Question

An experiment has a single factor with six groups and five values in each group. In determining the among-group variation, there are 5 degrees of freedom. In determining the within-group variation, there are 24 degrees of freedom. In determining the total variation, there are 29 degrees of freedom. Also, note that SSA = 120, SSW = 192, SST = 312, MSA = 24, MSW = 8, and FSTAT = 3. Complete parts (a) through (d). 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. 4 a. Construct the ANOVA summary table and fill in all values in the table. Mean Square (Variance) Sum of Squares Source Degrees of Freedom Among groups Within groups Total (Simplify your answers.) b. At the 0.005 level of significance, what is the upper-tail critical value from the F distribution? F0.005 = (Round to two decimal places as needed.) c. State the decision rule for testing the null hypothesis that all six groups have equal population means. Reject Ho if (1) d. What is your statistical decision? Since Fstat is (2) than the upper-tail critical value, (3) – - Ho. There is (4) evidence to conclude there is a difference in the population means for the six groups.

Answer 1

a. According to the provided data. ANOVA table can be completed as: Source Degrees of freedom Sum of Squares Mean Square (Variance) Among groups 120 Within groups 312 Total w b. The upper tail F-critical value for degrees of freedom, 5 and 24 and the level of significance, 0.005 is obtained as 4.49 using F-distribution table. c. The decision rule can be stated as: If Fstatistic > Fcritical, then reject the null hypothesis. If Fstatistic <Fcritical, then fail to reject the null hypothesis. So, Reject Ho if FSTAT > 4.49. d. Since, FSTAT is less than the upper tail critical value, do not reject Ho. There is insufficient evidence to conclude there is a difference in the population means for the six groups.