AMS394.01
Practice Midterm
Fall 2015
Name: ____________________________ ID: ___________________ Signature: ___________________
Instruction: This is an open book exam. However no communication is allowed
between students. Please provide complete solutions for full credit. Good luck!
1. We want to test the relative durability of 4 different surface coatings for optical lenses. The
durability test involves subjecting a coated lens to 150 cycles of abrasion. The response
variable is a measure of the increase in lens haziness. Please write up the SAS code, and
the R code to do the following. In addition, please provide the out and summary of your
tests/plots using one of these two programs:
(1) We are testing the two hypotheses
H0: 1 = 2 = 3= 4
vs.
Ha: At least one of the means differs from the others.
(2) Please include the follow-up tests for detecting specific differences among the means.
(3) Please also include the side-by-side boxplot to check for homogeneity of variances, and, a
residual plot.
(4) Please conduct a usual t-test to compare the mean haziness between coatings 1 and 2.
Solution:
data one;
input coating haziness;
label coating = "Lens Surface Coating"
haziness = "Lens Haziness after Abrasion";
datalines;
1 8.52
1 9.21
1 10.45
1 10.23
1 8.75
1 9.32
1 9.65
2 12.50
2 11.84
2 12.69
2 12.43
2 12.78
2 13.15
2 12.89
3 8.45
3 10.89
3 11.49
3 12.87
3 14.52
3 13.94
3 13.16
1
4 10.73
4 8.00
4 9.75
4 8.71
4 10.45
4 11.38
4 11.35
;
Run;
proc boxplot data=one;
plot haziness*coating;
title "Side-by-Side Boxplots of Response Variable";
title2 "by Levels of Treatment";
Run;
Proc glm data=one;
class coating;
model haziness = coating;
lsmeans coating /out=outmns;
means coating / cldiff bon;
output out=resout p=preds rstudent=exstdres;
title "Analysis of Variance for Optical Lens Surface Coatings";
title2 "With Follow-Up Tests";
Run;
Quit;
title 'Profile Plot';
symbol i=j;
proc gplot data=outmns;
where coating ne .;
plot lsmean*coating;
run;
quit;
goptions reset=all;
title 'Residual Plot';
proc gplot data=resout;
plot exstdres*preds;
run; quit;
data two;
input coating haziness;
label coating = "Lens Surface Coating"
2
haziness = "Lens Haziness after Abrasion";
datalines;
1 8.52
1 9.21
1 10.45
1 10.23
1 8.75
1 9.32
1 9.65
2 12.50
2 11.84
2 12.69
2 12.43
2 12.78
2 13.15
2 12.89
;
Run;
Proc univariate data=two normal;
Class coating;
Var haziness;
Title ‘check for normality’;
Run;
Proc ttest data=two;
Class coating;
Var haziness;
Title ‘Independent samples t-test’;
Run;
3
Selected output and summary:
(1)
The GLM Procedure
Dependent Variable: haziness
Lens Haziness after Abrasion
Sum of
Source
DF
Squares
Mean Square
F Value
Pr > F
Model
3
51.06744286
17.02248095
10.12
0.0002
Error
24
40.35205714
1.68133571
Corrected Total
27
91.41950000
Summary: we reject the ANOVA null hypothesis.
(2)
The GLM Procedure
Bonferroni (Dunn) t Tests for haziness
NOTE: This test controls the Type I experimentwise error rate, but it generally has a higher Type
II error rate than Tukey's for all pairwise comparisons.
Alpha
0.05
Error Degrees of Freedom
Error Mean Square
24
1.681336
Critical Value of t
2.87509
Minimum Significant Difference
1.9927
Comparisons significant at the 0.05 level are indicated by ***.
