Question

You are a nurse working in a distress center and would like to study the self-esteem of domestic violence victims. You would like to know if self-esteem is associated with education level. You know from the existing literature that self-esteem scores are generally normally distributed with homogeneous variance across education groups. The table below shows the data that you collected for a random sample of clients that recently visited your center. Let the probability of committing a type I error be 0.05. Can you conclude that there is a difference in self-esteem across the education groups? If an overall significant difference is found, which pairs of individual sample means are significant different?

Less than High School Diploma	High School Diploma	Some College	Bachelor’s Degree and Above
17	22	24	26
15	23	25	27
14	24	26	28
16	25	24	29
17	26	28	30
26	27	29	31
15	28	27	32
18	20	26	33
19	18	25	34
21	20	23	35

Answer 1

	Less than High School Diploma	High School Diploma	Some College	Bachelor’s Degree and Above	Total
Sum	178	233	257	305	973
Count	10	10	10	10	40
Mean, sum/n	17.8	23.3	25.7	30.5
Sum of square, Ʃ(xᵢ-x̅)²	113.6	98.1	32.1	82.5

Null and Alternative Hypothesis:

Ho: µ1 = µ2 = µ3 = µ4

H1: At least one mean is different

Number of treatment, k = 4

Total sample Size, N = 40

df(between) = k-1 =3

df(within) = k(n-1) =36

df(total) = N-1 =39

SS(between) = (Sum1)²/n1 + (Sum2)²/n2 + (Sum3)²/n3 + (Sum4)²/n4 - (Grand Sum)²/ N =836.475

SS(within) = SS1 + SS2 + SS3 + SS4 =326.3

SS(total) = SS(between) + SS(within) =1162.775

MS(between) = SS(between)/df(between) =278.825

MS(within) = SS(within)/df(within) =9.0639

F = MS(between)/MS(within) =30.7622

p-value = F.DIST.RT(30.7622, 3, 36) = 0.0000

Decision:

Reject the null hypothesis

Yes, we can conclude that there is a difference in self-esteem across the education groups at 0.05 significance level.

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Group	Sample Mean	Size
x̅1	17.8	10
x̅2	23.3	10
x̅3	25.7	10
x̅4	30.5	10

At α = 0.05, k = 4, N-K = 36, Q value =    3.81

Critical Range, CV = Q*√(MSW/n) = 3.81*√(9.0639/10) = 3.6273  

Comparison	Absolute Diff. = |xi - xj|	CR	Results
x̅1-x̅2	-5.5	3.6273	Means are different
x̅1-x̅3	-7.9	3.6273	Means are different
x̅1-x̅4	-12.7	3.6273	Means are different
x̅2-x̅3	-2.4	3.6273	Means are not different
x̅2-x̅4	-7.2	3.6273	Means are different
x̅3-x̅4	-4.8	3.6273	Means are different