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Here are the conviction rates with the "stand your ground" data mentioned in the previous exercise. "White shooter on nonwhite" means that a white assailant shot a minority victim. $$ \begin{array}{lr} & \text { Conviction Rate } \\ \hline \text { White shooter on white } & 35 / 97=36.1 \% \\ \text { White shooter on nonwhite } & 3 / 21=14.3 \% \\ \text { Nonwhite shooter on white } & 8 / 22=36.4 \% \\ \text { Nonwhite shooter on nonwhite } & 11 / 45=24.4 \% \end{array} $$ a. Which has the higher conviction rate: white shooter on nonwhite or nonwhite shooter on white? b. Create a two-way table using White Shooter on Nonwhite and NonWhite Shooter on White across the top and Convicted and Not Con- victed on the side. c. Test the hypothesis that race and conviction rate (for these two groups) are independent at the \(0.05\) level. d. Because some of the expected counts are pretty low, try a Fisher's Exact Test with the data, reporting the p-value (two-tailed) and the conclusion.

Short Answer

Expert verified
a. Nonwhite shooter on white has higher conviction rate. b. Use ratios to construct the table. c. Perform a chi-square test for hypothesis testing. d. For small samples, Fisher's Exact Test can be used. This test also provides a p-value that indicates the probability of observed data if variables were independent.

Step by step solution

01

Compare Conviction Rates

Evaluate the conviction rates of a white shooter on nonwhite versus a nonwhite shooter on white. The conviction rate is higher for a nonwhite shooter on a white victim with \( 36.4% \) compared to a white shooter on a nonwhite victim with \( 14.3% \).
02

Create a Two-way table

Generate a two-way table. Here, the number of convictions and non-convictions need to be calculated for both cases - White shooter on nonwhite and Nonwhite shooter on white. The sums can be calculated based on the provided ratios.
03

Hypothesis Testing (Race and Conviction rate independence)

For this hypothesis testing, use a chi-square test for independence. The null hypothesis, \(H_0\), would be that race and conviction rate are independent. The alternative hypothesis, \(H_1\), would be that they are not independent. A statistical software or scientific calculator can be used to compute the Chi-square statistic and the p-value.
04

Fisher's Exact Test

Since some expected counts are low, apply Fisher's Exact Test. This test is often used when sample sizes are small. It computes the probability of obtaining a distribution of values in a 2*2 contingency table, given the marginal totals. This test will provide a p-value indicating the probability that the observed data would occur if the shooting race and conviction rate were independent of each other.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Chi-Square Test
The Chi-Square Test, a cornerstone in statistics, is a method to determine if there is a significant association between two categorical variables. In the context of conviction rates, such as in our exercise, this test helps us to assess whether the race of the shooter and conviction rates are related.

When conducting a Chi-Square Test, the first step is to establish the null hypothesis, which typically posits that there is no association between the variables—race and conviction rate in this example. On the contrary, the alternative hypothesis claims that an association does exist. To carry out the test, one must observe the frequencies in each category of the two-way table and compare them to the frequencies one would expect if there were no association - these are called expected frequencies.

Calculating the Chi-Square statistic involves summing the squared differences between observed and expected frequencies, divided by the expected frequencies. A high Chi-Square value indicates that the differences between what we observe and what we would expect are more than just due to chance, leading us to reject the null hypothesis.

Interpreting Chi-Square Results

If the p-value obtained from the test is less than the chosen significance level, commonly 0.05, it suggests there is evidence to reject the null hypothesis, indicating a potential relationship between the two variables.
Fisher's Exact Test
Fisher's Exact Test is another powerful statistical tool used when dealing with categorical data, especially when sample sizes are small, as mentioned in our exercise. Unlike the Chi-Square Test that relies on approximations, Fisher's Exact Test calculates the exact probability of observing a table as extreme or more extreme than the one observed, given the fixed marginal totals.