Difference
coating
Comparison
Between
Simultaneous 95%
Means
Confidence Limits
2 - 3
0.4229
-1.5699
2.4156
2 - 4
2.5586
0.5659
4.5513 ***
2 - 1
3.1643
1.1716
5.1570 ***
3 - 2
-0.4229
-2.4156
3 - 4
2.1357
0.1430
4.1284 ***
3 - 1
2.7414
0.7487
4.7341 ***
4 - 2
-2.5586
-4.5513
1.5699
-0.5659
***
4
4 - 3
-2.1357
-4.1284
-0.1430
***
4 - 1
0.6057
-1.3870
2.5984
1 - 2
-3.1643
-5.1570
-1.1716
***
1 - 3
-2.7414
-4.7341
-0.7487
***
1 - 4
-0.6057
-2.5984
1.3870
Summary: the pairwise comparisons show that coatings 1/2, 1/3, 2/4, 3/4
are significantly different at the familywise error rate of 0.05. Note,
although we used the Bonferroni method here as an example, the Tukey method
is less conservative and in general, better.
(3)
Side-by-Side Boxplots of Response Variable
by Levels of Treatment
Lens Haziness after Abrasion
16
14
12
10
8
1
2
3
4
Lens Surface Coating
5
Residual Plot
exstdres
3
2
1
0
-1
-2
-3
-4
9
10
11
12
13
preds
Summary: The box-plots make us worry about the equal variance assumptions.
The residual plot shows some concern of unequal variance too.
(4)
The UNIVARIATE Procedure
Variable:
haziness (Lens Haziness after Abrasion)
coating = 1
Tests for Normality
Test
--Statistic---
Shapiro-Wilk
W
0.953597
-----p Value------
Pr < W
0.7623
Kolmogorov-Smirnov
D
0.148529
Pr > D
>0.1500
Cramer-von Mises
W-Sq
0.027158
Pr > W-Sq
>0.2500
Anderson-Darling
A-Sq
0.196002
Pr > A-Sq
>0.2500
The UNIVARIATE Procedure
Variable:
haziness (Lens Haziness after Abrasion)
coating = 2
Tests for Normality
Test
Shapiro-Wilk
Kolmogorov-Smirnov
--Statistic---
W
D
0.949828
0.188846
-----p Value------
Pr < W
0.7281
Pr > D
>0.1500
6
Cramer-von Mises
W-Sq
0.036567
Pr > W-Sq
>0.2500
Anderson-Darling
A-Sq
0.251295
Pr > A-Sq
>0.2500
Summary: The Shapiro-Wilk test shows that both samples are normal and
thus we can continue with the independent samples t-test.
The TTEST Procedure
Variable:
coating
haziness
N
(Lens Haziness after Abrasion)
Mean
Std Dev
Std Err
Minimum
Maximum
1
7
9.4471
0.7162
0.2707
8.5200
10.4500
2
7
12.6114
0.4169
0.1576
11.8400
13.1500
Diff (1-2)
coating
-3.1643
Method
0.5860
Mean
0.3132
95% CL Mean
Std Dev
95% CL Std Dev
1
9.4471
8.7848
10.1095
0.7162
0.4615
1.5771
2
12.6114
12.2259
12.9970
0.4169
0.2686
0.9180
Diff (1-2)
Pooled
-3.1643
Diff (1-2)
Satterthwaite
-3.1643
Method
Variances
Pooled
Equal
Satterthwaite
Unequal
-3.8467
-3.8657
-2.4818
0.5860
0.4202
0.9673
-2.4629
DF
t Value
Pr > |t|
12
-10.10
<.0001
9.6468
-10.10
<.0001
Equality of Variances
Method
Folded F
Num DF
6
Den DF
6
F Value
2.95
Pr > F
0.2135
Summary: The F-test shows that the variances can be considered equal.
Therefore we adopted the pooled-variance t-test and found significant
mean differences (in terms of haziness of lenses) between coatings 1 and
2.
