Question

The following data were obtained in a study using three separate samples to compare three different treatments.

Treatment 1	Treatment 2	Treatment 3
4	3	8
3	1	4
5 4	3 1	6 6
M = 4	M = 2	M = 6
SS = 2	SS = 4	SS = 8

a. State the null and alternative hypotheses.

b. Fill in the ANOVA table below.

Source	SS	df	MS	F
Between treatments
Within treatments
Total

c. If α = .05 and the Fcritical value = 4.26., what decision would you make?

d. Calculate Tukey’s HSD, with a q = 3.95.

e. Interpret which means are significantly different from each other based on Tukey’s HSD.

f. Interpret the results in paragraph form as if you were reporting them for a journal article. Attempt to use APA style.

Answer 1

Solution

Final answers in the required format are given below. Back-up Theory and Details of calculations follow at the end.

Part (a)

Null and alternative hypotheses:

H₀: α₁ = α₂ = α₃   0 Vs Alternative: H₁: at least one αi is different from other αi’s. Answer 1

Part (b)

ANOVA Table

Source	SS	df	MS	F
Between treatments	32	2	16	10.28571
Within treatments	14	9	1.5556
Total	46	11

Answer

Part (c)

Since F > Fcritical value, H₀ is rejected. Hence we conclude that the mean effect is different among the three treatments. Answer 3

Part (d)

Tukey’s HSD

Test Statistic

Q = (M_L - M_S)/√(MSw/n),

Where

M_L = Larger of the two means being compared

M_S = Smaller of the two means being compared

MSw = Mean Sum of Squares of Within (Error) in ANOVA

n = Number of observations per treatment in ANOVA (must be the same for all treatments)

Critical Value

The Studentized range statistic (q)* whose value is given to be 3.95

Decision

Reject null hypothesis of equal means if Qobs > qcrit.

Calculations

MSw (MSE)	1.5556	df	14	n	5	k	4
SE =√(MSw/n)	0.5578	qcrit(14,3,0.05)	3.85	α	0.05
Comparison	Larger Mean	Smaller Mean	D = Difference	Q = D/SE	Significance
1 vs 2	4	2	2	3.585635	no
1 vs 3	6	4	2	3.585635	No
2 vs 3	6	2	4	7.171269	yes

Answer 4

Part (e)

The difference between only Treatment 2 and Treatment 3 is significant. Answer 5

Back-up Theory and Details of calculations

1. 1-WAY CLASSIFICATION EQUAL # OBSNS PER CELL

Suppose we have data of a 1-way classification ANOVA, with r rows, and n observations per cell.

Let xij represent the jth observation in the ith row, j = 1,2,…,n; i = 1,2,……,r

Then the ANOVA model is: xij = µ + αi + εij, where µ = common effect, αi = effect of ith row, and εij is the error component which is assumed to be Normally Distributed with mean 0 and variance σ².

Hypotheses:

Null hypothesis: H₀: α₁ = α₂ = ….. = α_r = 0 Vs Alternative: H₁: at least one αi is different from other αi’s.

Now, to work out the solution,

Terminology:

Row total = xi.= sum over j of xij

Grand total = G = sum over i of xi.

Correction Factor = C = G²/N, where N = total number of observations = r x n

Total Sum of Squares: SST = (sum over i,j of xij²) – C

Row Sum of Squares: SSR = {(sum over i of xi.²)/n} – C

Error Sum of Squares: SSE = SST – SSR

Mean Sum of Squares = Sum of squares/Degrees of Freedom

Degrees of Freedom:

Total: N (i.e., rn) – 1;

Rows: (r - 1);

Error: Total - Row

F_obs: MSR/MSE;

F_crit: upper α% point of F-Distribution with degrees of freedom n1 and n2, where n1 is the DF for the numerator MS and n2 is the DF for the denominator MS of F_obs

Significance: F_obs is significant if F_obs > F_crit

Calculations

i	xij				xi.	sumxij^2
1	4	3	5	4	16	66
2	3	1	3	1	8	20
3	8	4	6	6	24	152

G	48
C	192
SST	46
SSR	32
SSE	14

DONE