 1 group-Mean of 2nd group))/pooled standard deviation

Pooled standard deviation =sqrt((n₁-1)*S₁^2+(n₂-1)*S₂^2)/(n₁+n₂-2))

Pooled Standard deviations:		Effect size
A and B:	7.822332701	0.843738083
B and C	9.907628935	0.010093232
C and D	8.478731561	-1.462482909
A and C:	8.690032604	0.770998258
B and D:	7.586903921	-1.621214678
A and D:	5.909032634	-0.964624898

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