Question

10\. Recidivism Recidivism refers to the return to prison of a prisoner who has been released or paroled. The data that follow reports the group median age at which a prisoner was released from a federal prison and the percentage of those arrested for another crime. \({ }^{7}\) Use the MS Excel printout to answer the questions that follow. $$ \begin{array}{l|lllllll} \text { Group Median Age }(x) & 22 & 27 & 32 & 37 & 42 & 47 & 52 \\ \hline \text { \% Arrested }(y) & 64.7 & 59.3 & 52.9 & 48.6 & 44.5 & 37.7 & 23.5 \end{array} $$ $$ \begin{aligned} &\text { SUMMARY OUTPUT }\\\ &\begin{array}{ll} \hline \text { Regression Statistics } & \\ \hline \text { Multiple R } & 0.9779 \\ \text { R Square } & 0.9564 \\ \text { Adjusted R Square } & 0.9477 \\ \text { Standard Error } & 3.1622 \\ \text { Observations } & 7.0000 \\ \hline \end{array} \end{aligned} $$ $$ \begin{aligned} &\text { ANOVA }\\\ &\begin{array}{llrrr} \hline & & & & {\text { Significance }} \\ & \text { df } & \text { SS } & \text { MS } & \text { F } & \text { F } \\ & & & & & \\ \hline \text { Regression } & 1 & 1096.251 & 1096.251 & 109.631 & 0.000 \\ \text { Residual } & 5 & 49.997 & 9.999 & & \\ \text { Total } & 6 & 1146.249 & & & \\ \hline \end{array} \end{aligned} $$ $$ \begin{array}{lrrrrrr} \hline& {\text { Coeffi- Standard }} \\ & \text { cients } & \text { Error } & \text { tStat } & \text { P-value } & \text { Lower } 95 \% & \text { Upper } 95 \% \\ \hline \text { Intercept } & 93.617 & 4.581 & 20.436 & 0.000 & 81.842 & 105.393 \\ \mathrm{x} & -1.251 & 0.120 & -10.471 & 0.000 & -1.559 & \- \\ \hline \end{array} $$ a. Find the least-squares line relating the percentage arrested to the group median age. b. Do the data provide sufficient evidence to indicate that \(x\) and \(y\) are linearly related? Test using the \(t\) statistic at the \(5 \%\) level of significance. c. Construct a \(95 \%\) confidence interval for the slope of the line. d. Find the coefficient of determination and interpret its significance.

Short answer

Answer: The least-squares line relating the percentage arrested to the group median age is y = 93.617 - 1.251x. The coefficient of determination (R Square) is 0.9564, which indicates that 95.64% of the variation in the percentage arrested can be explained by the linear relationship with the group median age. This suggests that group median age is a strong predictor of the percentage arrested.

Step-by-step solution

Formula for least-squares line

The least-squares line can be written as: \(y = a + bx\). In this context, a is the intercept and b is the slope.

Intercept and Slope from the table

Looking at the provided output, we have: - Intercept (a) = 93.617 - Slope (b) = -1.251

Least-squares line

So, the least-squares line for the given data is: \(y = 93.617 - 1.251x\) #b. Test if the data provides sufficient evidence to indicate that the variables are linearly related using the t statistic at the 5% level of significance#

Null Hypothesis

H0: b = 0 (no linear relationship) Ha: b ≠ 0 (linear relationship)

t-Statistic and P-value

From the provided output: - tStat = -10.471 - P-value = 0.000

Compare P-value to Significance Level

Since the P-value (0.000) is less than the 5% significance level (0.05), we reject the null hypothesis.

Conclusion

There is sufficient evidence to indicate that the variables are linearly related. #c. Construct a 95% confidence interval for the slope of the line#

Slope and Standard Error

From the provided output: - Slope (b) = -1.251 - Standard error = 0.120

Confidence Interval

Again from the provided output, the 95% confidence interval for the slope is given as: - Lower 95% = -1.559 - Upper 95% = -0.943 This means we can be 95% confident that the true slope of the line falls within the interval [-1.559, -0.943]. #d. Find the coefficient of determination and interpret its significance#

Coefficient of Determination

From the provided output, the coefficient of determination, R Square, is 0.9564.

Interpretation

The coefficient of determination (R Square) indicates that 95.64% of the variation in the percentage arrested can be explained by the linear relationship with the group median age. This means that the group median age is a strong predictor of the percentage arrested.

Key concepts

Least-Squares Line

The least-squares line is a foundational concept in linear regression analysis, which aims to find the best-fitting line through a set of data points. This line minimizes the sum of the squared differences between the observed values and the values predicted by the line.

In mathematical terms, if we denote the intercept by 'a' and the slope by 'b', the equation of the least-squares line can be expressed as \( y = a + bx \). The intercept represents the point where the line crosses the y-axis, and the slope indicates how much 'y' changes for a one-unit increase in 'x'.

For the recidivism dataset, the least-squares line is found by using the given intercept and slope from the Excel output: \( y = 93.617 - 1.251x \). This equation provides a model to predict the percentage of those arrested based on the group median age at which prisoners are released.

T Statistic

The t statistic is a crucial measure used to test hypotheses in statistics. Specifically, it helps determine whether there is a significant linear relationship between two variables in a simple linear regression model.

To test the significance of this relationship, we start by setting up a null hypothesis (\(H_0\)) that there is no linear relationship between the variables (i.e., the slope 'b' is equal to zero). The alternative hypothesis (\(H_a\)) claims the opposite, that there is a linear relationship (the slope 'b' is not equal to zero).

If the t statistic is large in absolute value and the associated p-value is less than the significance level (commonly 0.05 for a 5% level), we have evidence against the null hypothesis. In our exercise, with a t statistic of -10.471 and a p-value of 0.000, we reject the null hypothesis, confirming that there is a significant linear relationship between the group median age and the percentage arrested.

Coefficient of Determination

The coefficient of determination, often represented as \(R^2\), is a measure used in statistical analysis that assesses the proportion of variance in the dependent variable that can be explained by the independent variable(s) in a regression model.

A value of \(R^2\) close to 1 indicates that a large percentage of the variance in the outcome variable is predictable from the independent variables. Conversely, a value near 0 suggests that the regression model does not explain much of the variance in the outcome.

In the context of the recidivism example, an \(R^2\) value of 0.9564 means that approximately 95.64% of the variation in the percentage arrested is explained by the group median age. This high \(R^2\) value implies a strong predictive ability of the median age on the probability of recidivism.

Confidence Interval

A confidence interval provides a range of values within which we can expect the true parameter of the population to fall, with a certain level of confidence. In regression, we often construct a 95% confidence interval for the slope of the regression line. This interval offers a range of plausible values for the slope based on the sample data.

When we say we are '95% confident,' we imply that if we were to take many samples and compute a confidence interval for each, we expect about 95% of those intervals to contain the true slope.

For the given data on recidivism, the 95% confidence interval for the slope is from -1.559 to -0.943. This interval reinforces the evidence that the group median age is negatively related to the percentage arrested, and we can be 95% confident that this negative relationship exists in the population from which our sample was drawn.