Transcript Example 2: Fabric Flammability Tests

Lesson 5
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?
Fabric Flammability Tests
Lab 1
Lab 2
Lab 3
Lab 4
Lab 5
2.9
2.7
3.3
3.3
4.1
3.1
2.9
3.3
3.2
4.1
3.7
3.2
3.5
2.7
4.2
3.1
3.7
2.8
2.7
3.1
4.2
3.2
2.8
3.3
3.5
3.1
3.4
3.5
2.7
4.2
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
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.
Example 2 - continued
• Calculate Fisher LSD and use it to
determine where the differences lie.
• LSD = 0.471
Fisher LSD
Populations
Lab 1 & 2
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
Mean Dif
LSD
0.471
0.471
0.471
0.471
0.471
0.471
0.471
0.471
0.471
0.471
Difference
(Y/N) ??
No. 21
• K = 5; n = 7 N = 35
Source
DF
SS
MS
F
TR
4
300
75
75/5.33 =
14.07
ER
30
160
5.33
Total
34
460
p
P-value
• P-value = Fcdf(14.07,E99,4,30)
• = .0000014 < alpha
• Conclusion: Reject the null hypothesis
No. 24
• n1 = 12; n2 = 15; n3 = 20; N = 47
Source
DF
TR
2
ER
44
Total
46
SS
MS
F
p
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
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:
A1 ANOVA Table
Source
Treatment
(k-1)
Blocks
(b-1)
Error
(k-1)(b-1)
Total
DF
SS
MS
F
p
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.
System A
System B
System C
Operator 1
16
16
24
Operator 2
19
17
22
Operator 3
14
13
19
Operator 4
13
12
18
Operator 5
18
17
22
• Enter data into an N x 3 matrix [D] as
• Obs.
• 16
• 16
•
…
• 22
Factor
1
2
3
Block
1
1
5
ANOVA TABLE: Ho vs. Ha
Source
DF
SS
MS
F
p
Treatment
(A)
2
103
51.5
.000
Blocks
(B)
4
65
16.25
56.36
MSTR/
MSE
17.64
MSBL/
MSE
Error
8
7.3
Total
14
175.3
.9125
.000