Question

An operations manager of a large firm is studying the monthly sales (in $’000) of three subsidiary companies over a six-month period as given in the table below.

Company A	Company B	Company C
152	148	150
135	158	148
130	136	126
142	138	128
125	140	140
128	127	135

At the 0.01 significance level, can we conclude that there is a difference in the means of monthly sales of the three subsidiary companies over a six-month period?

d)  Compute F test statistics.

	A.	Source of Variation Sum of Squares Degree of Freedom Mean of Square F Test Statistic Between treatments 120.8778 2 53.1889 0.7846 Within treatments 1851 15 150.4 Total 1971.8778 17
	B.	Source of Variation Sum of Squares Degree of Freedom Mean of Square F Test Statistic Between treatments 111.7778 2 48.211 0.4611 Within treatments 1145 15 105.4 Total 1256.7778 17
	C.	Source of Variation Sum of Squares Degree of Freedom Mean of Square F Test Statistic Between treatments 102.7778 2 51.3889 0.4876 Within treatments 1581 15 105.4 Total 1683.7778 17
	D.	Source of Variation Sum of Squares Degree of Freedom Mean of Square F Test Statistic Between treatments 110.5624 2 49.5477 0.5086 Within treatments 2014 15 115.8 Total 2124.5624 17

Answer 1

Solution

At the 0.01 significance level, we can conclude that there is NO difference in the means of monthly sales of the three subsidiary companies over a six-month period. Answer 1

[because F test Statistic < Fcrit or equivalently, p-value > 0.01 – refer to thw ANOVA Table]

F-test statistic

Optioin C Answer 2

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

ANOVA TABLE		α	0.05
Source	df	SS	MS	F	Fcrit	p-value
Row	2	102.78	51.3889	0.4876	6.3589	0.6235
Error	15	1581.00	105.4000
Total	17	1683.78

Steps

Obsn #	Company A		Company B		Company C
1	152		148		150
2	135		158		148
3	130		136		126
4	142		138		128
5	125		140		140
6	128		127		135
R = 3   K = 6 N = 18
i		Company A		Company B	Company C
xi.		812		847	827
xi.2		659344		717409	683929
Σxij^2		110402		120137	114489
G		2486
n		18
C		343344.2222
ΣΣxij^2		345028
SST		1683.7778
Σxi.^2		2060682
SSTr		102.7778

DONE