In: Statistics and Probability
A pool of participants was randomly divided into FIVE treatment groups. The groups were administered daily doses of vitamin C over a 12-month period. The data in the table represents the number of cold and flu viruses reported by the participant as a function of their vitamin C dosage. Using α = .05, analyze the data using the correct statistical procedure.
0mg |
250mg |
500mg |
1000mg |
2000mg |
6 |
3 |
3 |
4 |
1 |
5 |
4 |
3 |
1 |
0 |
3 |
4 |
5 |
2 |
1 |
4 |
5 |
6 |
1 |
0 |
3 |
4 |
5 |
3 |
2 |
2 |
3 |
2 |
1 |
0 |
3 |
2 |
4 |
2 |
1 |
5 |
4 |
1 |
1 |
1 |
3 |
5 |
3 |
2 |
0 |
3 |
5 |
4 |
0 |
2 |
Be sure to use a Post Hoc test (I recommend Tukey). Also, click on options and select descriptive statistics when you run your test.
I performed the analysis using basic commands in MS Excel and I am giving the steps to be followed-
a) H0: There is no difference between the five groups(All the group has the same effect).
b) To test the difference between the more than two groups, we use ANOVA.
c) F-value=14.46
d) Yes, since the F-value is greater than the cut-off value of F-distribution.
e) Though it is a very long and time taking process to calculate Tukey for each pair of the level. Using pair-wise t-test with unequal-variances we have
n=5*(5-1)/2=10 numbers of groups out which only 6 are significantly different.