1. Lesson 5

2. Example 2: Fabric Flammability Tests
Flammability tests were conducted on children’s sleep wear. The Vertical Semi-restrained Test was used, in which pieces of fabric were burned under controlled conditions. After the burning stopped, the length of the charred portion was measured and recorded. Random samples using the same material was obtained from each of 5 testing labs. Because the same fabric was used, the different labs should have obtained the same results. Is there sufficient evidence to support the claim that the mean lengths for the different labs are the same?

3. Fabric Flammability Tests
Lab 1
Lab 2
Lab 3
Lab 4
Lab 5
2.9
2.7
3.3
4.1
3.1
3.2
3.7
3.5
4.2
2.8
3.4

4. ANOVA Table Source DF SS MS F p Factor or Treatments 4 2.68 0.667 4.26
0.009
Error 25
3.93
0.157
Total 29
6.61

5. Conclusion Reject at H (sub zero) < alpha
The evidence in the data is strong enough to suggest at least one of the mean charred lengths for the 5 labs is significantly different.

6. Example 2 - continued
Calculate Fisher LSD and use it to determine where the differences lie.
LSD = 0.471

7. Fisher LSD Populations Mean Dif LSD Difference (Y/N) ?? Lab 1 & 2
0.471
Lab 1 & 3
Lab 1 & 4
Lab 1 & 5
Lab 2 & 3
Lab 2 & 4
Lab 2 & 5
Lab 3 & 4
Lab 3 & 5
Lab 4 & 5

8. No. 21 K = 5; n = 7 N = 35 Source DF SS MS F p TR 4 300 75
75/5.33 = 14.07 ER 30
160
5.33
Total 34
460

9. P-value P-value = Fcdf(14.07,E99,4,30) = .0000014 < alpha
Conclusion: Reject the null hypothesis

10. No. 24 n1 = 12; n2 = 15; n3 = 20; N = 47 Source DF SS MS F p TR 2 ER44
Total 46

11. Randomized Block Design
Extraneous factors (other than the treatment that we are testing for) cause MSE to become large and so cause F = MSTR/MSE to become small which leads to a conclusion of …..
We can control the source of some of this extraneous variation by removing it from the MSE. We can separate the extraneous effects of individual differences into BLOCKS to remove variation from MSE and give a clearer view of the differences in treatments

12. Completely Randomized Blocks
SSE ---- SSBL (blocks: variation due to operator differences) and SSE (other variation)
SST = SSTR + SSBL + SSE
Partition the total sum of squares into 3 parts:

13. A1 ANOVA Table Source DF SS MS F p Treatment (k-1) Blocks (b-1) Error
Total

14. Example 1
An important factor in selecting software for word processing and database management systems is the time required to learn how to use the system. To evaluate three file management systems, a firm designed an experiment involving five word-processing operators. Since operator variability was believed to be a significant factor, each of the 5 operators were trained on each of the three systems. The data collected are times in hours and are given in the chart. Use a 5% significance level to test for any difference in the mean training times for the three systems.

15. System A
System B
System C
Operator 1 16 24
Operator 2 19 17 22
Operator 3 14 13
Operator 4 12 18
Operator 5

16. Enter data into an N x 3 matrix [D] as
Obs. Factor Block …

17. ANOVA TABLE: Ho vs. Ha Source DF SS MS F p Treatment (A) 2 103 51.5
56.36 MSTR/MSE
.000
Blocks
(B) 4 65
16.25
17.64 MSBL/
MSE
Error 8
7.3
.9125
Total 14
175.3