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Murders and poverty, Part II. D Exercise 8.25 presents regression output from a model for predicting annual murders per million from percentage living in poverty based on a random sample of 20 metropolitan areas. The model output is also provided below. \begin{tabular}{rrrrr} \hline & Estimate & Std. Error & t value & \(\operatorname{Pr}(>|\mathrm{t}|)\) \\\ \hline (Intercept) & -29.901 & 7.789 & -3.839 & 0.001 \\ poverty\% & 2.559 & 0.390 & 6.562 & 0.000 \\ \hline \end{tabular} \(s=5.512 \quad R^{2}=70.52 \% \quad R_{a d j}^{2}=68.89 \%\) (a) What are the hypotheses for evaluating whether poverty percentage is a significant predictor of murder rate? (b) State the conclusion of the hypothesis test from part (a) in context of the data. (c) Calculate a \(95 \%\) confidence interval for the slope of poverty percentage, and interpret it in context of the data. (d) Do your results from the hypothesis test and the confidence interval agree? Explain.

Step by step solution

01

Define Hypotheses

To determine if poverty percentage is a significant predictor of murder rate, we formulate the hypotheses as follows: \( H_0: \beta_1 = 0 \) (Poverty percentage is not a predictor) and \( H_a: \beta_1 eq 0 \) (Poverty percentage is a predictor). Here, \( \beta_1 \) is the coefficient of the poverty percentage.

02

Conclusion from Hypothesis Test

The p-value associated with the coefficient of poverty percentage is given as \( 0.000 \). Since this value is less than the typical significance level (\( \alpha = 0.05 \)), we reject \( H_0 \) and conclude that the poverty percentage is a significant predictor of the murder rate.

03

Confidence Interval for Slope

To calculate the 95% confidence interval for the slope (\( \beta_1 \)), use the formula: \[ \text{CI} : \hat{\beta}_1 \pm t^{*} \times \text{SE}(\hat{\beta}_1) \] where \( \hat{\beta}_1 = 2.559 \), \( \text{SE}(\hat{\beta}_1) = 0.390 \), and \( t^{*} \) is the critical value from the t-distribution for \( 18 \) degrees of freedom (20 - 2) at 95%, approximately 2.101. Thus: \[ 2.559 \pm 2.101 \times 0.390 \approx [1.833, 3.285] \].

04

Interpretation of Confidence Interval

The 95% confidence interval for the slope of the poverty percentage ([1.833, 3.285]) suggests that for every additional percentage point increase in the poverty rate, the murder rate is expected to increase by between 1.833 and 3.285 murders per million. This interval does not contain zero, supporting the significance of poverty percentage as a predictor.

05

Comparing Hypothesis Test and Confidence Interval Results

Both the hypothesis test and the confidence interval for the slope agree. The p-value from the hypothesis test indicates significance, and the confidence interval does not include zero, reinforcing that poverty percentage is a significant predictor of murder rate.

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Key Concepts

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

Hypothesis Testing

When conducting regression analysis, hypothesis testing helps us determine if a certain variable is a meaningful predictor. In this exercise, we look at whether the poverty percentage can predict murder rates. The hypotheses are:

• **Null hypothesis (H鈧�):** The poverty percentage is not a predictor, \( \beta_1 = 0 \). This means there's no relationship between poverty level and murder rate.
• **Alternative hypothesis (H鈧�):** The poverty percentage is a predictor, \( \beta_1 eq 0 \). This suggests some level of association.

A p-value helps decide whether to accept or reject the null hypothesis. If the p-value is less than a significance level (\( \alpha = 0.05 \)), we reject H鈧�. In the given solution, the p-value is \(0.000\), which shows that poverty is indeed a significant predictor of murder rates.

Confidence Interval

A confidence interval gives us a range within which the true parameter鈥攈ere, the slope of poverty percentage鈥攊s likely to lie. For our exercise, a 95% confidence interval tells us where we might expect the true slope to be found with 95% certainty.

• Formula: \[ \text{CI} : \hat{\beta}_1 \pm t^{*} \times \text{SE}(\hat{\beta}_1) \]
• Substitute the values: \[ 2.559 \pm 2.101 \times 0.390 \approx [1.833, 3.285] \]

This result means that for each additional percentage point in poverty, the murder rate increases between 1.833 and 3.285 murders per million. Importantly, this interval doesn't include zero, further supporting the variable's significance in prediction.

Statistical Significance

Statistical significance indicates whether the results observed (like regression coefficients) in the data are likely due to an actual effect or merely by chance. In regression, the p-value and confidence interval are tools to assess this. If a p-value is below a chosen threshold (often \( \alpha = 0.05 \)), it suggests significance.

In our exercise, a p-value of \(0.000\) signals strong significance because it's far below \(0.05\).

Furthermore, the CI for the slope of the poverty percentage doesn't bracket zero, bolstering that poverty percentage is statistically significant.

Thus, both these tools help confirm that the poverty percentage significantly affects the murder rate, not just random chance.

Linear Regression

Linear regression allows us to model the relationship between two or more variables. In the given exercise, we use it to see how changes in poverty percentage might explain differences in murder rates. The core components of a regression model include:

• **Intercept**: This is where the line crosses the y-axis. It's the predicted value when all predictors are zero.
• **Slope (Coefficient)**: Tells us how much the murder rate changes with each additional unit of poverty percentage.

In our model, the intercept estimate is -29.901, and the slope for poverty percentage is 2.559, indicating a positive relationship. This suggests that as poverty increases, so does the murder rate, making poverty a key variable to consider when predicting or understanding murder rates in these metropolitan areas.