Question

State laws vary in the degree to which employers are allowed to have access to information about the criminal history records of potential employees. In particular, Florida allows employers very open access, Michigan provides access for only some employers, and Tennessee provides access to no employers.  Employers sometimes use the criminal history records to avoid hiring ex-offenders.  As a result, ex-offenders in states where it is easier to find out who is an ex-offender have worse employment outcomes (lower wages) than ex-offenders in states with more restrictive laws.  To test this hypothesis, a researcher surveys 50 ex-offenders each in Florida, Michigan and Tennessee.  The average hourly wages for these ex-offenders are listed below by state.

= $7.94                                                = $8.20                                  = $9.40

n_Florida= 50                                          n_Michigan= 50                                        n_Tennessee= 50

The grand mean is $8.51

TSS (Total Sum of Squares) = 635.54

Calculate the Between Sum of Squares

1. What is the independent variable and what is the dependent variable?                                 

IV:

DV:

2. Test the null hypothesis that the three population means are not different against the alternative that they are different.  Use an alpha of .01

Anova	Sum of Squares	df	Variance	F
Between
Within
Total	635.54

1. If they are different, conduct the Tukey’s HSD test to determine which ones are different.

2. What is the strength of the relationship?

                        

Answer 1

(a) The Independent Variable is offender and the dependent variable is wages

(b) The Hypothesis:

H0: There is no difference between the population mean wages earned.

Ha: At least one mean is different from the others.

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The Test Statistic: The calculation are given at the end

Source	SS	DF	Mean Square	F
Between	60.66	2.00	30.33	7.76
Within/Error	574.88	147.00	3.91
Total	635.54	149.00

F = 7.76

The Critical Value: For Alpha = 0.01, df numerator = 2, df denominator = 147, F critical = 4.753

The p value for F = 7.76, df numerator = 2, df denominator = 147, p value = 0.0006

The Decision Rule:

If Ftest is > F critical, Then Reject H0.

Also if p-value is < , Then reject H0.

The Decision:

Since Ftest (7.76) is > F critical (4.753), We Reject H0.

Also since p-value (0.0006) is < (0.01), We reject H0.

The Conclusion: There is sufficient evidence at the 95% level of significance to conclude that there is a difference between the means of at least 2 groups.

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Calculations For the ANOVA Table:

Overall Mean = 8.51

SS treatment = SUM n* ( - overall mean)² = 50 * (7.94 - 8.51)² + 50 * (8.2 - 8.51)² + 50 * (9.4 - 8.51)² = 60.66

df1 = k - 1 = 3 - 1 = 2

MSTR = SS treatment / df1 = 60.655 / 2 = 30.33

SSerror = Total Sum of Squares - SS Treatment = 635.54 - 60.66 = 574.88

df2 = N - k = 150 - 3 = 147

Therefore MS error = SSerror / df2 = 574.88 / 147 = 3.91

F = MSTR/MSE = 30.33 / 3.91 = 7.76

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(c) To find which locations differ in the means we use TUKEYS HSD

It is given by

where Mi and Mj are the 2 means being compared and their positive difference is taken. n = number of replicates in each sample. Here n = 50, MSerror = 3.91

The Rule is that if Tukeys Observed is > Tukeys Critical, then there is a significant difference between the groups.

Tukeys critical is found from the critical values table for = 0.01, for k (no. of columns) = 3 on the horizontal and df error = 147 on the vertical. The Value = 4.18 (Using an online calculator)

M1 = 7.94, M2 = 8.2, M3 = 9.4

Group 1 and Group 2 : M1 - M2 = Absolute (7.94- 8.2) = 0.26

Therefore HSD = which is 0.26 / 0.28 = 0.93 which is < 4.18. Therefore there is no significant difference.

Group 1 and Group 3 : M1 - M2 = Absolute (7.94- 9.4) = 1.46

Therefore HSD = which is 1.46 / 0.28 = 5.21 which is > 4.18. Therefore there is a significant difference.

Group 2 and Group 3 : M1 - M2 = Absolute (8.2- 9.4) = 1.2

Therefore HSD = which is 1.2 / 0.28 = 4.29 which is > 4.18. Therefore there is a significant difference.

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(d) The Strength of The relationship = = SS between / SS Total = 60.66 / 635.54 = 0.095

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