STATISTICS 402B
Spring 2016
Homework Set#4 Solution
1. ANOVA Table
Source of Variation d.f.
SS
MS
F p − value
Catalyst
4 141.44 35.36 11.309
.0005
Batch
4 15.44
3.86
Day
4 12.24
3.06
Error
12 37.52 3.1267
Total
24 206.64
F -statistic=11.309, p-value=.0005 Since p-value < .05, reject H0 : τ1 = . . . = τ5 = 0 vs.
at least one τi 6= 0 at α = .05, where τi , i = 1, . . . = 5 are catalyst effects.
H1 :
Arrange the sample means in increasing order of magnitude. The underscoring procedure gives (from
JMP):
Catalyst
Means
E
D
B
3.2 3.4
5.6
———————
A
8.4
C
8.8
—————Conclusions:
• Catalysts E, D, and B provide the fastest reaction times that are not significantly different from each
other.
• Catalysts A and C have reaction times that are significantly slower that E, D and B and are not
significantly different from each other
2. (a) The sum of squares for factor A is 0.165.
(b) The number of degrees of freedom for factor A is 2.
(c) The number of degrees of freedom for factor B is 1.
(d) The mean square for error is 1.667.
(e) The p-value (or its upper bound) for the intercation test is (.05, .1).
(f) The engineers used 3 levels of factor A in the experiment.
(g) The engineers used 2 levels of factor B in the experiment.
(h) There are 2 replicates in this experiment.
(i) Would you conclude that effect of factor B depends on the level of factor A?(yes/no) N0.
(j) Give a reason for your answer in the above question:
p-value for interaction>.05.
(k) An estimate of the standard deviation of the response variable is 1.29.
1
3. (a) .
Method
1
2
3
µ.j.
0.5
0.3630
0.4940
0.5630
0.4733
Dosage
1.0
1.5
0.4820 0.5835
0.5465 0.5840
0.6095 0.6315
0.5460 0.5997
2.0
0.6410
0.6660
0.7230
0.6767
µi..
0.517375
0.572625
0.631750
(b) ANOVA Table
Source of Variation d.f.
SS
MS
F p − value
Method
2 0.05234658 0.026173 11.7019
.0015
Dosage
3 0.13270183 0.044234 19.7767
< .0001
Method*Dosage
6 0.01529742 0.002550 1.1399
0.3971
Error
12 0.02684000 0.002237
Total
23 0.22718583
(c) The graph does indicate that most of the time the differences in dosage effects are not affected by
the method, except for Method 1, the means appear to be higher than expected for dosage 1.5 mg.
and 2.0 mg., if there was no interaction. However, since the interaction is not significant, we elect to
consider the main effects of Dosage and Method to be independent.
(d) The Method means are 0.63175 0.572625 0.517375. The LSD (letters or underscoring) procedure and
the 95% confidence intervals both confirm that all pairs of differences are significantly different from
zero at the 5% level. This implies that the three Method means (effects) are different from each other,
with Method 1 mean being the smallest
For the Dosage means, the underscoring procedure gives (from JMP):
Dosage
Means
.5
.4733
1
1.5
2
.5460 .5997 .6767
——————(e) The four levels of Dosage are equally spaced. Thus we can use the contrast −3 − 1 + 1 + 3
to test whether there is a linear relationship between the levels of Dosage and the Dosage means
0.4733 0.5460 0.5997 0.6767. Using the contrast method for Dosage in JMP (sse the JMP otput) we
have the t-statistic is 7.979 with a very small p-value. Thus we see that there is a linear relationship
as hypothesized.
√
p
(f) (0.363 − 0.494) ± t.25,12 · .002237 2/2 = −.131 ± 2.179 · 0.04729 = (−.234, .0279)
(g) See attached JMP output for the plots.
