Question

A researcher studying four different populations proposes the following:

                 H0 : μ1=μ2=μ3=μ4

                 H1 : not all μi are equal (i=1,2,3,4 )

            He takes samples from each of the populations, obtaining the following data:

	Group 1	Group 2	Group 3	Group 4
Sample mean	110	105	91	102
Sample size	8	9	5	6

            The within-group variation of the data (SSW) is equal to 108.

1. Using the Tukey-Kramer procedure, calculate the critical range for making a comparison of groups 1 and 4.      
2. Using the data and your answer to part (a), explain whether or not the null hypothesis should be rejected.

Answer 1

We have total number of observations, n= 28

Number of groups = t=4

So, the error degree of freedom = 28-4-1

Error df= 23

Also SSW= 108

Hence the mean square error = SSW/error df

MSE = 108/23

MSE = 4.695652

Standard error =

s= 2.167

Answer(a):

For obtaining critical range for comparison we need to calculate table value of at 95% confidence level for given data

q(0.95,4,23) = 3.91356

Critical range for making a comparison of group 1 and group 4 can be obtained as below:

Answer(b):

We have to test:

H0: μ1=μ4

H1: μ1≠μ4

The difference between sample 1 and sample 4 is

The difference sample means of group 1 and group 4 is 8 which is greater than the critical range calculated above.

i.e.

The above expression indicates that we have enough evidence against null hypothesis to reject it, so we reject the null hypothesis and conclude that means of group 1 and group 4 is significantly different.