Question

The following data are from a completely randomized design. In the following calculations, use α = 0.05 (level of significance)

	Treatment 1	Treatment 2	Treatment 3
	63	82	69
	47	72	54
	54	88	61
	40	66	48
sample mean	51	77	58
sample variance	96.67	97.34	81.99

a. Use analysis of variance to test for a significant difference among the means of the three treatments.

b. Use Fisher’s LSD procedure to determine which means are different.

Answer 1

	treatment 1	treatment 2	treatment 3
count, ni =	4	4	4
mean , x̅ i =	51.00	77.00	58.00
std. dev., si =	9.83	9.87	9.06
sample variances, si^2 =	96.667	97.333	82.000
total sum	204	308	232		744	(grand sum)
grand mean , x̅̅ =	Σni*x̅i/Σni =				62.00
square of deviation of sample mean from grand mean,( x̅ - x̅̅)²	121	225	16
					TOTAL
SS(between)= SSB = Σn( x̅ - x̅̅)² =	484	900	64		1448
SS(within ) = SSW = Σ(n-1)s² =	290	292	246		828

no. of treatment , k =   3
df between = k-1 =    2
N = Σn =   12
df within = N-k =   9
  
mean square between groups , MSB = SSB/k-1 =    724.000
  
mean square within groups , MSW = SSW/N-k =    92.000
  
F-stat = MSB/MSW =    7.870

a)

anova table
	SS	df	MS	F	p-value
Between:	1448.00	2	724.00	7.87	0.0106
Within:	828.00	9	92.00
Total:	2276.00	11

Ho: there is no significant difference among the means of the three treatments

H1: there is a significant difference among the means of the three treatments

p-value=0.0106

conclusion :    p-value<α=0.05 , reject null hypothesis    

so, there is enough evidence that there is a significant difference among the means of the three treatments at α=0.05

b)

Level of significance	0.05
no. of treatments,k	3
DF error =N-k=	9
MSE	92.00
t-critical value,t(α/2,df)	2.262

	treatment 1	treatment 2	treatment 3
count, ni =	4	4	4
mean , x̅ i =	51.00	77.00	58.00

Fishers LSD critical value=tα/2,df √(MSE(1/ni+1/nj)) = 15.3427

if absolute difference of means > critical value,means are significnantly different ,otherwise not                      
                      

population mean difference		critical value			result
µ1-µ2	26.000	15.3427			means are different
µ1-µ3	7.000	15.3427			means are not different
µ2-µ3	19.000	15.3427			means are different

so, mean (1,2) and ( 2,3) are different.