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 the treatment means are equal at the 0.05 significance level.

Treatment 1	Treatment 2	Treatment 3
8	3	3
11	2	4
10	1	5
	3	4
	2

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

Null hypothesis

• H₀: μ₁ = μ₂

• H₀: μ₁ = μ₂ = μ₃

1. (a-2). Alternate hypothesis.

• H₁: Treatment means are all the same

• H₁: At least one pair of treatment means is not the same

2. What is the decision rule? (Round your answer to 2 decimal places.)

3. Compute SST, SSE, and SS total. (Round your answers to 2 decimal places.)

4. Complete an ANOVA table. (Round F, SS to 2 decimal places and MS to 3 decimal places.)

	Source SS df MS F Treatments Error Total

5. State your decision regarding the null hypothesis.

Answer 1

	treatment 1	treatment 2	treatment 3
count, ni =	3	5	4
mean , x̅ i =	9.67	2.20	4.00
std. dev., si =	1.53	0.84	0.82
sample variances, si^2 =	2.333	0.700	0.667
total sum	29	11	16		56	(grand sum)
grand mean , x̅̅ =	Σni*x̅i/Σni =				4.67
square of deviation of sample mean from grand mean,( x̅ - x̅̅)²	25	6.08	0.444444444
					TOTAL
SS(between)= SSB = Σn( x̅ - x̅̅)² =	75	30.42	1.777777778		107.2
SS(within ) = SSW = Σ(n-1)s² =	4.666666667	2.80	2		9.466667

a)

H₀: μ₁ = μ₂ = μ₃

H₁: At least one pair of treatment means is not the same

b)

no. of treatment , k =   3
df between = k-1 =    2
N = Σn =   12
df within = N-k =   9
F--critical value (0.05,2,9) = 4.26

if F-stat > 4.26  , then reject Ho, otherwise not

c)

SS(error ) = SSE= Σ(n-1)s² = 9.47

SStreatment = Σn( x̅ - x̅̅)² =107.2

SST=SStreatment + SS error = 116.67

d)

anova table
	SS	df	MS	F
Between:	107.20	2	53.600	50.96
Within:	9.47	9	1.052
Total:	116.67	11

e)

F-stat = 50.96 >4.26, so, reject Ho

hence, there is enough evidence to conlcude that At least one pair of treatment means is not the same at α=0.05