This test is particularly useful when the data in the two-way table includes small numbers, which can happen for rare events or in studies with limited samples. In our case, we noticed some low expected counts for the conviction rates, which makes the Fisher's Exact Test more appropriate than the Chi-Square Test.

Using a two-tailed version of the test will assess the hypothesis that the race of the shooter and the conviction rate are independent in both directions—whether the effect is more or less than expected. By computing the p-value, we find the probability of the observed association occurring by random chance. A very low p-value leads to rejecting the null hypothesis.
Two-way Tables
Two-way tables, also known as contingency tables, are a fundamental part of analyzing categorical data. They allow us to observe the relationship between two categories by displaying the frequencies of different combinations of outcomes. In the exercise, the table categorized the data based on the race of the shooter and the conviction outcome.

To create a two-way table, as performed in the exercise, one lists one category on each axis. These can be 'White shooter on nonwhite' and 'Nonwhite shooter on white' on the top, and 'Convicted' and 'Not Convicted' on the side, which creates four different combinations or cells in the table.

The frequencies in each cell tell us how many cases fall into each combination. For instance, a two-way table enables us to directly compare the conviction rates for different racial groups and thereby tackle questions on equality and fairness in the justice system.

Utilizing Two-way Tables for Tests

After constructing a two-way table, we can apply statistical tests, like the Chi-Square Test or Fisher's Exact Test, to interpret the data. Results from these tests, paired with insight from the table, help us draw conclusions about the independence or the potential association between the categories explored.

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Most popular questions from this chapter

Fill in the blank by choosing one of the options given: Chi-square goodness- of-fit data are often summarized with (one row or one column of observed counts-but not both, or at least two rows and at least two columns of observed counts).

Suppose a polling organization asks a random sample of people if they are Democrat, Republican, or Other and asks them if they think the country is headed in the right direction or the wrong direction. If we wanted to test whether party affiliation and answer to the question were associated, would this be a test of homogeneity or a test of independence? Explain.

Professional musicians listened to five violins being played, without seeing the instruments. One violin was a Stradivarius, and the other four were modern-day violins. When asked to pick the Stradivarius (after listening to all five), 39 got it right and 113 got it wrong. a. Use the chi-square goodness-of-fit test to test the hypothesis that the experts are not simply guessing. Use a significance level of \(0.05\). b. Perform a one-proportion \(z\) -test with the same data, using a one-tailed alternative that the experts should get more than \(20 \%\) correct. Use a significance level of \(0.05\). c. Compare your p-values and conclusions.

The Perry Preschool Project discussed in exercises \(10.39\) to \(10.41\) found that 8 of the 58 students who attended preschool had at least one felony arrest by age 40 and that 31 of the 65 students who did not attend preschool had at least one felony arrest (Schweinhart et al. 2005 ). a. Compare the percentages descriptively. What does this comparison suggest? b. Create a two-way table from the data and do a chi-square test on it, using a significance level of \(0.05 .\) Test the hypothesis that preschool attendance is associated with being arrested. c. Do a two-proportion \(z\) -test. Your alternative hypothesis should be that preschool attendance lowers the chances of arrest. d. What advantage does the two-proportion \(z\) -test have over the chi-square test?

Randomly chosen people were observed for about 10 seconds in several public places, such as malls and restaurants, to see whether they smiled during that time. The table shows the results for different age groups. $$ \begin{array}{lccccc} & & & \text { Age Group } \\ & \mathbf{0 - 1 0} & \mathbf{1 1 - 2 0} & \mathbf{2 1 - 4 0} & \mathbf{4 1 - 6 0} & \mathbf{6 1 +} \\ \hline \text { Smile } & 1131 & 1748 & 1608 & 937 & 522 \\ \hline \text { No Smile } & 1187 & 2020 & 3038 & 2124 & 1509 \end{array} $$ a. Find the percentage of each age group that were observed smiling, and compare these percentages. b. Treat this as a single random sample of people, and test whether smiling and age group are associated, using a significance level of \(0.05 .\) Comment on the results.

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