Question

A criminal justice researcher is interested in studying the relationship between the religious affiliation of offenders and sentence (in years) handed out. Here is the data she is working with: CatholicsProtestantsJewsMuslims10101212161012121616161012131612121318131312131614121313121811131211121112121412.

Calculate the within-group sum of squares. Calculate the between-group sum of squares.  Calculate the mean square between groups (MSbetween). Calculate the mean square within groups (MSwithin). Calculate the F ratio. Are your results statistically significant, and if so, at what level? What are your conclusions about the null hypothesis?

Calculate the within-group sum of squares.

Answer 1

Answer:

The following table is obtained: Group 1 Group 2 Group 3 Group 4 10 10 12 12 16 10 12 12 16 16 16 10 12 13 16 12 12 13 18 13 13 12 13 16 14 12 13 13 12 18 11 13 12 11 12 11 12 12 14 12 Sum = 129 127 137 124 Average - 12.9 12.7 13.7 12.4 1697 1671 1923 1560 St. Dev. = 1.912 2.541 2.263 1.578 SS = 32.9 58.1 46.1 22.4 n = 10 10 10 10 The total sample size is N = 40. Therefore, the total degrees of freedom are: dfrotal = 40 -1= 39 Also, the between-groups degrees of freedom are dfbetween = 4 -1 = 3, and the within-groups degrees of freedom are: dfwithin = df total - dfbetween = 39 - 3 = 36

First, we need to compute the total sum of values and the grand mean. The following is obtained Xij = 129 + 127 + 137 + 124 = 517 Also, the sum of squared values is x} = 1697 +1671 + 1923 + 1560 = 6851 Based on the above calculations, the total sum of squares is computed as follows 2 Stotal = x I 13-15 (3x3) 5172 = 6851 - 40 = 168.775 The within sum of squares is computed as shown in the calculation below: SS within SS withingroups = 32.9 + 58.1 + 46.1 – 22.4 = 159.5 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: MS within SS within S Steen 9.275 M Steen = = 3.092 df between 3 159.5 = 4.431 dfcithin 36 Now that sum of squares are computed, we can proceed with computing the mean sum of squares S Steen 9.275 M Sbetween = 3.092 3 between MS within SS within dfithin 159.5 36 = 4.431

Finally, with having already calculated the mean sum of squares, the F-statistic is computed as follows: M Stuen F= 3.092 4.431 = 0.698 MS within (1) Null and Alternative Hypotheses The following null and alternative hypotheses need to be tested: Ho: M1 = P2 = M3 = 4 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 a = 0.05, and the degrees of freedom are df1 = 3 and df2 = 3, therefore, the rejection region for this F-test is R = {F: F> Fi = 2.866) (3) Test Statistics M Set ween F= 3.092 4.431 = 0.698 MS within (4) Decision about the null hypothesis Since it is observed that F = 0.698<F. = 2.866, it is then concluded that the null hypothesis is not rejected. Using the P-value approach: The p-value is p = 0.5595, and since p = 0.5595 > 0.05, it is concluded that the null hypothesis is not rejected. (5) Conclusion It is concluded that the null hypothesis Ho is not rejected. Therefore, there is not enough evidence to claim that not all 4 population means are equal, at the a = 0.05 significance level.