Problem 35 - An Introduction to Mathematical Statistics and Its Applications

Question

In some criminal cases, the judge and the defendant's lawyer will enter into a plea bargain, where the accused pleads guilty to a lesser charge. The proportion of time this happens is called the mitigation rate. A Florida Corrections Department study showed that Escambia County had the state's fourth highest rate, \(61.7 \%\) (1033 out of 1675 cases). Concerned that the guilty were not getting appropriate sentences, the state attorney put in new policies to limit the number of plea bargains. A followup study (143) showed that the mitigation rate dropped to \(52.1 \%\) (344 out of 660 cases). Is it fair to conclude that the drop was due to the new policies, or can the decline be written off to chance? Test at the \(\alpha=0.01\) level.

Step-by-Step Solution

Answer

To answer the question, calculations need to be done based on the provided data according to the described steps. Depending on the compared P-value and the significance level (0.01), a conclusion can be drawn regarding whether the drop in the mitigation rate is due to the new policies or chance.

Step 1: Define Null and Alternative Hypotheses

The null hypothesis (H0) states that the mitigation rate before and after the policy intervention are the same (P1=P2), while the alternative hypothesis (H1) states a decline (P1>P2). Thus, \(H_0: P1=P2\) and \(H_1: P1>P2\). Where P1 and P2 denote the mitigation rate before and after respectively.

Step 2: Compute Test Statistic

A Z-test will be used here because the sample size is large. First, find the combined sample proportion (P): \(P = (X1+X2) / (n1+n2)\), where X denotes the number of successful outcomes (mitigation rate), and n denotes the number of trials (total cases). Then calculate the test statistic using the formula: Z = (P1 - P2) / sqrt [ (P(1-P)) *((1/n1)+(1/n2)) ]. Using the provided data, calculate P = (1033+344) / (1675+660), then compute the Z score using the returned P-value in the above formula.

Step 3: Calculate P-value

P-value is the probability of obtaining a result at least as extreme as the observed result under the null hypothesis. By using the standard normal distribution table or a statistical software, the P-value corresponding to the absolute Z-value can be found.

Step 4: Conclusion

By comparing the calculated P-value with the significance level (α = 0.01), if P ≤ α, reject the null hypothesis. Hence, it can be concluded that the drop in the mitigation rate is statistically significant due to the new policies. If P > α, fail to reject the null and the drop could be due to chance.

Key Concepts

Null Hypothesis

The null hypothesis is a statement of 'no difference' or 'no effect.' In statistical hypothesis testing, it serves as a starting assumption that there is no significant relationship between two measured phenomena. For instance, in the exercise provided, the null hypothesis (\(H_0\)) suggests that the mitigation rate before and after the implementation of new policies in the Escambia County is the same (\(P1 = P2\)). It is only rejected if the evidence is strong enough to prove otherwise, which is determined by the p-value obtained from a test statistic.

Alternative Hypothesis

The alternative hypothesis (\(H_1\)) is the assertion opposing the null hypothesis. It proposes that there is a statistical difference or effect that exists between the two compared groups. In our exercise, the alternative hypothesis hypothesizes that the mitigation rate has decreased after the policy change (\(P1 > P2\)). The goal of statistical hypothesis testing is to determine whether there is enough evidence to favor the alternative hypothesis over the null.

Z-test

A Z-test is a type of statistical test used when the data is approximately normally distributed and the sample size is large. It is especially useful for comparing sample means to see if there's a significant difference between them. Given the exercise context, a Z-test is appropriate to assess whether the policy changes impacted the mitigation rate because of the large number of cases involved. The test statistic measures the number of standard deviations the sample mean is from the null hypothesis.

P-value

The p-value is a crucial concept in statistical hypothesis testing. It conveys the strength of the evidence against the null hypothesis. It tells us the probability of observing the sample data, or something more extreme, if the null hypothesis is true. If the obtained p-value is less than the predetermined significance level (in this case, \(\alpha = 0.01\)), the null hypothesis is rejected. This concept is paramount because it helps decide if the observed data is statistically significant or a result of mere chance.

Significance Level

The significance level, denoted as \(\alpha\), is a threshold below which the p-value must fall for us to reject the null hypothesis. It represents the degree of risk we are willing to take of rejecting the null hypothesis when it's actually true (a Type I error). A common significance level is 5%, but a more stringent level of 1% (\(\alpha = 0.01\)) is used in the exercise to minimize the chance of incorrectly concluding that the new policies impacted the mitigation rate.

Mitigation Rate Statistics

Mitigation rate statistics involve analyzing the frequency at which legal outcomes like plea bargains occur. The statistical test conducted in the exercise helps to understand whether policy interventions had a genuine effect on the mitigation rate or not. Given that the mitigation rate in Escambia County dropped after the policy implementation, the statistical test aims to clarify whether this decrease is statistically significant or could merely be due to random variation.