Question

An experiment was performed to determine the effect of four different chemicals on the strength of a fabric. These chemicals are used as part of the permanent press finishing process. Five fabric samples were selected, and a randomized complete block design was run by testing each chemical type once in random order on each fabric sample. The data are shown in Table below. test for differences in means using an ANOVA with α=0.01 Fabric Sample Chemical Type 1 2 3 4 5 1 1.3 1.6 0.5 1.2 1.1 2 2.2 2.4 0.4 2.0 1.8 3 1.8 1.7 0.6 1.5 1.3 4 3.9 4.4 2.0 4.1 3.4

Answer 1

Solution

ANOVA Table

Source	SS	df	MS	Fobs	Fcrit	p-value
Row	18.044	3	6.01466667	75.89484753	5.9525447	4.51831E-08
Column	6.693	4	1.67325	21.11356467	5.4119514	2.31891E-05
Error	0.951	12	0.07925
Total	25.688	19

Decision:

Since Fobs > Fcrit or equivalently since p-value < α (=0.01), both null hypotheses of no chemical effect and no fabric effect, are rejected.

Conclusion:

There is sufficient evidence to conclude that

mean fabric strength is different among the 4 chemicals and also among the five fabric samples. Answer

Details of calculations

Data

Chemical	Fabric Sample
	1	2	3	4	5
1	1.3	1.6	0.5	1.2	1.1
2	2.2	2.4	0.4	2.0	1.8
3	1.8	1.7	0.6	1.5	1.3
4	3.9	4.4	2.0	4.1	3.4

Calculations

r	4	c	5	N	20			sumxij^2
	xij					xi.	xi.^2
1	1.3	1.6	0.5	1.2	1.1	5.7	32.49	7.15
2	2.2	2.4	0.4	2.0	1.8	8.8	77.44	18
3	1.8	1.7	0.6	1.5	1.3	6.9	47.61	10.43
4	3.9	4.4	2.0	4.1	3.4	17.8	316.84	66.94
sum	9.2	10.1	3.5	8.8	7.6	39.2	474.38	102.52
						sum/c	94.876
x.j^2	84.64	102.01	12.25	77.44	57.76
sumx.j^2			334.1
sumx.j^2/r			83.53

G	39.2
C	76.832
SST	25.688
SSR	18.044
SSC	6.693
SSE	0.951

Back-up Theory

1. 2-WAY CLASSIFICATION ANOVA with ONE OBSNS PER CELL

Suppose we have data of a 2-way classification ANOVA, with r rows, c columns and 1 observation per cell.

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

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

Now, to work out the solution,

Terminology:

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

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

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

Column Sum of Squares: SSC = {(sum over j of x.j²)/(r)} – C

Error Sum of Squares: SSE = SST – SSR - SSC

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

Degrees of Freedom:

Total: N (i.e., rc) – 1;

Rows: (r - 1);

Columns: (c - 1);

Error: DF for Total – DF for Rows – DF for Columns;

F_obs:

for Rows: MSSR/MSSE;

for Columns: MSSC/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

DONE