Question

The following are three observations collected from treatment 1, five observations collected from treatment 2, and four observations collected from treatment 3. Test the hypothesis that are equal at the 0.05 significance level.

Treatment 1__ Treatment 2__ Treatment 3

8 __________ 3________ 3

11__________ 2________ 4

10__________ 1________ 5

0___________3_________ 4

0___________ 2________ 0

a-1. State the null hypothesis and the alternate hypothesis.

b. What is the decision rule?

c. Compute SST, SSE, and SS total.

d. Complete an ANOVA table.

e. State your decision regarding the null hypothesis

f. If H0 is rejected, can we conclude that treatment 1 and treatment 2 differ? Use the 95 percent level of confidence.

Answer 1

a)

Ho: µ1=µ2=µ3
H1: not all means are equal

b)

no. of treatment , k =   3
df between = k-1 =    2
N = Σn =   12
df within = N-k =   9

F(0.05,2,9) = 4.26

reject Ho if F>4.26

c)

	1	2	3
count, ni =	3	5	4
mean , x̅ i =	9.667	2.20	4.00
std. dev., si =	1.528	0.837	0.816
sample variances, si^2 =	2.333	0.700	0.667
total sum	29	11	16		56
grand mean , x̅̅ =	Σni*x̅i/Σni =				4.67
square of deviation of sample mean from grand mean,( x̅ - x̅̅)²	25.000	6.084	0.444
					TOTAL
SS(between)= SSB = Σn( x̅ - x̅̅)² =	75.000	30.422	1.778		107.2
SS(within ) = SSW = Σ(n-1)s² =	4.667	2.800	2.000		9.467

SST=107.2

SSE=9.467

SStotal = SST+SSE=116.67

d)

ANOVA
Source of Variation	SS	df	MS	F	P-value	F crit
Between Groups	107.20	2	53.60	50.96	0.0000	4.256
Within Groups	9.47	9	1.05
Total	116.67	11

e)

since, F>4.26, reject Ho

f)

Level of significance	0.0500
no. of treatments,k	3
DF error =N-k=	9
MSE	1.052
t-critical value,t(α/2,df)	2.2622

confidence interval = mean difference ± critical value  
  
if confidence interval contans zero, then means are not different.  

			confidence interval
population mean difference		critical value	lower limit	upper limit	result
µ1-µ2	7.47	1.69	5.77	9.16	means are different