Question

The ANOVA summary table for an experiment with six groups, with five values in each group, is shown to the right. Complete parts (a) through (d) below.

Source Degrees of Freedom Sum of Squares Mean Square (Variance)

   F

Among groups C −1 =55    SSA=150 MSA =3030    FSTAT =3.003.00

Within groups n- c = 2424 SSW =240 MSW =1010   

Total N −1 =2929 SST = 390

a. At the 0.05 level of significance, state the decision rule for testing the null hypothesis that all six groups have equal population means. Determine the hypotheses. Choose the correct answer below.

A.

H0​: μ1 = μ2 =• • =μ5

H1​: μ1 ≠ μ2 ≠•• ≠μ5

B.

H0​: μ1 = μ2 =• • =μ6

H1​: Not all the means are equal.

C.

H0​: μ1 = μ2 =• • =μ6

H1​: μ1 ≠ μ2 ≠•• ≠μ6

D.

H0​: μ1 = μ2 =• • =μ5

H1​: Not all the means are equal.

Find the test statistic.

FSTAT =

​(Round to two decimal places as​ needed.)

Determine the critical value.

Fα =

​(Round to two decimal places as​ needed.)

b.

What is your statistical​ decision?

c.

At the 0.05 level of​ significance, what is the​ upper-tail critical value from the Studentized range​ distribution?

Qα =

​(Round to two decimal places as​ needed.)

d.To perform the​ Tukey-Kramer procedure, what is the critical​ range?

Critical range =

​(Round to three decimal places as​ needed.)

Answer 1

a) option B

H0​: μ1 = μ2 =• • =μ6

H1​: Not all the means are equal

FSTAT =3.00

critical value F α = 2.62

c)

critical value of q with 0.05 level and at k=6 and N-k=24 degree of freedom=

4.37

d)

Tukey's (HSD) for group i and j =                  (q/√2)*(sp*√(1/ni+1/nj)         =

6.180