Question

The following data are from a completely randomized design.

	Treatment
Observation	A	B	C
1	162	142	126
2	142	156	122
3	165	124	138
4	145	142	140
5	148	136	150
6	174	152	128
Sample Mean	156	142	134
Sample Variance	164.4	131.2	110.4

1. Compute the sum of squares between treatments.
2. Compute the mean square between treatment
3. Compute the sum of squares due to error.
4. Compute the mean square due to error.
5. Set up the ANOVA table for this problem.
6. At the ∝ = .05 level of significance, test whether the means for the three treatments are equal.

Answer 1

For the given data

Null Hypothesis , the means of the three treatments are equal.

Alternative Hypothesis : Atleast two of the treatments means are different

Correction Factor (CF)

a)

Sum of Squares between Treatments

Degrees of freedom for Treatments = 2

b)

Mean Sum of Squares due to treatments = 1488/2 = 744

c)

Sum of Squares Due to error

Total Sum of Squares(TSS)

Raw Sum of Squares (RSS)

TSS = RSS-CF = 376766-373248 = 3518

Error Sum of Squares(SSE) = TSS-SST = 3518 -1488 = 2030

d)

Mean Error due to square

degrees of freedom for error = N-k = 18-3 =15

MSE = 2030/15 = 135.33

e)

ANOVA Table

Source	df	SS	MS	F
Between Treatment	k-1 = 3-1 =2	1488	744	744/135.33 = 5.4976
Within Treatment (Error)	N-k =18-3 = 15	2030	135.33
Total	N-1 =17	3518

f)

Critical value of F at 0.05 level of significance for (2,15) df is 3.68

Since Calculated value of F is greater than the tabulated value of F , we reject null hypothesis at 5% level of significance and conclude that the treatments A, B and C differ significantly.

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