Question

) A nuclear power facility produces a vast amount of heat which is usually discharged into aquatic systems. This heat raises the temperature of the aquatic system, resulting in a greater concentration of chlorophyll, which in turn extends the growing season. To study this effect, water samples were collected monthly at 3 stations for a period of 10 months. The associated data is available in a sheet named ‘Problem 2’

. (a) Write down the model form of the complete model.

(b) Write down the model form of the reduced model.

(c) Perform an analysis of variance to test the hypothesis at 0.1 level of significance, that there is no difference in the color density of a fabric for the three levels of dye.

		Station
	A	B	C
Month 1	9.8	3.8	4.4
Month 2	14.03	8.4	11.1
Month 3	15.2	20.8	4.46
Month 4	13.8	9.2	8.01
Month 5	5.8	3.8	34.1
Month 6	10.56	14.1	9
Month 7	11.1	6.9	17.3
Month 8	4.46	14.03	11.1
Month 9	14.03	15.2	4.46
Month 10	15.2	12.5	16.5

Step by step

Answer 1

Full Model form is; y_i,j = M_i + e

Reduced form is ;  y_i,j = M + e

M is the total mean, M_i is the group mean, e is the error.

c) Let null hypothesis be that there is no significant difference between group means.

Total Mean M = 11.438

Month	A	B	C
Month 1	9.8	3.8	4.4
Month 2	14.03	8.4	11.1
Month 3	15.2	20.8	4.46
Month 4	13.8	9.2	8.01
Month 5	5.8	3.8	34.1
Month 6	10.56	14.1	9
Month 7	11.1	6.9	17.3
Month 8	4.46	14.03	11.1
Month 9	14.03	15.2	4.46
Month 10	15.2	12.5	16.5
Mean	11.398	10.873	12.043
SS(Regression)	0.016	3.19225	3.66025

SS (Regression) is the sum of squared errors between the Total Mean and group Means.

SS (Regression) = 10*( Ma - M )^2 + 10*( Mb - M )^2 + 10*( Mc - M )^2 = 0.016+3.19225+3.66025

SS (Regression) = 6.8685

SS(Mean)				SS(Error)
A	B	C		A	B	C
2.7	58.3	49.5		2.6	50.0	58.4
6.7	9.2	0.1		6.9	6.1	0.9
14.2	87.6	48.7		14.5	98.5	57.5
5.6	5.0	11.8		5.8	2.8	16.3
31.8	58.3	513.6		31.3	50.0	486.5
0.8	7.1	5.9		0.7	10.4	9.3
0.1	20.6	34.4		0.1	15.8	27.6
48.7	6.7	0.1		48.1	10.0	0.9
6.7	14.2	48.7		6.9	18.7	57.5
14.2	1.1	25.6		14.5	2.6	19.9

SS(Mean) is the Sum of squared error around total mean.

SS(Mean) =

SS(Mean) = 1138.005

SS(Error) is the Sum of squared error between group mean and groups values.

SS(Error) =

SS(Error) = 1131.01

F = SS(Regression)/SS(Error) *DF(Error)/DF(Regression) = 6.8685/1131.01 * 27/2 = 0.0819

At 5% significance, critical F value = 3.35.

SO we fail to reject the null hypothesis that there is no difference between the 3 samples