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

The ANOVA summary table for an experiment with four Degrees of Sum of Mean Square groups, with five values in each group, is shown to the right. Source Freedom Squares (Variance) F Complete parts (a) through (d) below. Among C - 1 =3 SSA = 240 MSA = 80 FSTAT = 4.00 groups Within n - c = 16 SSW = 320 MSW = 20 groups Total n - 1 = 19 SST = 560 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 0.01 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. OA. Ho: My = H2= . . . = H4 OB. Ho: H1 = H2= . . . =15 Hy: My $ 12 # . . . $ HA H,: Not all the means are equal. O C. Ho: Hy = 12= . . . = H5 OD. Ho: My = H2= . . . = HA Hy: My * H2 # . . . $ 15 H: Not all the means are equal. Find the test statistic. FSTAT= (Round to two decimal places as needed.) Determine the critical value at the 0.01 level of significance Fa =(Round to two decimal places as needed.) b. What is your statistical decision? Ho. There is evidence to conclude that there is a difference in the population means of the groups