Solution
Case 1
There are two variables in the question, shape and number of trees. Number of trees is the dependent
variable. The dependent variable is quantitative. The shape is the independent variable. The
independent variable is categorical. There are three categories: square, round, rectangular. The model
Y~X now suggests ANOVA test.
Assumption checking:
1.The four samples are independent.
2.Are they normally distributed?
The QQ Plots suggest good normality.
SAS code
data treeshape;
infile "H:\stat350\treeshape.csv" dlm=',' firstobs=2;
input numberoftrees shape $;
run;
proc univariate data=treeshape normal;
var numberoftrees;
class shape;
QQplot/ncol=1 nrow=3;
run;
3. Are variances homogeneous in the four groups?
The boxplots suggest the variances are not equal. To make sure this, we run levene’s test for
homogeneity.
example of one-way ANOVA
The ANOVA Procedure
Levene's Test for Homogeneity of numberoftrees Variance
ANOVA of Squared Deviations from Group Means
Source DF Sum of Squares Mean Square F Value Pr > F
shape
2
223.4
111.7
Error
36
1802.6
50.0735
2.23 0.1220
Welch's ANOVA for numberoftrees
Source
DF F Value Pr > F
shape
2.0000
Error
23.3913
Since the p-value (0.122) is big enough, the
0.28 0.7550
 s for each group are the same.
SAS code
proc boxplot data=treeshape;
plot numberoftrees * shape/boxstyle=skeletal;
run;
proc ANOVA data=treeshape;
title example of one-way ANOVA;
class shape;
model numberoftrees = shape;
means shape / hovtest welch;
run;
The suggested method now is ANOVA. Output and the source code are shown below.
The ANOVA Procedure
Dependent Variable: numberoftrees
Source
DF Sum of Squares Mean Square F Value Pr > F
Model
2
3.1282051
1.5641026
Error
36
255.8461538
7.1068376
Corrected Total 38
258.9743590
0.22 0.8035
SAS code
proc anova data=treeshape;
class shape;
model numberoftrees=shape;
Conclusion: According to the p-value (0.8035), the ANOVA test is not significant. Hence, the shape does not
effect on the number of trees.
Case 2
Y(categorical)~X(categorical), the model suggests contingency test.
data case2;
input improvement $ drug $ count;
datalines;
yes clozapine 61
yes haloperidol 51
no clozapine 102
no haloperidol 108
;
proc freq data=case2;
tables improvement*drug/chisq nocol norow nopercent expected;
weight count;
run;
The FREQ Procedure
Table of improvement by drug
improvement
drug
Frequency‚
Expected ‚clozapin‚haloperi‚
ƒƒƒƒƒƒƒƒƒˆƒƒƒƒƒƒƒƒˆƒƒƒƒƒƒƒƒˆ
no
‚
102 ‚
108 ‚
‚ 106.3 ‚ 103.7 ‚
ƒƒƒƒƒƒƒƒƒˆƒƒƒƒƒƒƒƒˆƒƒƒƒƒƒƒƒˆ
yes
‚
61 ‚
51 ‚
‚ 56.696 ‚ 55.304 ‚
ƒƒƒƒƒƒƒƒƒˆƒƒƒƒƒƒƒƒˆƒƒƒƒƒƒƒƒˆ
Total
163
159
Total
210
112
322
The expected frequency is 106.3, 103.7, 56.696 and 55.304, they are all greater than 5. Other wise
Fisher’s exact test should be used.
Statistics for Table of improvement by drug
Statistic
DF
Value
Prob
ƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒ
Chi-Square
1
1.0148
0.3138
Likelihood Ratio Chi-Square
1
1.0158
0.3135
Continuity Adj. Chi-Square
1
0.7927
0.3733
Mantel-Haenszel Chi-Square
1
1.0116
0.3145
Phi Coefficient
-0.0561
Contingency Coefficient
0.0560
Cramer's V
-0.0561
The p-value = 0.3138 >0.5, the contingency test suggests there is an insignificant association between
patient’s improvement and the drug used.
Case 3
There are two variables group and hour of pain relief. Group is the independent variable. The
independent variable is categorical. There are two categories: one is demerol, the other is experimental
drug. The hour of pain relief is dependent variable. The dependent variable is quantitative. We should
begin with two sample t-test for this case.
