/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 14 Does prison really deter violent... [FREE SOLUTION] | 91Ó°ÊÓ

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Does prison really deter violent crime? Let \(x\) represent percent change in the rate of violent crime and \(y\) represent percent change in the rate of imprisonment in the general U.S. population. For 7 recent years, the following data have been obtained (Source: The Crime Drop in America, edited by Blumstein and Wallman, Cambridge University Press). $$ \begin{array}{r|rrrrrrr} \hline x & 6.1 & 5.7 & 3.9 & 5.2 & 6.2 & 6.5 & 11.1 \\ \hline y & -1.4 & -4.1 & -7.0 & -4.0 & 3.6 & -0.1 & -4.4 \\ \hline \end{array} $$ Complete parts (a) through (e), given \(\Sigma x=44.7, \Sigma y=-17.4, \Sigma x^{2}=315.85\), \(\Sigma y^{2}=116.1, \Sigma x y=-107.18\), and \(r \approx 0.084 .\) (f) Critical Thinking: Considering the values of \(r\) and \(r^{2}\), does it make sense to use the least-squares line for prediction? Explain.

Short Answer

Expert verified
Weak correlation suggests using the least-squares line for prediction is not sensible.

Step by step solution

01

Understand the Correlation Coefficient

The correlation coefficient, denoted by \( r \), measures the strength and direction of a linear relationship between two variables. Here, \( r \approx 0.084 \), indicating a very weak linear correlation between the percent change in violent crime and imprisonment rates.
02

Calculate the Coefficient of Determination

The coefficient of determination, \( r^2 \), represents the proportion of the variance for a dependent variable that's explained by an independent variable. Calculate it by squaring the correlation coefficient: \( r^2 = (0.084)^2 \approx 0.007056 \). This implies that approximately 0.7% of the variability in the change of violent crime can be explained by changes in imprisonment rates.
03

Analyze the Critical Thinking Question

Given that \( r \) is close to zero and \( r^2 \) is very small, the linear relationship between the variables is extremely weak. Therefore, it does not make sense to use the least-squares regression line for prediction as it would not provide reliable or useful estimates.

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Key Concepts

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

Least-Squares Regression
The Least-Squares Regression method is a way of finding the line that best fits the data points on a graph. This line is also called the regression line. It's like drawing a line through a cloud of points such that the distance between the line and each point is minimized. This is done by minimizing the sum of the squares of the vertical distances of the points from the line. This method is very popular in statistics for predicting the value of a dependent variable based on the value of one independent variable.
In our exercise, the Least-Squares Regression line would attempt to show the relationship between the percent change in the rate of violent crime (\(x\)) and the percent change in the rate of imprisonment (\(y\)). However, given that our correlation coefficient (\(r\)) is close to zero, indicating a very weak relationship, the regression line might not be very practical for predictions. Therefore, even though we can calculate it using the least-squares method, it may not offer useful insights or reliable predictions in this context.
Variance
Variance measures how much the data points in a set differ from the mean of the data set. It's essentially a way of quantifying the spread or dispersion of a set of data points. A high variance indicates that the data points are spread out over a wide range of values, while a low variance indicates that they are clustered close to the mean.
In the case of our exercise, variance helps us understand the distribution of the changes in violent crime rates and imprisonment rates over the years. The variance provides insight into the consistency of these changes. If one or both have high variance, it would mean the rates change significantly from year to year, indicating instability in those rates. For the correlation and regression analysis to yield meaningful insights, low variances in the variables would usually be preferable, as they suggest more consistent data.
Coefficient of Determination
The Coefficient of Determination is denoted by \(r^2\) and is a crucial tool in statistics. It tells us how well the regression line represents the data. Specifically, it indicates the proportion of variance in the dependent variable (\(y\)) that can be predicted from the independent variable (\(x\)).
In our exercise, when we calculated \(r^2\), we obtained a value of approximately 0.007 or 0.7%. This suggests that only 0.7% of the variability in the violent crime rate changes can be explained by changes in imprisonment rates. This is an incredibly small percentage, implying that other factors outside our model might influence changes in the violent crime rate more significantly. When \(r^2\) is this low, it indicates that the model does not explain much of the variability in the data, making any predictions from the model quite unreliable.
Linear Relationship
A Linear Relationship describes a straight-line connection between two variables. If one variable increases as the other increases or decreases in a consistent manner, they are said to have a linear relationship. The strength and direction of this relationship are often measured by the correlation coefficient \(r\).
In the exercise given, we observe a correlation coefficient \(r\) of approximately 0.084. This figure is very close to zero, implying an extremely weak linear relationship between the percent change in violent crime and the percent change in imprisonment rates. When \(r\) is this low, it typically indicates that the relationship is too weak to make accurate predictions or derive meaningful insights from the data. It suggests that the two variables do not have a strong enough relationship to justify the use of linear modeling methods, such as least-squares regression, for prediction purposes.

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Most popular questions from this chapter

When drawing a scatter diagram, along which axis is the explanatory variable placed? Along which axis is the response variable placed?

