In: Statistics and Probability
A medical researcher wishes to try three different techniques to lower the 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. Find the test statistic F to test the claim that there is no difference between the means.
Group 1 Group 2 Group 3
11 8 6
12 2 12
9 3 4
15 5 8
13 4 9
8 0 4
11.33 3.67 7.17 (Mean)
2.58 2.73 3.13(SD)
1. Identify the null hypothesis
2. Identify the alternative hypothesis
3. Determine the F value of the F test statistics
4. Identify the P-value
5. State the conclusion of the null hypothesis
6. State the final conclusion that addresses the original claim
7. Write the result in APA format.
Group 1 | Group 2 | Group 3 | Total | |
Sum | 68 | 22 | 43 | 133 |
Count | 6 | 6 | 6 | 18 |
Mean, Sum/n | 11.33333 | 3.666667 | 7.166667 | |
Standard deviation | 2.581989 | 2.73252 | 3.125167 |
Null and Alternative Hypothesis:
Ho: µ1 = µ2 = µ3
H1: At least one mean is different.
Number of treatment, k = 3
Total sample Size, N = 18
df(between) = k-1 = 2
df(within) = N-k = 15
df(total) = N-1 = 17
SS(between) = (x̅1)²*n1 + (x̅2)²*n2 + (x̅3)²*n3 - (Grand Mean)²*N = 176.7778
SS(within) = (n1-1)*s1² + (n2-1)*s2² + (n3-1)*s3² = 119.5
SS(total) = SS(between) + SS(within) = 296.2778
MS(between) = SS(between)/df(between) = 88.3889
MS(within) = SS(within)/df(within) = 7.9667
Test statistic:
F = MS(between)/MS(within) = 11.0948
p-value = F.DIST.RT(11.0948, 2, 15) = 0.0011
ANOVA | |||||
Source of Variation | SS | df | MS | F | P-value |
Between Groups | 176.7778 | 2 | 88.3889 | 11.0948 | 0.0011 |
Within Groups | 119.5000 | 15 | 7.9667 | ||
Total | 296.2778 | 17 |
Decision:
P-value < α, Reject the null hypothesis.
There is enough evidence to reject the claim that there is no difference between the means.