Question

Solve as an anova two-way model (the second factor is in the rows and has levels A and B))

The null hypothesES are:

H₀: μ_a = μ_b = μ_c (all Factor means for the column-wise factor are equal)

H₀: μ_Amy = μ_Bert (all Factor means for the row-wise factor are equal)

H₀: the column-wise Factor and the row-wise Factor do not interact

Can we reject them at the alpha = 0.05 level?

Use the Excel data below:

	A	B	C
Amy	23	21	20
	25	21	22
Bert	27	24	26
	28	27	27

Answer 1

The Hypothesis:

FACTOR A

H0:

Ha:

FACTOR B

H0: .

Ha:Not all the means are equal

INTERACTION

H0: There is no interaction between the two factors. They are independent.

Ha: There is an interaction between the two factors. They are not independent.

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The ANOVA Table is given below.

The calculation and procedures are mentioned after the hypothesis Test

	SS	df	MS	F	F critical	P value
A - Rows	60.75	1	60.75	38.45	5.99	0.0008
B - columns	14.00	2	7.00	4.43	5.14	0.0658
A x B	2.00	2	1.00	0.63	5.14	0.5645
Error	9.50	6	1.58
Total	86.25	11

The Decision Rule:

Critical Value Method: If F obtained is > F critical, then Reject H0

P Value Method: If p value is < Then Reject H0.

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The Decision:

For Factor A

F obtained (38.45) is > F critical. We Reject H0

P value (0.0008) is < (0.05). We Reject H0

The Conclusion: There is sufficient evidence to conclude that .

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For Factor B

F obtained (4.43) is < F critical (5.14). We Fail to Reject H0

P value (0.0658) is > (0.05). We Fail to Reject H0

The Conclusion: There is sufficient evidence to conclude that at least one mean differs from the others.

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For Interaction

F obtained (0.63) is < F critical (5.14). We Fail to Reject H0

P value (0.5645) is > (0.05). We Fail to Reject H0

The Conclusion: There is sufficient evidence to conclude there is an interaction between the 2 factors.

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ANOVA CALCULATIONS

The Marginal Totals Tables is as below

	A	B	C	Total
Amy	48	42	42	132
Bert	55	51	53	159
Total	103	93	95	291

(1) (Overall Total)² / N = (291)² / 12 = 7056.75

(2) Sum of squares of individual observations = 7143

(3) SUM (Row Total)² / (b * n)

b * n = 2 * 3 = 6

Therefore (132)² / 6 + (159)² / 6 = 7117.50

(4) SUM (Column Total)² / (a * n)

a * n = 2 * 2 = 4

Therefore (103)² / 4 + (93)² / 4 + (95)² / 4 = 7070.75

(5)   SUM(Cell)² / n; n = 2 = (48)² / 2 + (42)² / 2 + (42)² / 2 + (55)² / 2 + (51)² / 2 + (53)² / 2 = 7113.50

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SS A = (3) – (1) = 60.75

SS B = (4) – (1) = 14

SS AB = (5) + (1) – (3) – (4) = 2

SS Error = (2) – (5) = 9.5

SS Total = 86.25

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df A = a – 1 = 2 - 1 = 1

df B = b – 1 = 3 - 1 = 2

df AB = df A * df B = 1 * 2 = 2

df Error = N – (a * b) = 12 - (2*3) = 6

df Total = 1 + 2 + 2 + 6 = 11

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MS A = SS A / df A = 60.75 / 1 = 60.75

MS B = SS B / df B = 14 / 2 = 7

MS AB = SS AB / df AB = 2 / 2 = 1

MS Error = SS Error / df Error = 9.5 / 6 = 1.58

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F A = MS A / MS Error = 60.75 / 1.58 = 38.45

F B = MS B / MS Error = 7 / 1.58 = 4.43

FAB = MS AB / MS Error = 1 / 1.58 = 0.63

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