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When comparing proportions between two groups, we aim to determine if there is a significant difference in the occurrence of an event in each group. In the exercise regarding Teen Court (TC) and the Department of Juvenile Services (DJS), the goal is to compare recidivism rates.

To perform this comparison, we calculate the sample proportion for each group:

• For Teen Court (TC), the recidivism proportion is calculated as \( \hat{p}_{TC} = \frac{18}{56} \).
• For the traditional method (DJS), the recidivism proportion is \( \hat{p}_{DJS} = \frac{12}{51} \).

These calculations yield sample proportions that serve as estimates of the true population proportions.

However, to determine if these differences are substantial, we employ a statistical test, which assesses the likelihood that any observed differences are due to random sampling variability rather than a true difference between the populations.

The significance level, often denoted as \( \alpha \), is a critical part of hypothesis testing. It represents the threshold at which we decide whether to reject the null hypothesis.

In the context of our problem, a significance level of 0.10 is selected. This implies that we are willing to accept a 10% chance of incorrectly rejecting the null hypothesis, which means there's a one in ten chance to conclude that the groups differ when they actually do not.

Choosing the correct significance level is essential, as it balances the risk of Type I errors (false positives) against making a correct decision. In the case of teen court recidivism, using a 0.10 level suggests a relatively light consequence of making an incorrect conclusion, which may vary depending on the practical implications of the study.

A z-test is used to determine whether there is a significant difference between the proportions of two independent groups. This test is appropriate when dealing with large sample sizes and known variances, or as in this case, when comparing proportions.

For two proportions, the test statistic \( z \) is calculated as:\[ z = \frac{\hat{p}_{TC} - \hat{p}_{DJS}}{\sqrt{\hat{p}(1-\hat{p})\left(\frac{1}{n_1}+\frac{1}{n_2}\right)}} \]Where \( \hat{p} \) is the pooled sample proportion and \( n_1 \) and \( n_2 \) are the sample sizes of the two groups.

In our case, the calculated \( z \)-value is approximately 0.933. This statistic allows us to compare it to critical values from the z-distribution to determine if the observed difference in proportions is statistically significant.

The simplicity of the z-test lies in its systematic approach: a single calculation reveals if the observed difference could likely be due to chance.

The null hypothesis \( (H_0) \) is a statement that there is no effect or difference, and it serves as the starting assumption for any hypothesis test. In this case, the null hypothesis is that the proportions of recidivism between Teen Court (TC) and the traditional method (DJS) are the same: \( p_{TC} = p_{DJS} \).

Setting up the null hypothesis is crucial because it provides a baseline to test against. If the data provide enough statistical evidence, we may reject the null hypothesis, suggesting a difference in the underlying populations.

However, in this exercise, the null hypothesis was not rejected at the 0.10 significance level since the calculated \( z \)-value of 0.933 fell within the critical value range of \(-1.645\) to 1.645. Thus, we conclude there is not enough statistical evidence to claim a difference between the two groups' recidivism rates.

The practical implication of not rejecting \( H_0 \) means the intervention of TC does not significantly differ in reducing recidivism compared to the traditional method under the conditions of this study.