Question

6. A medical researcher wishes to try three different techniques to lower blood pressure of patients with high blood pressure. The subjects are randomly selected and assigned to one of three groups. Group 1 is given medication, Group 2 is given an exercise program, and Group 3 is assigned a diet program. At the end of six weeks, each subject’s blood pressure is recorded. Alpha=.05

Group 1	Group 2	Group 3
13	8	8
12	5	12
11	2	4
15	3	6
9	4	9
8	0	4

(a) What will an ANOVA test do in this situation?

(b) State the hypotheses for this ANOVA test.

(b) Create an ANOVA summary table for the test statistic F. You can use Stat-Test-ANOVA (or other technology) to create the chart.

(c) Using either the rejection region or the p-value, decide if you should reject the null hypothesis.

Answer 1

A medical researcher wishes to try three different techniques to lower blood pressure of patients with high blood pressure. The subjects are randomly selected and assigned to one of three groups. Group 1 is given medication, Group 2 is given an exercise program, and Group 3 is assigned a diet program. At the end of six weeks, each subject’s blood pressure is recorded.

Group 1	Group 2	Group 3
13	8	8
12	5	12
11	2	4
15	3	6
9	4	9
8	0	4

a)

Here we want to test equality of three means of blood pressure for group 1, group 2, group 3. i.e we have one factor which is group.

Answer: So, we will perform a One-way ANOVA test in this situation.

b)

Null and Alternative hypotheses for this ANOVA test:-

[ Where, Population mean for Group 1,   Population mean for Group 2, Population mean for Group 3 ]

We perform this one-way ANOVA test in statistical software MINITAB.

Steps for MINITAB:-

1) Open minitab-> Enter the data in three different column C1, C2, C3.

2) Click on " Stat " -> Click on " ANOVA " -> Click on " One-Way "

3) Select " Response data are in a separate column for each factor level " -> " In Responses Box select " Group 1, Group 2, Group 3 "

4) Click on " OK " -> And you get the " Output "

From whole output we copy only the " Analysis of Variance " part.

Output:

Analysis of Variance

Source DF Adj SS Adj MS F-Value P-Value
Factor 2 176.8 88.389 11.09 0.001
Error 15 119.5 7.967
Total 17 296.3

So from MINITAB output we get,

ANOVA table:-

Source	DF	SS	MS	F-Value	P-Value
Factor	2	176.8	88.389	11.09	0.001
Error	15	119.5	7.967
Total	17	296.3

c)

Rejection Rule Base on P-value:-

We reject the null hypothesis at level 0.05 (given) if we get,

For this problem we get , p-value= 0.001 and 0.05 (given)

i.e we get,

Result:- We have sufficient evidence to reject the null hypothesis.