2. The following SAS data step inputs a two-way ANOVA data set examining the relationship
between crop density, amount of fertilizers, and crop yield. Please write up the SAS code,
and the R code to do the following. In addition, please provide the out and summary of your
tests/plots using one of these two programs:
7
(1) We are testing the ANOVA hypotheses of (a) no interaction, (b) density main effect, and (c)
fertilizer main effect.
(2) Please include the follow-up tests for detecting specific differences among the means.
(3) Please also include the side-by-side boxplot to check for homogeneity of variances, and, a
residual plot.
PROC FORMAT;
VALUE den 1='regular' 2='thick';
VALUE fert 1='low' 2='medium' 3='high';
RUN;
DATA soybean(DROP=rep);
FORMAT density den. fertilizer fert.;
DO fertilizer = 1 TO 3;
DO density = 1 TO 2;
DO rep = 1 TO 4;
INPUT yield @@;
OUTPUT;
END;
END;
END;
DATALINES;
37.5 36.5 38.6 36.5 37.4 35.0 38.1 36.5
48.1 48.3 48.6 46.4 36.7 36.4 39.3 37.5
48.5 46.1 49.1 48.2 45.7 45.7 48.0 46.4
;
Run;
Proc sort data=soybean;
By fertilizer;
Run;
proc boxplot data=soybean;
plot yield*fertilizer;
title "Side-by-Side Boxplot of Response Variable";
title2 "by Levels of fertilizer";
Run;
Proc sort data=soybean;
By density;
Run;
proc boxplot data=soybean;
plot yield*density;
8
title "Side-by-Side Boxplot of Response Variable";
title2 "by Levels of density";
Run;
TITLE3 'Tests for Interaction & Main Effects';
PROC GLM DATA=soybean ORDER=INTERNAL;
CLASS density fertilizer;
MODEL yield = density | fertilizer;
lsmeans density fertilizer density*fertilizer /out=outmns;
means density fertilizer /cldiff bon;
output out=resout p=preds rstudent=exstdres;
RUN;
Quit;
title 'Profile/Interaction Plots';
symbol i=j;
proc gplot data=outmns;
where fertilizer ne . and density ne .;
plot lsmean*density=fertilizer;
plot lsmean*fertilizer=density;
run; quit;
goptions reset=all; *resets PROC GPLOT options;
title 'Residual Plot';
proc gplot data=resout;
plot exstdres*preds;
run; quit;
9
Selected output and summary:
(1)
Source
DF
Type III SS
Mean Square
F Value
Pr > F
density
1
102.9204167
102.9204167
74.01
<.0001
fertilizer
2
417.7733333
208.8866667
150.20
<.0001
42.27
<.0001
density*fertilizer
2
117.5633333
58.7816667
Summary: we see significant interaction and main effects.
(2)
Bonferroni (Dunn) t Tests for yield
NOTE: This test controls the Type I experimentwise error rate, but it generally has a higher Type
II error rate than Tukey's for all pairwise comparisons.
Alpha
0.05
Error Degrees of Freedom
18
Error Mean Square
1.390694
Critical Value of t
2.10092
Minimum Significant Difference
1.0115
Comparisons significant at the 0.05 level are indicated by ***.
Difference
density
Between
Simultaneous 95%
Comparison
Means
Confidence Limits
regular - thick
4.1417
3.1302
-4.1417
-5.1531
thick
- regular
5.1531 ***
-3.1302
***
Bonferroni (Dunn) t Tests for yield
NOTE: This test controls the Type I experimentwise error rate, but it generally has a higher Type
II error rate than Tukey's for all pairwise comparisons.
Alpha
Error Degrees of Freedom
Error Mean Square
Critical Value of t
0.05
18
1.390694
2.63914
10
Minimum Significant Difference
1.5561
Comparisons significant at the 0.05 level are indicated by ***.
