Four different types of solar energy collectors were tested. Each was tested at

Question

Four different types of solar energy collectors were tested. Each was tested at five randomly chosen times, and the power (in watts) was measured. The results were as follows. (a) Using alpha = 0.05, can you conclude that the mean power differs for different collectors? (b) Check the assumptions of the procedure used. (c) Use Fisher LSD method to analyze the mean power for the four collectors. Use alpha = 0.05. The effect of curing pressure on bond strength (in MPa) was tested for two different adhesives. There were three levels of curing pressure. Three replications were performed for each combination of curing pressure and adhesive. The results are presented in the following table. (a) Test the appropriate hypotheses and draw conclusions using alpha = 0.05. (b) Graphically analyze the interaction.

Explanation / Answer

The grand mean of power = 1.495

Mean of power for collector A = 1.78

Mean of power for collector B = 1.76

Mean of power for collector C = 1.12

Mean of power for collector D = 1.32

Total sum of squares, SST = (1.9 - 1.495)^2 + (1.6 - 1.495)^2 + ..... + (1.4 - 1.495)^2 = 2.1095

Treatment Sum of Squares (SSTR)  = [5*(1.78-1.495)^2] + [5*(1.76-1.495)^2] + [5*(1.12-1.495)^2] + [5*(1.32-1.495)^2]

= 1.6135

Error Sum of Squares (SSE) = SST - SSTR = 2.1095 - 1.6135 = 0.496

Total Mean Squares (MST) = SST/N-1 = 2.1095/ (20-1) = 0.111

Mean Square Treatment (MSTR)  = SSTR/ c-1 = 1.6135/(4-1) = 0.5378

Mean Square Error (MSE) = SSE/N-c = 0.496/(20-4) = 0.031

F = MSTR/MSE = 0.5378/0.031 = 17.3484

Critical value of F at significance level of 0.05 and numerator df = 3 and denominator df = 16 is 3.239

As, F (observed value) > F (critical value), we reject the null hypothesis of anova test and conclude that mean power differs for different collectors.

(b) Assumptions of ANOVA test:

(i) All populations involved follow a normal distribution.
(ii) All populations have the same variance (or standard deviation).
(iii) The samples are randomly selected and independent of one another.

(c) The t-critical value for =0.05, dfw = 16 is 1.746

LSD = t-critical value * sqrt(MSE * (1/nA + 1/nB))

= 1.746 * sqrt(0.031 * (1/5 + 1/5))

= 0.0615

Mean of power for collector A - Mean of power for collector B = 1.78 - 1.76 = 0.02 < 0.0615

Mean of power for collector A - Mean of power for collector C = 1.78 - 1.12 = 0.66 > 0.0615

Mean of power for collector A - Mean of power for collector D = 1.78 - 1.32 = 0.46 > 0.0615

Mean of power for collector B - Mean of power for collector C = 1.76 - 1.12 = 0.64 > 0.0615

Mean of power for collector B - Mean of power for collector D = 1.76 - 1.32 = 0.44 > 0.0615

Mean of power for collector C - Mean of power for collector D = 1.12 - 1.32 = 0.2 > 0.0615

So, as LSD = 0.0615, Mean power of A and B is only significant same. And Mean power of A,C; A;D ; B,C; B,D; C,D are not significantly same.

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