Question

In: Statistics and Probability

A pool of participants was randomly divided into FIVE treatment groups. The groups were administered daily...

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.

  1. State your null hypothesis 3pts
  1. What statistical test would you perform? 3pt
  1. What is the value of F? 3 pts.
  1. Is F significant? 3pts
  1. How many of the groups are significantly different from the others? 3pts.

Solutions

Expert Solution

I performed the analysis using basic commands in MS Excel and I am giving the steps to be followed-

  1. Select the Data tab and choose the Data Analysis in the top right-hand corner
  2. In the Data Analysis menu choose ANOVA: Single Factor and click OK
  3. In the ‘Input Range’ box, select all the data in the columns you created, including the variable names
  4. Check the ‘Labels in First Row’ box
  5. In the ‘Output Range’ box, enter a cell range where Excel will place the output and click OK
  6. If the p-value were less than 0.05, you would reject the null hypothesis that says the means of all categories are equal. If the p-value were greater than 0.05, then you would fail to reject the null.

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.


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