Question

An experimenter randomly allocated 125 male turkeys to five treatment groups: control and treatments A, B, C, and D. There were 25 birds in each group, and the mean results were 2.16, 2.45, 2.91, 3.00, and 2.71, respec- tively. The sum of squares for experimental error was 153.4. Test the null hypothesis that the five group means are the same against the alternative that one or more of the treatments differs from the control.

Answer 1

Ho: µ1=µ2=µ3=µ4=µ5
H1: one or more of the treatments differs from the control.

treatment	control	A	B	C	D
count, ni =	25	25	25	25	25
mean , x̅ i =	2.160	2.45	2.91	3.000	2.71

grand mean , x̅̅ =    Σni*x̅i/Σni =                     2.65

	control	A	B	C	D
square of deviation of sample mean from grand mean,( x̅ - x̅̅)²	0.236	0.038	0.070	0.125	0.004
						TOTAL
SS(between)= SSB = Σn( x̅ - x̅̅)² =	5.905	0.960	1.742	3.133	0.102	11.843

SSE=153.4

no. of treatment , k =   5
df between = k-1 =    4
N = Σn =   125
df within = N-k =   120

mean square between groups , MSB = SSB/k-1 =    2.9608
  
mean square within groups , MSW = SSW/N-k = 153.4/120 = 1.278
F-stat = MSB/MSW = 2.32

anova table
	SS	df		MS	F	p-value	F-critical
Between:	11.84	4		2.96	2.32	0.0612	2.447
Within:	153.40	120		1.278
Total:	165.24	124
						α =	0.05

Decision:   p-value>α , do not reject null hypothesis    
conclusion :        there is not enough evidence to conclude that one or more of the treatments differs from the control.