ST3900/4950HOMEWORK/LAB 2 Solutions Spring, 2018

Question 1

For REACT:

Stem-and-leave plots of REACT for two doses:

react Stem-and-Leaf Plot for dose= 1.00

Frequency Stem & Leaf

1.00 4 . 8

2.00 5 . 49

1.00 6 . 9

1.00 Extremes (>=15.8)

react Stem-and-Leaf Plot for dose= 2.00

Frequency Stem & Leaf

1.00 4 . 9

2.00 5 . 05

1.00 6 . 7

1.00 Extremes (>=18.2)

(SAS 9.4 does not produce stem-and-leave plots, so 4950 students need to draw the plots by other software or by hand)

Both stem-and-leaf plots show extreme outliers, so we cannot assume REACT is normally distributed for both Doses, also supported by normality tests (p-values less than .05) and normal Q-Q plots (far off from a line; not shown here); output shown below:

Therefore, we cannot do a t test. In addition, both stem-and-leaf plots show of a similar shape. We will do a Wilcoxon rank sum test.

SAS output:

Wilcoxon Two-Sample Test
Exact Test
One-Sided Pr >= S / 0.5000
Two-Sided Pr >= |S - Mean| / 1.0000
Zincludesacontinuitycorrectionof0.5.

SPSS output:

The p value is extremely large (~1.00!!), so the distributions of REACT of the two doses are not significantly different.

i. Stem-and-leave plots of LIVER_WT for two doses:

for dose= 1.00

Frequency Stem & Leaf

1.00 9 . 8

2.00 10 . 29

1.00 11 . 8

1.00 12 . 2

For dose= 2.00

Frequency Stem & Leaf

1.00 9 . 9

1.00 10 . 5

1.00 11 . 9

1.00 12 . 0

1.00 13 . 8

We can use t tests since the data of liver_wt pass the normality tests for both doses:

All p values (sig. in the table) are larger than .05. Now we conduct 2-sample t tests:

SAS output:

SPSS output:

The p-value 0.46 (in SAS) or 0.47 (in SPSS) of the test of the equality of variances indicates that it is reasonable to assume equal variances. Therefore, we can use the pooled t-test. The p-value is 0.456 and so we cannot reject the null hypothesis. In conclusion, we have insufficient evidence to claim there is a significant difference between LIVER_WT for DOSE 1 and LIVER_WT for DOSE 2.

Question 2

Create two lists of assignments of 8 subjects; one for men and the other for women. For SPSS students, check the SPSS random number handout for details. Here is the SAS code:

** Assign each subject a random number from Unif[0,1];

DATA RANDOM;

DO SUBJ= 1TO8;

GRP=RANUNI(123);

OUTPUT;

END;

RUN;

** Sort the random numbers into 2 groups AND print the results;

PROCRANKDATA=RANDOM GROUPS=2OUT=ASGNMNT;

VAR GRP;

RUN;

PROCPRINTDATA=ASGNMNT NOOBS;

* NOOBS = take off the column of OBS;

TITLE"SUBJECT TREATMENT ASSIGNMENTS";

VAR SUBJ GRP;

RUN;

Repeat the same code again for the list for women. But change the “seed” 123 in the function RANUNI if you want a different list. For example, use RANUNI(234).

For men:

subjranduniftreatment

10.686636Aspirin

20.090040Tylenol

30.848049Aspirin

40.331484Tylenol

50.076618Tylenol

60.740453Aspirin

70.237824Tylenol

80.780284Aspirin

For women:

subjranduniftreatment

10.149149Tylenol

20.299081Tylenol

30.440395Aspirin

40.652699Aspirin

50.417568Aspirin

60.543076Aspirin

70.096551Tylenol

80.033844Tylenol

SAS CODE FOR QUESTION 1

DATA LIVEREXP;

INPUTSUBJECT $

DOSE $

REACT

LIVER_WT;

DATALINES;

1 1 5.4 10.2

2 1 5.9 9.8

3 1 4.8 12.2

4 1 6.9 11.8

5 1 15.8 10.9

6 2 4.9 13.8

7 2 5.0 12.0

8 2 6.7 10.5

9 2 18.2 11.9

10 2 5.5 9.9

;

**NUMERICAL AND GRAPHICAL SUMMARIES FOR ALL VARIABLES;

PROCUNIVARIATEDATA=LIVEREXP NORMALPLOT;

TITLE"DESCRIPTIVE STATISTICS FOR REACT";

VAR REACT;

HISTOGRAM REACT / MIDPOINTS=7.0 TO 17.0 BY 5NORMAL;

HISTOGRAM REACT / MIDPOINTS=6.0 TO 18.0 BY 3NORMAL;

RUN;

PROCSORTDATA=LIVEREXP;

BY DOSE;

RUN;

PROCUNIVARIATEDATA=LIVEREXP NORMALPLOT;

TITLE"DESCRIPTIVE STATISTICS FOR REACT AND DOSE";

BY DOSE;

VAR REACT;

RUN;

PROCNPAR1WAY WILCOXONDATA=LIVEREXP;

CLASS DOSE;

VAR REACT;

EXACTWILCOXON;

RUN;

PROCUNIVARIATEDATA=LIVEREXP NORMALPLOT;

TITLE"DESCRIPTIVE STATISTICS FOR LIVER_WT AND DOSE";

BY DOSE;

VAR LIVER_WT;

RUN;

PROCTTESTDATA=LIVEREXP;

TITLE"T-TEST FOR LIVER_WT AND DOSE";

CLASS DOSE;

VAR LIVER_WT;

RUN;