Question

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.  Based on her own experiences in court, a prosecutor believes that some judges provide more severe punishments than other judges for people convicted of domestic violence.  Five of the most recent domestic violence sentences (in years) handed down by three judges are recorded below.

Judge 1	Judge 2	Judge 3
1	3	1
1	2	5
3	4	2
2	3	1
2	4	1

Using this information, test the null hypothesis at the .05 level of significance that judges do not vary in the sentence lengths imposed on individuals convicted of domestic violence.  In so doing, please: (1) identify the research and null hypotheses, (2) the critical value needed to reject the null, (3) the decision that you made upon analyzing the data, and (4) the conclusion you have drawn based on the decision you have made.

Answer 1

The Hypothesis:

H0: The lengths of the sentences are equal:

Ha: At least one of the sentence lengths differs from the rest.

The Critical Value:

The critical value for df1 = 2 and df2 = 12 at = 0.05 is 3.8853

Reject H0, if F > 3.8853

The Test Statistic:

F = 1.955 (Calculations are given after the conclusion)

The Decision:

Since F (1.955) is < Fc (3.8853), Fail to reject H0.

The Conclusion:

There is not sufficient evidence at the 5% significance level to conclude that there is a difference in the sentence length passed by the 3 judges.

_________________________________________________________

The summary statistics are as below

	A	B	C
Total	9	16	10
n	5	5	5
Mean	1.800	3.200	2
Sum Of Squares	2.8	2.8	12
Variance	0.7	0.7	3.000

Calculations For the ANOVA Table:

Overall Mean = [(9) + (16) + (10) ] / 15 = 2.33

SS treatment = SUM n* ( - overall mean)² = 5 * (1.8 - 2.33)² + 5 * (3.2 - 2.33)² + 5 * (2 - 2.33)² = 5.73

df1 = k - 1 = 3 - 1 = 2

MSTR = SS treatment/df1 = 5.73 / 2 = 2.87

SSerror = SUM [Sum of Squares] = 2.8 + 2.8 + 12 = 17.6

df2 = N - k = 15 - 3 = 12

Therefore MS error = SSerror/df2 = 17.6/12 = 1.4667

F = MSTR/MSE = 2.87/1.4667 = 1.955