Assumption checking:
1. The two samples are independent.
2. Are the two samples normally distributed?
First, we plot a QQ plot.
The UNIVARIATE Procedure
The QQ plots do not suggest the normality, so we use the tests for normality.
Group 1:
Tests for Normality
Test
Statistic
p Value
W
0.764281 Pr < W
0.0178
Kolmogorov-Smirnov D
0.247528 Pr > D
>0.1500
Shapiro-Wilk
Cramer-von Mises
W-Sq 0.121989 Pr > W-Sq
0.0447
Anderson-Darling
A-Sq 0.733319 Pr > A-Sq
0.0307
Since the p-value (0.0178) is smaller than 0.05, there is moderate evidence for nonormality.
Group 2:
Tests for Normality
Test
Statistic
p Value
W
0.72105 Pr < W
0.0039
Kolmogorov-Smirnov D
0.372958 Pr > D
<0.0100
Shapiro-Wilk
Cramer-von Mises
W-Sq 0.216307 Pr > W-Sq <0.0050
Anderson-Darling
A-Sq 1.100942 Pr > A-Sq <0.0050
Since the p-value (0.0039) is smaller than 0.01, there is strong evidence for nonnormality.
Conclusion: Since the distributions of the two samples do not follow normal distribution and the two
samples are independent, we should use WMW test.
SAS code
data relief;
infile "H:\stat350\relief.csv." dlm=',' firstobs=2;
input group $ hour;
run;
proc univariate data=relief normal plot;
var hour;
class group;
qqplot/ ncol=2 nrow=1;
run;
proc univariate data=relief normal;
qqplot hour / normal (mu=est sigma=est color=red l=1);
by group;
run;
Relevant WMW test output
Wilcoxon Two-Sample Test
Statistic (S)
53.0000
Normal Approximation
Z
-0.2933
One-Sided Pr < Z
0.3847
Two-Sided Pr > |Z|
0.7693
t Approximation
One-Sided Pr < Z
0.3868
Two-Sided Pr > |Z|
0.7736
Wilcoxon Two-Sample Test
Exact Test
One-Sided Pr <= S
0.3731
Two-Sided Pr >= |S - Mean|
0.7465
Z includes a continuity correction of
0.5.
SAS code
proc npar1way data=relief wilcoxon;
title "nonparametric test to compare the hours between two groups";
class group;
var hour;
exact wilcoxon;
run;
Conclusion: Look at the Exact Test. Since the p value (0.7465) of the two sided test are bigger than
0.05, we conclude that the number of pain hour between the two groups are insignificantly different.
So, Demerol is insignificantly different than the experimental drug on the effects.
Case 4
Research question: Is forearm length affected by body height?
There are two variables. One is body height, the other is forearm. Both are quantitative. Body height is independent
variable. Forearm is dependent variable.
The model Y~X now suggests linear regression.
Assumption checking:
1.
Are body height and forearm associated linearly?
Based on the picture above, body height and forearm are linear associated.
2.
Are the body height homoscedastic?
The scatter plot shows randomness, no patter, and approximately the same number of points above and below the
e=0 line.
SAS code
data forearm;
infile "H:\stat350\forearm.csv." dlm="," firstobs=2;
input height forearm;
run;
proc reg data=forearm;
model forearm=height;
plot forearm * height;
output out=forearm1 p=red r=resid;
run;
proc gplot data=forearm1;
plot resid * height/vref=0 cvref=red;
run;
3.
Are the body heights normally distributed?
QQ plot suggests the sample is normally distributed.
SAS code
proc univariate data=forearm1;
qqplot resid / normal (mu=est sigma=est);
run;
Relevant Output for linear regression is shown below.
Parameter Estimates
Variable DF Parameter Standard t Value Pr > |t|
Estimate
Error
Intercept
1
10.33121
4.80988
2.15 0.0497
height
1
0.09463
0.02929
3.23 0.0060
SAS code
proc reg data=forearm;
model forearm=height;
plot forearm * height;
output out=forearm1 p=red r=resid;
run;
proc gplot data=forearm1;
Conclusion: The linear regression function is calculated as body forearm=10.333121+0.09463*height, where
test for slope is significant with a ts=3.23 and p-value=0.006. Hence, the height has a significant positive linear
effect on weight.