Question

The ANOVA summary table for an experiment with four groups, with seven values in each group, is shown to the right. Complete parts (a) through (d) below. Degrees of Freedom C-1 =3 Sum of Squares SSA = 120 Mean Square (Variance) MSA = 40 F FSTAT = 2.00 Source Among groups Within groups Total n-c= 24 SSW = 480 MSW = 20 n-1 = 27 SST = 600 Click here to view page 1 of the F table. Click here to view page 2 of the F table. Click here to view page 3 of the F table. Click here to view page 4 of the F table. a. At the 0.05 level of significance, state the decision rule for testing the null hypothesis that all four groups have equal population means. Determine the hypotheses. Choose the correct answer below. = 44 O A. Ho: 41 = H2 = • • • = 14 Hj: Not all the means are equal. B. Ho: My = 42 Hy: My427 •••14 OD. Ho: M1 = H2= H:44* uz 7 ...=47 0 C. Ho: H=H2=:::=17 Hy: Not all the means are equal. #47 Find the test statistic. FSTAT (Round to two decimal places as needed.)
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Answer 1

Solution :

Degrees of Freedom C-1 = 3 Sum of Squares SSA= 120 Mean Square (Variance) MSA = 40 F FSTAT = 2.00 Source Among groups Within groups Total n-c=24 SSW = 480 MSW = 20 n-1 = 27 SST = 600

a.

A. HO: H = H2=..=H4 H: Not all the means are equal.

F_STAT = 2.00

Determine the critical value.

Fα = 3.01

b.

Fail to reject Ho. There is not sufficient evidence to conclude that there is a difference in the population means of the groups.