Question

Use the 95% level of confidence. - Given the following sample information, test the hypothesis that the treatment means are equal at the 0.05 significance level. Treatment 1 - 8, 11 and 10. Treatment 2 - 3, 2, 1, 3, and 2. Treatment 3 - 3, 4, 5 and 4. a) What is the decision rule? (Round your answer to 2 decimal places.) - Reject Ho if F > b) Compute SST, SSE, and SS total. (Round your answers to 2 decimal places.) SST- SSE- SS total - c) Complete an ANOVA table. (Round your answers to 2 decimal places.) Source SS df MS F Treatments Error Total

Answer 1

a)

Degree of freedom of Treatments = Number of level - 1 = 3 - 1 = 2

Degree of freedom of error = Number of observations - Number of level = (3 + 5 + 4) - 3 = 9

Critical value of F at DF = 2, 9 and 95% level of confidence is 4.256

Reject Ho if F > 4.256

b)

Let Ti be the total test score for group i, ni be number of observations of group i.

Let G be the total test scores of all observations and N be total number of observations.

ΣX2 is sum of squares if all observations

T1 = 29, T2 = 11 , T3 = 16

G = 29 + 11 + 16 = 56

ΣX2 = 285 + 27 + 66 = 378

SS total = ΣX2 - G2/N = 378 - 56^2/12 = 116.67

SST = ΣT2/n - G2/N = (29^2 /3 + 11^2 /5 + 16^2 /4 ) - 56^2/12 = 107.2

SSE = 116.67 - 107.2 = 9.47

c)

Source	SS	DF	MS	F
Treatments	107.2	2	53.6	51.05
Error	9.47	9	1.05
Total	116.67	11

For each source, MS = SS / DF

F = MS Treatments / MS Error = 53.6 / 1.05 = 51.05

Since F > 4.256, Reject H0.