Difference
fertilizer
Between
Simultaneous 95%
Comparison
Means
Confidence Limits
high
- medium
4.5500
2.9939
high
- low
10.2000
8.6439
11.7561 ***
medium - high
-4.5500
-6.1061
-2.9939 ***
medium - low
5.6500
4.0939
-10.2000
-11.7561
-8.6439 ***
-5.6500
-7.2061
-4.0939 ***
low
- high
low
- medium
6.1061
7.2061
***
***
Summary: the pairwise comparisons show that all pairs are significantly
different from each other in means.
(3)
Side-by-Side Boxplot of Response Variable
by Levels of fertilizer
50.0
47.5
yield
45.0
42.5
40.0
37.5
35.0
low
medium
high
fertilizer
11
Side-by-Side Boxplot of Response Variable
by Levels of density
50.0
47.5
yield
45.0
42.5
40.0
37.5
35.0
regular
thick
density
Residual Plot
exstdres
2
1
0
-1
-2
36
37
38
39
40
41
42
43
44
45
46
47
48
preds
Summary: The box-plots make us worry about the equal variance assumptions
for different fertilizers, but no worries for different density levels.
The residual plot seems okay.
12
3. The following dataset examines the relationship between time to
headache relief, and the three brands of pain killers. Please use the
REGRESSION procedures in SAS and R to analyze this data set.
(1) Please write down the program for both SAS and R, and use one of
these two programs to analyze the data.
(2) Please include necessary plots and analyses to verify the
underlying model assumptions.
(3) Please include your output and summary of results.
Data three;
Input BRAND RELIEF;
Dummy1= 0;
Dummy2= 0;
If brand=1 then dummy1=1;
If brand=2 then dummy2=1;
Datalines;
1 24.5
1 23.5
1 26.4
1 27.1
1 29.9
2 28.4
2 34.2
2 29.5
2 32.2
2 30.1
3 26.1
3 28.3
3 24.3
3 26.2
3 27.8
;
Run;
Proc print data=three;
Run;
proc boxplot data=three;
plot relief*brand;
title "Side-by-Side Boxplots of Response Variable";
title2 "by brands of Treatment";
Run;
Proc glm data=three;
13
class brand;
model relief = brand;
lsmeans brand /out=outmns;
means brand / cldiff bon;
output out=resout p=preds rstudent=exstdres;
title "Analysis of Variance for Pain Relief by Drug Brands";
title2 "With Follow-Up Tests";
Run;
Quit;
Proc reg data=three;
model relief = dummy1 dummy2;
Run;
Quit;
title 'Profile Plot';
symbol i=j;
proc gplot data=outmns;
where brand ne .;
plot lsmean*brand;
run;
quit;
goptions reset=all;
title 'Residual Plot';
proc gplot data=resout;
plot exstdres*preds;
run; quit;
14
Selected output and summary:
Dear students, the only difference of what is required in this problem
versus that in Problem 1, is that I need you to write down the general
linear model. This can be accomplished by you setting up the dummy
variables and then run the regression with the dummy variables directly.
There will be other approaches but we are showing the easiest one here.
So to save time, I will only show this different part.
Obs
BRAND
RELIEF
Dummy1
Dummy2
1
1
24.5
1
0
2
1
23.5
1
0
3
1
26.4
1
0
4
1
27.1
1
0
5
1
29.9
1
0
6
2
28.4
0
1
7
2
34.2
0
1
8
2
29.5
0
1
9
2
32.2
0
1
10
2
30.1
0
1
11
3
26.1
0
0
12
3
28.3
0
0
13
3
24.3
0
0
14
3
26.2
0
0
15
3
27.8
0
0
Parameter Estimates
Parameter
Standard
Variable
DF
Estimate
Error
t Value
Pr > |t|
Intercept
1
26.54000
0.96720
27.44
<.0001
Dummy1
1
-0.26000
1.36782
-0.19
0.8524
Dummy2
1
4.34000
1.36782
3.17
0.0080
Summary: Here you see the dataset with the two dummy variables. The
estimated general linear model is:
Yˆ  26.54  0.26* dummy1  4.34* dummy 2
15