STATISTICS 402B
Spring 2016
Homework Set#3 Solution
1. (a) ANOVA Table
Source of Variation d.f.
SS
MS
F p − value
Fiber %
4 6.2627 1.56567 12.9751
.0006
Error
10 1.2067 0.12067
Total
14 7.4693
F -statistic=12.98, p-value=.0006 Since p-value < .05, reject H0 at α = .05
Best estimates:
µ̂1 = ȳ1. = 7.4667, µ̂2 = ȳ2. = 7.5667, µ̂3 = ȳ3. = 6.8667, µ̂4 = ȳ4. = 6.7, µ̂5 = ȳ5. = 5.7667
σ̂ 2 = s2E = 0.12067
(b) Testing for linear relationship between factor means and the factor levels:
Number of factor levels=5; thus the coefficients of the orthogonal polynomial are: −2 − 1 0 + 1 + 2
Contrast Coefficients for the linear polynomial is :
Contrast Label
Coefficients
Estimate Std. Err. t-statistic p-value
Test H0 :
Linear
-2 -1 0 +1 +1
-1.422
0.2114
-6.728
0.0001
No linear relationship vs. H1 : Linear relationship exists
Since the p-value is small we conclude that there is a linear relationship exists between mean strength
of concrete (y) and the percentage of fiber (x)
Performing a straight line fit using JMP we get the fitted line as y = 7.7267 − 1.7067 ∗ x with an
R2 = 73% (see the attached JMP output).
2. (a) ANOVA Table
Source
Treatment
Block
Error
Total
d.f.
4
5
20
29
SS
1010.56
323.82
169.33
1503.71
MS
252.64
64.765
8.4665
F
29.84
P-value
0.0003
(b) Six blocks were used in this experiment.
(c) The statistic for testing the equality of the treatment means is F = 29.84 with p-value smaller than
0.05, then there is significant statistical evidence that at least one treatment is different from others.
3. (a) ANOVA Table
Source of Variation d.f.
SS
MS
F p − value
Chemical
3 12.95 4.3167 2.376
.1211
Bolt
4 157.00 39.25
Error
12 21.80 1.8167
Total
19 191.75
Since the p-value is not less than .05 we fail to reject H0 : µ1 = µ2 = µ3 = µ4 .
(b) See attached JMP output for analysis of residuals.
(c) Since the F-test associated with the ANOVA failed to find differences among the means we should
not proceed any further, because Fisher’s LSD procedure requires that H0 : µ1 = µ2 = µ3 = µ4 be
rejected first. However, some experimenter proceed with performing LSD procedure to find differeces
among the means.
Arrange the sample means in increasing order of magnitude. The underscoring procedure gives (from
JMP):
1
Chemical
Means
1
2
3
4
70.6 71.4
72.4
72.6
—————————————————-
4. (a) ANOVA Table
Source of Variation d.f.
SS
MS
F p − value
Tip
3 0.385
0.1283 14.4375
.0009
Coupon
3 0.825
0.275
Error
9 0.08 0.008889
Total
15 1.29
Since the p-value is less than .05 we reject H0 : µ1 = µ2 = µ3 = µ4 .
(b) See attached JMP output for analysis of residuals.
(c) Since the F-test associated with the ANOVA failed to find differences among the means we should
not proceed any further, because Fisher’s LSD procedure requires that H0 : µ1 = µ2 = µ3 = µ4 be
rejected first. However, some experimenter proceed with performing LSD procedure to find differeces
among the means.
Arrange the sample means in increasing order of magnitude. The underscoring procedure gives (from
JMP):
Chemical
Means
3
1
2
9.45 9.575
9.6
————————-
4
9.875
—————
5. (a) First randomly allocate the three Oils (A, B, C) to 3 piston rings of each type (using random permutations of the three letters):
Piston Ring
1
2
3
4
Brand of Oil
C A
B
C B
A
B A
C
A C
B
Then obtain a random permuation of integers 1 through 12 to determine the order of the runs:
Piston Ring
1
2
3
4
Run Order
6 12 1
10 7 9
11 3 4
5
8 2
So
Run 1 will be to test the third piston ring of Type 1 with Oil B
Run 2 will be to test the third piston ring of Type 4 with Oil B
Run 3 will be to test the second piston ring of Type 3 with Oil A
etc,;
(b) ANOVA Table
Source of Variation d.f.
SS
MS
F p − value
Brand
2 0.0374 0.01869 11.1225
0.0096
Piston Ring Type
3 1.296
0.432
Error
6 0.010 0.001681
Total
11 1.3434
Since the p-value is less than .05 we reject H0 : µ1 = µ2 = µ3 .
2
(c) See attached JMP output for analysis of residuals.
(d)
µ2 + µ 3
µ2 + µ 3
= 0 vsHa : µ1 −
>0
2
2
i.e. the contrast is c=(1,-0.5,-0.5) The t-statistics is
H0 : µ1 −
t0 =
sE
P3
i=1 ci µ̂i
qP
3
2
i=1 ci /4
= 4.7151
The p-value is 0.00165 which is smaller than 0.05. So there is statistical signifcant evidence to reject
null hypothesis.
JMP for Problem1
3
4
5
6
7