Statistics
Statistics
4th Edition
ISBN: 9780393929720
Author: David Freedman, Robert Pisani, Roger Purves
Publisher: Norton, W. W. & Company, Inc.
Question
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Chapter 27.6, Problem 7RE

­­(a)

To determine

State whether the income support reduces recidivism.

­­(a)

Expert Solution
Check Mark

Answer to Problem 7RE

There is enough evidence that the income did not support to reduce recidivism.

Explanation of Solution

In this scenario, out of 592 prisoners in the treatment group (A), 48.3% were rearrested within a year of release.

Out of 154 prisoners in the control group (B), 49.4% were rearrested within a year of release.

Based on the given information, the following values are known:

Percentage A=48.3%Percentage B=49.4%Sample size A=592Sample size B=154

The box corresponding to the treatment group contains a ticket per subject, where the value on the ticket is a 1, when he/she was rearrested within a year of release. Otherwise, the value on the ticket is 0.

The box corresponding to the control group contains a ticket per subject, where the value on the ticket is a 1, when he/she was rearrested within a year of release. Otherwise, the value on the ticket is 0.

The test hypotheses are given below:

Let μ be the population mean.

Null hypothesis: H0:μ=0

That is, the percentages of the two boxes are the same.

Alternative hypothesis: H1:μ0

That is, the percentages of the two boxes are different.

The standard deviation A is given below:

SEA=(Big numberSmall number){Fraction with big number×Fraction with small number}=(10)0.483×(10.483)=(10)0.483×0.517=0.4997

The standard error of Box A is given below:

SESumA=Sample size ×SD A=592×0.4997=12.1582

The standard error A of the average is the standard error of the sum divided by the number of draws is given below:

SE avearge A=SE sum ANumber of draws×100%=12.1582592×100%=2.05%

The standard deviation B is given below:

SEB=(Big numberSmall number){Fraction with big number×Fraction with small number}=(10)0.494×(10.494)=(10)0.494×0.506=0.5

The standard error of Box B is given below:

SESumB=Sample size ×SD B=154×0.5=6.2048

The standard error B of the average is the standard error of the sum divided by the number of draws as given below:

SE avearge B=SE sum BNumber of draws×100%=6.2048154×100%=4.03%

The standard error for the difference is shown below:

SE difference=(SE first quantity)2+((SE second quantity)2)=2.052+4.032=4.52%

The formula for test statistic is as follows:

z=ObservedExpectedSE

Known values:

μ=0σ=4.52

The observed value, x is given below:

x=Percentage APercentageB=48.3%49.4%=1.1%

The z-score is obtained as given below:

z=1.104.52=0.24

The P-value is given below:

P=P(Z<0.24 or Z>0.24)=1[P(Z<0.24)P(Z<0.24)]

Using Standard normal table, the value corresponding to P(Z<0.24) is 0.4052 and for P(Z<0.24) is 0.5948.

Remaining calculation:

P=1[P(Z<0.24)P(Z<0.24)]=1[0.59480.4052]=10.1896=0.8104

Since the P-value is greater than any level of significance, it is not unusual to obtain a difference in the sample percentages of 1.1% when there is no difference in the population percentages, and thus, the difference appears to be due to chance variation.

Therefore, this implies that the income support did not reduce recidivism.

(b)

To determine

Delineate whether the income support reduces the amount that ex-convicts worked.

(b)

Expert Solution
Check Mark

Answer to Problem 7RE

There is enough evidence that the income support reduces the amount that ex-convicts worked.

Explanation of Solution

Based on the given information, the following values are known:

Average A=16.8 weeksSD A=15.9 weeksAverage B=24.3 weeksSD B=17.3 weeksSample size A=592Sample size B=154

The box corresponding to the treatment group contains a ticket per subject, where the value on the ticket is the number of weeks of paid work after release from the prison.

The box corresponding to the control group contains a ticket per subject, where the value on the ticket is the number of weeks of paid work after release from the prison.

The test hypotheses are given below:

Let μ be the population mean.

Null hypothesis: H0:μ=0

Alternative hypothesis: H1:μ0

The standard error of Box A is given below:

SESumA=Sample size ×SD A=592×15.9=386.8637

The standard error of the average is the standard error of the sum divided by the number of draws is given below:

SE avearge A=SE sum ASample size=386.8637592=0.6535

The standard error of Box B is given below:

SESumB=Sample size ×SD B=154×17.3=214.6874

The standard error of the average is the standard error of the sum divided by the number of draws as given below:

SE avearge B=SE sum BSample size=214.6874154=1.3941

The standard error for the difference is as shown below:

SE difference=(SE first quantity)2+((SE second quantity)2)=0.65352+1.39412=1.5397

The formula for test statistic is as follows:

z=ObservedExpectedSE

Known values:

μ=0σ=1.5397

The observed value, x is given below:

x=Average AAveargeB=16.824.3=7.5

The z-score is obtained as given below:

z=7.501.5397=4.87

The P-value is given below:

P=P(Z<4.87 or Z>4.87)=1[P(Z<4.87)P(Z<4.87)]

Using the Standard normal table, the value corresponding to P(Z<4.87) is 0 and for P(Z<4.87) is 1.

Remaining calculation:

P=1[P(Z<4.87)P(Z<4.87)]=1[10]=11=0

Since the P-value is less than any of level of significance, it is unusual to obtain a difference in sample averages of 7.5 when there is no difference in the population averages, and thus, the difference does not appear to be due to chance variation.

Therefore, this implies that the income support reduces the amount that the ex-convicts worked.

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