Question

Teen Court is a juvenile diversion program designed to circumvent the formal processing of first-time juvenile offenders within the juvenile justice system. The article "An Experimental Evaluation of Teen Courts" (J. of Experimental Criminology, 2008: 137-163) reported on a study in which offenders were randomly assigned either to Teen Court or to the traditional Department of Juvenile Services method of processing. Of the \(56TC\) individuals, 18 subsequently recidivated (look it up!) during the 18 -month follow-up period, whereas 12 of the 51 DJS individuals did so. Does the data suggest that the true proportion of TC individuals who recidivate during the specified follow-up period differs from the proportion of DJS individuals who do so? State and test the relevant hypotheses using a significance level of 0.10.

Short answer

There is not sufficient evidence to support the claim that the true proportion of TC individuals who recidivate during the specified follow-up period differs from the proportion of DJS individuals who do so.

Step-by-step solution

Step 1: Given information

\(\begin{array}{l}{\rm{Sample size of first sample: }}{n_1} = 56{\rm{ and }}{x_1} = 18\\{\rm{Sample size of second sample:}}{n_2} = 51{\rm{ and }}{x_2} = 12\\{\rm{Level of significance }}\alpha = 0.10\end{array}\)

The claim is either the null hypothesis or the alternative hypothesis. The null hypothesis and the alternative hypothesis state the opposite of each other. The null hypothesis needs to contain an equality.

\(\begin{array}{l}{H_0}:{p_1} = {p_2}\\{H_a}:{p_1} \ne {p_2}\end{array}\)

Test statistic

\({\rm{ The sample proportion is the number of successes divided by the sample size: }}\)

\(\begin{array}{c}{{\hat p}_1} = \frac{{{x_1}}}{{{n_1}}} = \frac{{18}}{{56}} \approx 0.3214\\{{\hat p}_2} = \frac{{{x_2}}}{{{n_2}}} = \frac{{12}}{{51}} \approx 0.2353\\{{\hat p}_p} = \frac{{{x_1} + {x_2}}}{{{n_1} + {n_2}}}\\ = \frac{{18 + 12}}{{56 + 51}}\\ = \frac{{30}}{{107}}\\ \approx 0.2804\end{array}\)

Determine the value of the test statistic:

\(\begin{array}{c}z = \frac{{{{\hat p}_1} - {{\hat p}_2}}}{{\sqrt {{{\hat p}_p}\left( {1 - {{\hat p}_p}} \right)} \sqrt {\frac{1}{{{n_1}}} + \frac{1}{{{n_2}}}} }}\\ = \frac{{0.3214 - 0.2353}}{{\sqrt {0.2804(1 - 0.2804)} \sqrt {\frac{1}{{56}} + \frac{1}{{51}}} }}\\ \approx 0.99\end{array}\)

Finding P-value

The P-value is the probability of obtaining the value of the test statistic, or a value more extreme, assuming that the null hypothesis is true. Determine the P-value using table the normal probability table in the appendix:

\(\begin{array}{c}P = P(Z < - 0.99{\rm{ or }}Z > 0.99)\\ = 2P(Z < - 0.99)\\ = 2(0.1611)\\ = 0.3222\end{array}\)

If the P-value is smaller than the significance level, then reject the null hypothesis:

\(P > 0.10 \Rightarrow {\rm{ Fail to reject }}{H_0}\)

The P-value is the probability of obtaining the value of the test statistic, or a value more extreme, assuming that the null hypothesis is true. Determine the P-value using table the normal probability table in the appendix:

\(\begin{array}{c}P = P(Z < - 0.99{\rm{ or }}Z > 0.99)\\ = 2P(Z < - 0.99)\\ = 2(0.1611)\\ = 0.3222\end{array}\)

If the P-value is smaller than the significance level, then reject the null hypothesis:

\(P > 0.10 \Rightarrow {\rm{ Fail to reject }}{H_0}\)

Final conclusion

There is not sufficient evidence to support the claim that the true proportion of TC individuals who recidivate during the specified follow-un period differs from the proportion of DJS individuals who do so.