Question

A component manufacturer wishes to measure the quality of five different
methods of production of components. An experiment to test this was
conducted with a one-way layout, where the response being measured was
the lifetime of the components when tested to destruction, measured in
units of 10⁶ seconds. The data is shown in Dataset. Carry out an
appropriate statistical test to see if anyone of the five methods has
produced better components than the others, as measured by their lifetime.

Dataset:

Method 1	Method 2	Method 3	Method 4	Method 5
102.57	101.97	101.48	100.69	100.80
101.52	102.13	102.53	101.45	100.33
100.21	99.25	99.06	99.45	98.69
99.35	99.73	100.85	101.77	101.85
101.46	100.36	99.40	100.31	101.07
100.32	100.70	100.23	99.89	100.18
99.21	100.36	99.33	98.94	97.88
97.42	98.15	98.82		98.60
98.19	97.23	98.34		99.82
		98.67

Answer 1

Solution

The solution is based on One-way Analysis of Variance with unequal number of observations per treatment.

Final answer is given below. Back-up Theory and Details of Calculations follow at the end.

ANOVA Table		Alpha = 0.05
Source	DF	SS	MS	F	Fcrit	p-value
Method	4	1.1283	0.28207108	0.138013	2.612306	0.9671866
Error	39	79.708	2.04380483
Total	43	80.837	1.87992262
Decision: Since F < Fcrit or equivalently, since p-value < alpha, null hupothesis of no difference in means between the five methods is accepted. Conclusion: We conclude that the different methods do not differ in terms of the life term of the components. Answer

Back-up Theory

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

Let xij represent the jth observation in the ith row/column, 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: H₀₁: α₁ = α₂ = ….. = α_r = 0 Vs Alternative: H₁₁: at least one αi is different from other αi’s.

Now, to work out the solution,

Terminology:

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

Details of Calculations

alpha	0.05
#treat	5
n1	9	n4	7
n2	9	n5	9
n3	10
N	44
x1.	900.25	x4.	702.50
x2.	899.88	x5.	899.22
x3.	998.71
G = x..	4400.56
C	440112.0071
Sx1j^2	90072.1725	Sx4j^2	70507.2458
Sx2j^2	89996.8978	Sx5j^3	89857.5516
Sx3j^2	99758.9761
Sxij^2	440192.8438
Sxi.^2/ni	440113.1354
SST	80.83667273
SSR	1.128284314
SSE	79.70838841

DONE