Question

Determine the test criteria:

Specify the level of significance (Type I error associated with the null hypothesis),

Determine the test statistic (the appropriate statistical test as mentioned under point 5 above),

Determine the critical values (and region(s) if applicable),

Calculate the value of the test statistic (obtained value).

Make a decision about the null and research hypothesis by comparing the obtained value to the critical value and interpret the results of the data. You can pay attention also to the p-values.

ANSWER IN EXCEL FORM

Shift	Stress
1	2
1	5
1	7
1	8
1	5
1	4
1	3
1	6
1	5
1	4
2	4
2	5
2	3
2	6
2	16
2	11
2	9
2	8
2	12
3	7
3	13
3	10
3	16
3	11
3	11
3	7
3	3
3	10
3	15
3	12

Answer 1

Solution

The solution is obtained through ANOVA (Analysis of Variance) – One-way with three shifts as factor.

Let xij be the jth stress observation in the ith shift, i = 1, 2, 3 and j = 1 to n_i, n₁ = 10, n₂ = 9, n₃ = 11.

Then the ANOVA model is: xij = µ + αi + εij, where µ = common effect, αi = effect of ith shift, and εij is the

error component which is assumed to be Normally Distributed with mean 0 and variance σ².

Null hypothesis: H₀: α₁ = α₂ = α₃ Vs Alternative: H_A: H₀ is false [at least one αi is different from others]

Final ANOVA Table and decisions and conclusions are given below. Detailed Back-up Theory and Excel

calculations follow at the end.

ANOVA	Table	alpha	0.05
Source	DF	SS	MS	Fobs	Fcrit	p-value
Shift	2	162.68	81.34192	7.012619	3.354131	0.0035255
Error	27	313.18	11.59936
Total	29	475.87	16.4092

Decision: Since Fobs > Fcrit or equivalently, p-value < alpha, H₀ is rejected.

Conclusion: There is enough evidence to suggest that the mean stress levels are different for the 3 shifts. Answer

Now, to work out the solution,

Back-up Theory

Terminology:

Shift 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 = n₁ + n₂ + n₃

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

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

Error Sum of Squares: SSE = SST – SSR

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

F_crit: upper α% point of F-Distribution with degrees of freedom υ₁ and υ₂, where υ₁ is the DF for shift

and υ₁ is the DF for Error

Significance: F_obs is significant if F_obs > F_crit

Calculations

Data

j	i
	Sh1	Sh2	Sh3
1	2	4	7
2	5	5	13
3	7	3	10
4	8	6	16
5	5	16	11
6	4	11	11
7	3	9	7
8	6	8	3
9	5	12	10
10	4	-	15
11	-	-	12

Excel Calculations

alpha	0.05
#treat	3
n1	10
n2	9
n3	11
n	30
x1.	49
x2.	74
x3.	115
G = x..	238
C	1888.13
Sx1j^2	269
Sx2j^2	752
Sx3j^2	1343
Sxij^2	2364
Sxi.^2/ni	2050.82
SST	475.867
SSR	162.684
SSE	313.183

DONE