91Ó°ÊÓ

Some straightforward but slightly tedious algebra shows that $$ \text { SSResid }=\left(1-r^{2}\right) \sum(y-\bar{y})^{2} $$ from which it follows that $$ s_{e}=\sqrt{\frac{n-1}{n-2}} \sqrt{1-r^{2}} s_{y} $$ Unless \(n\) is quite small, \((n-1) /(n-2) \approx 1\), so $$ s_{e} \approx \sqrt{1-r^{2}} s_{y} $$ a. For what value of \(r\) is \(s\), as large as \(s_{y} ?\) What is the least- squares line in this case? b. For what values of r will se be much smaller than \(s_{y} ?\) c. A study by the Berkeley Institute of Human Development (see the book Statistics by Freedman et al., listed in the back of the book) reported the following summary data for a sample of n 5 66 California boys: \(r \approx .80\) At age 6 , average height \(\approx 46\) in., standard deviation \(\approx\) \(1.7\) in. At age 18 , average height \(\approx 70\) in., standard deviation \(\approx\) \(2.5\) in. What would \(s_{e}\) be for the least-squares line used to predict 18-year-old height from 6-year-old height? d. Referring to Part (c), suppose that you wanted to predict the past value of 6 -year-old height from knowledge of 18 -year-old height. Find the equation for the appropriate least-squares line. What is the corresponding value of \(s_{e} ?\)

Step by step solution

01

Part a: When s_e = s_y

The standard deviation of residuals s_e equals the standard deviation of y (s_y) when r equals 0 because s_e = sqrt(1-r^2) s_y. When r = 0, it means there is no linear relationship between the variables, thus, the best least-squares line is horizontal at y-bar.

02

Part b: When s_e

The standard deviation of residuals s_e would be much smaller than the standard deviation of y (s_y) if the value of r is close to 1 or -1. This is because s_e = sqrt(1-r^2) s_y, meaning the value of s_e is directly affected by the square of r. If r is close to -1 or 1, it means there is a strong negative or positive linear correlation.

03

Part c: Formulate s_e

In this case, we know that r = 0.8 and s_y(i.e., at age 18) = 2.5. So, we can substitute these values into the equation for s_e to obtain: s_e = sqrt(1 - (0.8)^2) * 2.5 = 1.5 in.

04

Part d: Predicting 6-year-old height from 18-year-old height

To construct the regression line, use the formula: y = mx + b, where m = r(s_y/s_x), and b = y_mean - m*x_mean. Given that r = 0.8, s_y(at age 6) = 1.7, s_x(at age 18) = 2.5, y_mean(at age 6) = 46, and x_mean(at age 18) = 70, we substitute these values to the formulas to predict 6-year-old height from 18-year-old height. We then calculate s_e using the formula s_e = sqrt(1-r^2) s_y for the new values of r and s_y, which are the correlation and standard deviation for the 6-year-old height.

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Key Concepts

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

SSResid (Sum of Squares of Residuals)

The SSResid, or the Sum of Squares of Residuals, is a critical measure in the world of statistics, specifically in regression analysis. It helps us understand how well our regression line fits the observed data. Residuals are the differences between the observed values and the values predicted by our regression model; thus, the SSResid calculates the overall discrepancy between our model and the actual data.

By squaring the residuals, we ensure that positive and negative differences do not cancel each other out, and summing these squares gives us a single comprehensive value. The formula as seen in the textbook exercise is \[\text{SSResid} = \left(1-r^2\right) \sum(y-\bar{y})^2\]. Here, \(r\) represents the correlation coefficient, and \(\bar{y}\) is the mean of the observed data. When the SSResid is small, it indicates that the model has a good fit with the data. Large values of SSResid suggest that the model is not adequately capturing the relationship present in the data.

Standard Deviation of Residuals (se)

The standard deviation of residuals, or \(s_e\), provides insight into the typical size of the residuals, showing the typical amount by which the observed data points deviate from the regression line. The smaller the value of \(s_e\), the closer the data points are to the line, indicating a more precise prediction of our regression model.

The formula derived from the textbook is \[s_e = \sqrt{\frac{n-1}{n-2}} \sqrt{1-r^2} s_y\] where \(s_y\) is the standard deviation of the observed outcomes, and \(r\) is the correlation coefficient. For large datasets, we can simplify the formula to \[s_e \approx \sqrt{1-r^2} s_y\] since \(\frac{n-1}{n-2}\) approaches 1. Understanding \(s_e\) is integral to evaluating how well our regression model works in practice, as it relates to how much our predictions deviate from reality, on average.

Linear Relationship

A linear relationship in statistics signifies that there is a straight-line relationship between two variables. This is the foundation of linear regression analysis, where we assume that as one variable increases or decreases, the other variable tends to also increase or decrease by a certain amount, consistently. This relationship is described by a linear equation of the form \[y = mx + b\], where \(y\) and \(x\) are the variables, \(m\) indicates the slope, and \(b\) represents the intercept.

In the context of the exercise, when the correlation coefficient \(r\) is zero, it is indicative that there is no linear relationship between the variables, leading to a horizontal line at \(\bar{y}\), the mean of the observed data. As \(r\) moves away from zero, indicating either a positive or negative linear relationship, the fit of the regression line improves, and the residuals decrease, which can lead to more accurate predictions.

Correlation Coefficient (r)

The correlation coefficient, denoted as \(r\), measures the strength and direction of a linear relationship between two variables. It is a value between -1 and 1 where a correlation of 1 signifies a perfect positive linear relationship, -1 indicates a perfect negative linear relationship, and 0 means there is no linear relationship at all.

In the exercise presented, \(r\) greatly influences the standard deviation of residuals \(s_e\). When \(r\) is close to 1 or -1, the residuals are typically smaller, showcasing a strong linear relationship between the variables. Consequently, the regression line is a good fit for predicting outcomes. On the contrary, when \(r\) is closer to 0, the prediction power diminishes. An example provided in the exercise shows an \(r\) of approximately 0.8, suggesting a strong positive linear relationship and thereby allowing for a more robust prediction of an 18-year-old's height based on their height at 6 years old.