When we take measurements of the same general type, a power law of the form \(y=\alpha x^{\beta}\) often gives an excellent fit to the data. A lot of research has been conducted as to why power laws work so well in business, economics, biology, ecology, medicine, engineering, social science, and so on. Let us just say that if you do not have a good straight-line fit to data pairs \((x, y)\), and the scatter plot does not rise dramatically (as in exponential growth), then a power law is often a good choice. College algebra can be used to show that power law models become linear when we apply logarithmic transformations to both variables. To see how this is done, please read on. Note: For power law models, we assume all \(x>0\) and all \(y>0\). Suppose we have data pairs \((x, y)\) and we want to find constants \(\alpha\) and \(\beta\) such that \(y=\alpha x^{\beta}\) is a good fit to the data. First, make the logarithmic transformations \(x^{\prime}=\log x\) and \(y^{\prime}=\log y .\) Next, use the \(\left(x^{\prime}, y^{\prime}\right)\) data pairs and a calculator with linear regression keys to obtain the least-squares equation \(y^{\prime}=a+b x^{\prime} .\) Note that the equation \(y^{\prime}=a+b x^{\prime}\) is the same as \(\log y=a+b(\log x)\). If we raise both sides of this equation to the power 10 and use some college algebra, we get \(y=10^{a}(x)^{b} .\) In other words, for the power law model \(y=\alpha x^{\beta}\), we have \(\alpha \approx 10^{a}\) and \(\beta \approx b\). In the electronic design of a cell phone circuit, the buildup of electric current (Amps) is an important function of time (microseconds). Let \(x=\) time in microseconds and let \(y=\) Amps built up in the circuit at time \(x\). $$ \begin{array}{c|ccccc} \hline x & 2 & 4 & 6 & 8 & 10 \\ \hline y & 1.81 & 2.90 & 3.20 & 3.68 & 4.11 \\ \hline \end{array} $$ (a) Make the logarithmic transformations \(x^{\prime}=\log x\) and \(y^{\prime}=\log y .\) Then make a scatter plot of the \(\left(x^{\prime}, y^{\prime}\right)\) values. Does a linear equation seem to be a good fit to this plot? (b) Use the \(\left(x^{\prime}, y^{\prime}\right)\) data points and a calculator with regression keys to find the least-squares equation \(y^{\prime}=a+b x^{\prime} .\) What is the correlation coefficient? (c) Use the results of part (b) to find estimates for \(\alpha\) and \(\beta\) in the power law \(y=\) \(\alpha x^{\beta .}\) Write the power law giving the relationship between time and Amp buildup. Note: The TI-84Plus calculator fully supports the power law model. Place the original \(x\) data in list \(\mathrm{L} 1\) and the corresponding \(y\) data in list \(\mathrm{L} 2 .\) Then press STAT, followed by CALC, and scroll down to option A: PwrReg. The output gives values for \(\alpha, \beta\), and the correlation coefficient \(r\).

Describe the relationship between two variables when the correlation coefficient \(r\) is (a) near \(-1\) (b) near 0 (c) near 1

If two variables have a negative linear correlation, is the slope of the least-squares line positive or negative?

In Section \(5.1\), we studied linear combinations of independent random variables. What happens if the variables are not independent? A lot of mathematics can be used to prove the following: Let \(x\) and \(y\) be random variables with means \(\mu_{x}\) and \(\mu_{y}\), variances \(\sigma_{x}^{2}\) and \(\sigma_{y}^{2}\) and population correlation coefficient \(\rho\) (the Greek letter rho). Let \(a\) and \(b\) be any constants and let \(w=a x+b y .\) Then $$ \begin{aligned} &\mu_{w}=a \mu_{x}+b \mu_{y} \\ &\sigma_{w}^{2}=a^{2} \sigma_{x}^{2}+b^{2} \sigma_{y}^{2}+2 a b \sigma_{x} \sigma_{y} \rho \end{aligned} $$ In this formula, \(\rho\) is the population correlation coefficient, theoretically computed using the population of all \((x, y)\) data pairs. The expression \(\sigma_{x} \sigma_{y} \rho\) is called the covariance of \(x\) and \(y\). If \(x\) and \(y\) are independent, then \(\rho=0\) and the formula for \(\sigma_{w}^{2}\) reduces to the appropriate formula for independent variables (see Section 5.1). In most real-world applications the population parameters are not known, so we use sample estimates with the understanding that our conclusions are also estimates. Do you have to be rich to invest in bonds and real estate? No, mutual fund shares are available to you even if you aren't rich. Let \(x\) represent annual percentage return (after expenses) on the Vanguard Total Bond Index Fund, and let \(y\) represent annual percentage return on the Fidelity Real Estate Investment Fund. Over a long period of time, we have the following population estimates (based on Morningstar Mutual Fund Report). $$ \mu_{x} \approx 7.32 \quad \sigma_{x} \approx 6.59 \quad \mu_{y} \approx 13.19 \quad \sigma_{y} \approx 18.56 \quad \rho \approx 0.424 $$ (a) Do you think the variables \(x\) and \(y\) are independent? Explain. (b) Suppose you decide to put \(60 \%\) of your investment in bonds and \(40 \%\) in real estate. This means you will use a weighted average \(w=0.6 x+0.4 y\). Estimate your expected percentage return \(\mu_{w}\) and risk \(\sigma_{w}\). (c) Repeat part (b) if \(w=0.4 x+0.6 y\). (d) Compare your results in parts (b) and (c). Which investment has the higher expected return? Which has the greater risk as measured by \(\sigma_{w} ?\)

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