Chapter 27.6, Problem 7RE

­­(a)

To determine

State whether the income support reduces recidivism.

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:

$\begin{array}{l}\text{Percentage }A=48.3\%\\ \text{Percentage }B=49.4\%\\ \text{Sample size }A=592\\ \text{Sample size }B=154\end{array}$

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 $\mu$ be the population mean.

Null hypothesis: $H_0:\mu =0$

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

Alternative hypothesis: $H_1:\mu \ne 0$

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

The standard deviation A is given below:

$\begin{array}{c}\text{SE}A=\left(\text{Big number}-\text{Small number}\right)\sqrt{\left\{\begin{array}{l}\text{Fraction with big number}\times \\ \text{Fraction with small number}\end{array}\right\}}\\ =\left(1-0\right)\sqrt{0.483\times \left(1-0.483\right)}\\ =\left(1-0\right)\sqrt{0.483\times 0.517}\\ =0.4997\end{array}$

The standard error of Box A is given below:

$\begin{array}{c}\text{SE}\text{Sum}A=\sqrt{\text{Sample size }}\times \text{SD }A\\ =\sqrt{592}\times 0.4997\\ =12.1582\end{array}$

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

$\begin{array}{c}\text{SE avearge }A=\frac{\text{SE sum }A}{\text{Number of draws}}\times 100\%\\ =\frac{12.1582}{592}\times 100\%\\ =2.05\%\end{array}$

The standard deviation B is given below:

$\begin{array}{c}\text{SE}B=\left(\text{Big number}-\text{Small number}\right)\sqrt{\left\{\begin{array}{l}\text{Fraction with big number}\times \\ \text{Fraction with small number}\end{array}\right\}}\\ =\left(1-0\right)\sqrt{0.494\times \left(1-0.494\right)}\\ =\left(1-0\right)\sqrt{0.494\times 0.506}\\ =0.5\end{array}$

The standard error of Box B is given below:

$\begin{array}{c}\text{SE}\text{Sum}B=\sqrt{\text{Sample size }}\times \text{SD }B\\ =\sqrt{154}\times 0.5\\ =6.2048\end{array}$

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

$\begin{array}{c}\text{SE avearge }B=\frac{\text{SE sum }B}{\text{Number of draws}}\times 100\%\\ =\frac{6.2048}{154}\times 100\%\\ =4.03\%\end{array}$

The standard error for the difference is shown below:

$\begin{array}{c}\text{SE difference}=\sqrt{{\left(\text{SE first quantity}\right)}^2+\left({\left(\text{SE second quantity}\right)}^2\right)}\\ =\sqrt{{2.05}^2+{4.03}^2}\\ =4.52\%\end{array}$

The formula for test statistic is as follows:

$z=\frac{\text{Observed}-\text{Expected}}{\text{SE}}$

Known values:

$\begin{array}{c}\mu =0\\ \sigma =4.52\end{array}$

The observed value, x is given below:

$\begin{array}{c}x=\text{Percentage }A-\text{Percentage}B\\ =48.3\%-49.4\%\\ =1.1\%\end{array}$

The z-score is obtained as given below:

$\begin{array}{c}z=\frac{-1.1-\text{0}}{\text{4}\text{.52}}\\ =-0.24\end{array}$

The P-value is given below:

$\begin{array}{c}P=P\left(Z<-0.24\text{ or }Z>0.24\right)\\ =1-\left[P\left(Z<0.24\right)-P\left(Z<-0.24\right)\right]\end{array}$

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

Remaining calculation:

$\begin{array}{c}P=1-\left[P\left(Z<0.24\right)-P\left(Z<-0.24\right)\right]\\ =1-\left[0.5948-0.4052\right]\\ =1-0.1896\\ =0.8104\end{array}$

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.

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:

$\begin{array}{l}\text{Average }A=16.8\text{ weeks}\\ \text{SD }A=15.9\text{ weeks}\\ \text{Average }B=24.3\text{ weeks}\\ \text{SD }B=17.3\text{ weeks}\\ \text{Sample size }A=592\\ \text{Sample size }B=154\end{array}$

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 $\mu$ be the population mean.

Null hypothesis: $H_0:\mu =0$

Alternative hypothesis: $H_1:\mu \ne 0$

The standard error of Box A is given below:

$\begin{array}{c}\text{SE}\text{Sum}A=\sqrt{\text{Sample size }}\times \text{SD }A\\ =\sqrt{592}\times 15.9\\ =386.8637\end{array}$

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

$\begin{array}{c}\text{SE avearge }A=\frac{\text{SE sum }A}{\text{Sample size}}\\ =\frac{386.8637}{592}\\ =0.6535\end{array}$

The standard error of Box B is given below:

$\begin{array}{c}\text{SE}\text{Sum}B=\sqrt{\text{Sample size }}\times \text{SD }B\\ =\sqrt{154}\times 17.3\\ =214.6874\end{array}$

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

$\begin{array}{c}\text{SE avearge }B=\frac{\text{SE sum }B}{\text{Sample size}}\\ =\frac{214.6874}{154}\\ =1.3941\end{array}$

The standard error for the difference is as shown below:

$\begin{array}{c}\text{SE difference}=\sqrt{{\left(\text{SE first quantity}\right)}^2+\left({\left(\text{SE second quantity}\right)}^2\right)}\\ =\sqrt{{0.6535}^2+{1.3941}^2}\\ =1.5397\end{array}$

The formula for test statistic is as follows:

$z=\frac{\text{Observed}-\text{Expected}}{\text{SE}}$

Known values:

$\begin{array}{c}\mu =0\\ \sigma =1.5397\end{array}$

The observed value, x is given below:

$\begin{array}{c}x=\text{Average }A-\text{Avearge}B\\ =16.8-24.3\\ =-7.5\end{array}$

The z-score is obtained as given below:

$\begin{array}{c}z=\frac{-7.5-\text{0}}{\text{1}\text{.5397}}\\ =-4.87\end{array}$

The P-value is given below:

$\begin{array}{c}P=P\left(Z<-4.87\text{ or }Z>4.87\right)\\ =1-\left[P\left(Z<4.87\right)-P\left(Z<-4.87\right)\right]\end{array}$

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

Remaining calculation:

$\begin{array}{c}P=1-\left[P\left(Z<4.87\right)-P\left(Z<-4.87\right)\right]\\ =1-\left[1-0\right]\\ =1-1\\ =0\end{array}$

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.