A medical researcher wishes to try three different techniques to lower cholesterol levels of patients with high cholesterol levels. 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 cholesterol level is recorded. Determine the test statistic F-value to test the hypothesis that there is no difference among the means. Use α = 0.05. (round to the nearest hundredth)
Group 1 | Group 2 | Group 3 |
---|---|---|
9 | 8 | 6 |
12 | 3 | 12 |
11 | 2 | 4 |
15 | 5 | 4 |
13 | 4 | 9 |
8 | 0 | 8 |
Using Excel. go to Data, select Data Analysis, choose Anova: Single Factor
ANOVA | ||||||
Source of Variation | SS | df | MS | F | P-value | F crit |
Between Groups | 176.78 | 2 | 88.39 | 11.09 | 0.00 | 3.68 |
Within Groups | 119.50 | 15 | 7.97 | |||
Total | 296.28 | 17 |
H0: μ1 = μ2 = μ3, All means are equal
H1: At least one mean is different from the others
Test statistic = 11.09
p-value = 0.00
Since p-value is less than 0.05, we reject the null hypothesis and conclude that at least one mean is different from the others.
Get Answers For Free
Most questions answered within 1 hours.