Question

ANONVA test. Perform the ANONVA test in Statistics and explain the group if any of the stores outperform the others. Explain how you arrived at your answer. Be sure to share the null and alternative hypotheses. The p-value is 0.966.

January-March Franchise Performances are: Store 1 (3), Store 2 (3), Store 3 (2.8) ,Store 4 (3.1), Store 5 (3.2).

April-June Franchise Performances are: Store 1 (2.4), Store 2 (2), Store 3 (3.2), Store 4 (2.3), Store 5 (3).

July-September Franchise Performances are: Store 1 (4), Store 2 (3.8), Store 3 (4.1), Store 4 (5), Store 5 (2).

October-Dec. Franchise Performances are: Store 1 (6), Store 2 (7.1), Store 3 (7.9), Store 4 (7), Store (6.6).

P-Value is 0.966.

Answer 1

The following table is obtained Group 1 Group 2 Group 3 2.8 3.2 Group 4 Group 5 3.2 2.3 2.4 4 6 2 2 3.8 6.6 15.4 3.85 66.76 1.578 7.47 4 15.9 3.975 77.85 2.21 14.6475 4 14.8 3.7 66.8 2.003 12.04 Sum- 17.4 4.35 88.9 2.098 13.21 4.5 97.3 2.331 16.3 Average- St. Dev. = The total sample size is N- 20. Therefore, the total degrees of freedom are dftotal 20-1-19 Also, the between-groups degrees of freedom are dfbetueen- 5-1-4, and the within-groups degrees of freedom are

dfwithin-dftotal-dfbetween-19-4 = 15 First, we need to compute the total sum of values and the grand mean. The following is obtained Σ x1-15.4-15.9 + 18 + 17.4 + 14.8 = 81.5 Also, the sum of squared values is 66.76+77.85+97.3+88.9+66.8 397.61 Based on the above calculations, the total sum of squares is computed as follows 81.52 397.61- 65.498 20 SStotalー The within sum of squares is computed as shown in the calculation below SwithinS Swithingroups7 7.47 +14.6475 16.3 + 13.21 +12.04 63.668 The between sum of squares is computed directly as shown in the calculation below The between sum of squares is computed directly as shown in the calculation below Now that sum of squares are computed, we can proceed with computing the mean sum of squares Sbetween-SSbetween -1.83 = 0.458 · between SSwithin63.668 MSuithind fwithin 4.245 15

Now that sum of squares are computed, we can proceed with computing the mean sum of squares 1.830.458 S Sbetween 1.83 M Sbetueendfbetween 63.668 = 4.245 MS within - dfwithin 15 Finally, with having already calculated the mean sum of squares, the F-statistic is computed as follows: MSbetuen 0.458 M Swithin 4.245 (1)Null and Alternative Hypotheses The following null and alternative hypotheses need to be tested Ha: Not all means are equal The above hypotheses will be tested using an F-ratio for a One-Way ANOVA. (2) Rejection Region Based on the information provided, the significance level is α , and the degrees of freedom are dfi 4 and df2 4, therefore, the rejection region for this F-test is R {F : F > E 1.2E + 308) (3) Test Statistics MSbetuen 0.458 M Swithin4.245

(4) Decision about the null hypothesis Since it is observed that F = 0.108 〈 F, = 1.2E + 308, it is then concluded that the null hypothesis is not rejected. 0.978, and since p 0.978 >, it is concluded that Using the P-value approach: Th the null hypothesis is not rejected. e p-Value is p