2
Two-Way Factorial (Method X Dosage) :
Side-Effects Experiment
Dosage
Least Squares Means Table
Level
Analysis of Variance
Source
DF
Model
Error
C. Total
11
12
23
Effect Tests
Source
Sum of
Squares
0.20034583
0.02684000
0.22718583
Nparm
DF
2
3
6
2
3
6
Method
Dosage
Dosage*Method
Mean Square
F Ratio
0.018213
0.002237
8.1430
Prob > F
0.0005*
Sum of
Squares
0.05234658
0.13270183
0.01529742
F Ratio
11.7019
19.7767
1.1399
Effect Details
Method
Least Squares Means Table
Level
Least Sq
Mean
0.51737500
0.57262500
0.63175000
1
2
3
Std Error
Mean
0.01672075
0.01672075
0.01672075
0.517375
0.572625
0.631750
LSMeans Differences Student's t
α=
0.050
t=
LSMean[i] By LSMean[j]
Mean[i]-Mean[j]
Std Err Dif
Lower CL Dif
Upper CL Dif
1
1
0
0
0
0
0.05525
0.02365
0.00373
0.10677
0.11438
0.02365
0.06285
0.1659
2
3
Level
3
2
1
2.17881
A
B
C
2
3
Least Sq
Mean
0.47333333
0.54600000
0.59966667
0.67666667
0.5
1
1.5
2
Std Error
Mean
0.01930745
0.01930745
0.01930745
0.01930745
0.473333
0.546000
0.599667
0.676667
LSMeans Differences Student's t
α=
0.050
t=
2.17881
LSMean[i] By LSMean[j]
0.5
1
Prob Mean[i]-Mean[j]
>F
Std Err Dif
Lower CL Dif
0.0015*
Upper CL Dif
<.0001*
0.5
0 -0.0727
0.3971
0 0.0273
0 -0.1322
0 -0.0132
1
0.07267
0
0.0273
0
0.01317
0
0.13216
0
1.5
0.12633 0.05367
0.0273 0.0273
0.06684 -0.0058
0.18583 0.11316
2
0.20333 0.13067
0.0273 0.0273
0.14384 0.07117
0.26283 0.19016
1.5
2
-0.1263
0.0273
-0.1858
-0.0668
-0.0537
0.0273
-0.1132
0.00583
0
0
0
0
0.077
0.0273
0.01751
0.13649
-0.2033
0.0273
-0.2628
-0.1438
-0.1307
0.0273
-0.1902
-0.0712
-0.077
0.0273
-0.1365
-0.0175
0
0
0
0
Level
-0.0553
0.02365
-0.1068
-0.0037
0
0
0
0
0.05913
0.02365
0.0076
0.11065
-0.1144
0.02365
-0.1659
-0.0629
-0.0591
0.02365
-0.1106
-0.0076
0
0
0
0
2
1.5
1
0.5
A
B
B
Least Sq
Mean
0.67666667
0.59966667
0.54600000
0.47333333
C
Levels not connected by same letter are significantly
different.
Contrast
Test Detail
Least Sq
Mean
0.63175000
0.57262500
0.51737500
Levels not connected by same letter are significantly
different.
0.5
1
1.5
2
Estimate
Std Error
t Ratio
Prob>|t|
SS
SS
0.132
-0.75
-0.25
0.25
0.75
0.1659
0.0216
7.6862
5.7e-6
0.1321
NumDF
1
DenDF
12
F Ratio
59.0772
Prob > F
<.0001*
Dosage*Method
Least Squares Means Table
Level
0.5,1
0.5,2
0.5,3
1,1
1,2
1,3
1.5,1
1.5,2
1.5,3
2,1
2,2
2,3
Least Sq
Mean
0.36300000
0.49400000
0.56300000
0.48200000
0.54650000
0.60950000
0.58350000
0.58400000
0.63150000
0.64100000
0.66600000
0.72300000
Residual Distribution
Std Error
0.03344149
0.03344149
0.03344149
0.03344149
0.03344149
0.03344149
0.03344149
0.03344149
0.03344149
0.03344149
0.03344149
0.03344149
Residual Amount vs. Method
0.075
0.06
-1.64 -1.28
0.0
0.67
1.28
0.02
0
-0.02
-0.04
-0.06
0.04
0.1
0.25
0.4 0.55 0.7 0.82
0.92
Normal Quantile Plot
Residual Amount vs. Dosage
0.050
0.025
0.025
0.000
0.000
-0.025
-0.025
-0.050
-0.050
1
2
Method
3
Residual by Predicted Plot
1.64
0.04
0.075
0.050
-0.075
-0.67
-0.075
0.5
1
Dosage
1.5
2