Question

Use SAS to answer the question:

A drug company tested three formulations of a pain relief medicine for migraine headache sufferers.
For the experiment 27 volunteers were selected and 9 were randomly assigned to one of three drug
formulations. The subjects were instructed to take the drug during their next migraine headache episode
and to report their pain on a scale of 1 to 10 (10 being most pain)

Drug A: 4 5 4 3 2 4 3 4 4
Drug B: 6 8 4 5 4 6 5 8 6
Drug C: 6 7 6 5 7 5 6 6 5

a) Test if the means for all three drug groups are equal.
(b) If the means are not equal, perform a multiple comparison at the 0.01 level.
(c) Create a contrast to compare group B against the mean of group A and group C at the 0.01 level.

Answer 1

We want to test the effect of drug A. B and C on headache pain.

a)

Testing of Hypothesis:

H0: The average effect of all three drugs is the same.

against,

H1: The average effect of all three drugs is different.

One way ANOVA in SAS:

data a;

input drug $ value;

datalines;

A 4

A 5

A 4

A 3

A 2

A 4

A 3

A 4

A 4

B 6

B 8

B 4

B 5

B 4

B 6

B 5

B 8

B 6

C 6

C 7

C 6

C 5

C 7

C 5

C 6

C 6

C 5

;

run;

proc anova data=a;

class drug;

model value=drug;

quit;

Decision Rule:

If p-value greater than 0.01 level of significance then accept the null hypothesis.

From the above ANOVA the p-value (0.0003) < 0.01

so we reject the null hypothesis.

i.e. Effects of all drugs are not the same.

b) For checking the individual pairwise effects of drugs we want to use paired t-test here by using SAS

The syntax for the t-test in SAS:

proc t test plots = all;

var drug $ value;

run;

The hypothesis for the t-test is,

H0: The effect of both drugs A and B is the same.

against,

H1: The effect of both drugs A and B are different.

Test Statistics:

for Drug A and B:

T = -6.0083

p-value = 0.000320

p-value < 0.01, so reject the null hypothesis at 1 % level of significance

i.e. The effect of both drugs A and B are different.

for Drug A and C:

T = -5.5470

p-value = 0.000542

the p-value < 0.01, to reject the null hypothesis at 1 % level of significance

i.e. The effect of both drugs A and C are different.

for Drug B and C:

T = -0.2062

p-value = 0.8417

the p-value > 0.01, to accept the null hypothesis at 1 % level of significance

i.e. The effect of both drugs B and C are the same.

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