Question

8,9,8,12,7	11,8,9,14,15
8,11,12,14,10	13,15,14,18,19
16,22,17,18,19	9,12,8,10,13

a=3levels b=2 levels n=5

a) Test whether or not the two factors interact.

b) Test whether or not Main effects for Factor A and Factor B are present

c)For each level of Factor A compare the levels of Factor B

d) For each level of Factor B compare the levels of Factor A

Answer 1

Solution

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

Part (a)

The two factors interact. Answer 1

Part (b)

Main effects for Factor A is present. Answer 2

Main effects for Factor B is not present. Answer 3

Part (c) and (d)

Means for each combination is given below:

Factor A	Factor B		Comparison
	Level 1	Level 2
Level 1	8.8	11.2	Level 2 is better
Level 2	11	15.8	Level 2 is better
Level 3	18.4	10.4	Level 1 is better
Comparison	Level 3 is best	Level 2 is best

Answer

Back-up Theory and Details of calculations

ANOVA 2-WAY CLASSIFICATION EQUAL # OBSNS PER CELL

Suppose we have data of a 2-way classification ANOVA, with r rows, c columns and n observations per cell.

Let xijk represent the kth observation in the ith row-jth column, k = 1,2,…,n; i = 1,2,……,r ; j = 1,2,…..,c.

Then the ANOVA model is: xijk = µ + αi + βj + γij + εijk, where µ = common effect, αi = effect of ith row, βj = effect of jth column, γij = row-column interaction and εijk is the error component which is assumed to be Normally Distributed with mean 0 and variance σ².

Hypotheses:

Null hypothesis: H₀₁: α₁ = α₂ = ….. = α_r = 0 Vs Alternative: H₁₁: at least one αi is different from other αi’s.

Null hypothesis: H₀₂: β₁ = β₂ = ….. = β_c = 0 Vs Alternative: H₁₂: at least one βi is different from other βi’s.   

Null hypothesis: H₀₃: γij = 0 for all i and j   Vs Alternative: H₁₃: at least one γij is not zero.

Now, to work out the solution,

Terminology:

Cell total = xij. = sum over k of xijk

Row total = xi..= sum over j of xij.

Column total = x.j. = sum over i of xij.

Grand total = G = sum over i of xi.. = sum over j of x.j.

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

Total Sum of Squares: SST = (sum over i,j and k of xijk²) – C

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

Column Sum of Squares: SSC = {(sum over j of x.j.²)/(rxn)} – C

Between Sum of Squares: SSB = {(sum over i and jof xij.²)/n} – C

Interaction Sum of Squares: SSI = SSB – SSR – SSC

Error Sum of Squares: SSE = SST – SSB

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

Degrees of Freedom:

Total: N (i.e., rcn) – 1;

Between: rc – 1;

Within(Error): DF for Total – DF for Between;

Rows: (r - 1);

Columns: (c - 1);

Interaction: DF for Between – DF for Rows – DF for Columns;

F_obs:

for Rows: MSSR/MSSE;

for Columns: MSSC/MSSE;

for Interaction: MSSI/MSSE

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

Significance: F_obs is significant if F_obs > F_crit

Calculations

i	j	xijk; k =					xij.	xijksquare	xij.square	Row sum	Row sum	Col sum	x.j.^2/9
		1	2	3	4	5		sum		xi..	sq/cn	x.j.
1	1	8	9	8	12	7	44	402	1936	101	1020.10	191	2432.067
	2	11	8	9	14	15	57	687	3249			188	2356.267
2	1	8	11	12	14	10	55	625	3025	134	1795.6
	2	13	15	14	18	19	79	1275	6241
3	1	16	22	17	18	19	92	1714	8464	144	2073.6
	2	9	12	8	10	13	52	558	2704
					Total		379	5261	25619

G	379
C	4788.03
SST	472.97
SSR	101.2667
SSC	0.30
SSB	335.77

ANOVA TABLE		α =	0.05
Source	DF	SS	MS	Fobs	Fcrit	p-value
Row(Factor A)	2	101.2667	50.6333	8.8571	3.4028	0.0013
Column (Factor B)	1	0.3000	0.3000	0.0525	4.2597	0.8207
A x B Interaction	2	234.2000	117.1000	20.4840	3.4028	0.0000
Between	5	335.7667	67.1533
Error	24	137.2000	5.7167
Total	29	472.9667	16.3092

Conclusion: Factor A is significant; Factor B is not significant; Interaction is significant.

DONE