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Chapter 12: Problem 6

Prey attracts predators Refer to Exercise 4 Computer output from the least- squares regression analysis on the perch data is shown below. $$ \begin{array}{lllll} \text { Predictor } & \text { Coef } & \text { Stdev. } & \text { t-ratio } & \text { p } \\ \text { Constant } & 0.12049 & 0.09269 & 1.30 & 0.215 \\ \text { Perch } & 0.008569 & 0.002456 & 3.49 & 0.004 \\ \mathrm{~S}=0.1886 & \mathrm{R}-\mathrm{Sq}=46.5 \% & \mathrm{R}-\mathrm{Sq}(\mathrm{adj}) & =42.7 \% \end{array} $$ The model for regression inference has three parameters: \(\alpha, \beta,\) and \(\sigma .\) Explain what each parameter represents in context. Then provide an estimate for each.

Short Answer

Expert verified
\( \alpha \) is 0.12049, \( \beta \) is 0.008569, \( \sigma \) is 0.1886.

Step by step solution

01

Identify the Context of Each Parameter

In the context of the least-squares regression, the three parameters are components of the regression equation: \( Y = \alpha + \beta X + \varepsilon \), where \( Y \) is the dependent variable, \( X \) is the independent variable, \( \alpha \) is the intercept, \( \beta \) is the slope of the line, and \( \varepsilon \) is the error term. \( \sigma \) represents the standard deviation of the residuals, measuring the variability of the data from the regression line.
02

Estimate the Intercept (\( \alpha \))

From the regression analysis output, the intercept, \( \alpha \), is represented by the constant term. The estimate for \( \alpha \) is given as 0.12049.
03

Estimate the Slope (\( \beta \))

The slope, \( \beta \), represents how much the predicted value of the dependent variable changes for each unit increase in the independent variable. From the output, \( \beta \) is estimated as 0.008569.
04

Estimate the Standard Deviation of Residuals (\( \sigma \))

The standard deviation of the residuals, \( \sigma \), indicates how much the points spread around the regression line. It is provided as \( S = 0.1886 \) in the output.

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Key Concepts

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

Least-Squares Regression
When we talk about least-squares regression, we refer to a statistical method used to discover the best-fitting line through a set of data points. By minimizing the sum of the squares of the vertical distances between the data points and the line, least-squares regression finds the line that best expresses the relationship between the independent variable (often called "X") and the dependent variable ("Y").
This technique helps in predicting the value of Y based on given values of X by fitting a simple linear equation to observed data. The equation of the line is generally expressed as \( Y = \alpha + \beta X \).
  • \( \alpha \) is the intercept of the line, representing the value of Y when X is zero.
  • \( \beta \) is the slope, indicating how much Y changes for a one-unit change in X.
This regression analysis is widely used due to its simplicity and ability to reveal trends in data.
Regression Parameters
The regression parameters are crucial elements that define the line of best fit in a regression analysis. These include the intercept (\(\alpha\)), the slope (\(\beta\)), and the standard deviation of the residuals (\(\sigma\)).

The intercept (\(\alpha\)) indicates the expected mean value of the dependent variable when all independent variables are zero. In our example from the exercise, it is 0.12049.

The slope (\(\beta\)) measures how much the dependent variable is expected to increase (or decrease) with a one-unit increase in the independent variable. Given the exercise data, it is calculated as 0.008569.

\(\sigma\), the standard deviation of the residuals, shows how much the actual data points deviate from the fitted line. Understanding these parameters allows better insights into the relationship and strength of association between variables.
Estimate of Parameters
Estimating parameters in regression involves calculating the values of \(\alpha\), \(\beta\), and \(\sigma\) that best fit the data. Through computer algorithms and matrix calculations, these estimations help build the regression equation.

In our step-by-step solution, the estimates are as follows:
  • The intercept \(\alpha\) is estimated at 0.12049, reflecting the expected value of Y when X is zero.
  • The slope \(\beta\) is assigned a value of 0.008569, showing a much tighter fit of Y based on varying X.
  • \(\sigma\), or the standard deviation of the residuals, is 0.1886. This measures the data's spread around the regression line and shows the degree to which individual data points deviate from the predicted values of Y.
Accurate estimation assists in drawing precise conclusions about the relationship between the independent and dependent variables.
Standard Deviation of Residuals
The standard deviation of residuals, often denoted by \(\sigma\), is a measure that indicates the average distance between the observed values and the values predicted by the regression line. It shows how scattered the observed data points are around the regression line.
A smaller standard deviation indicates that the data points are closer to the fitted line, suggesting a strong association between the independent variable and the dependent variable.
In contrast, a larger standard deviation implies that the data points are more spread out around the line, indicating a weaker fit. Based on the exercise, \(\sigma\) is 0.1886, meaning that, on average, the observed values deviate from the fitted values by approximately 0.1886 units.
Understanding the standard deviation of residuals helps to assess the accuracy of predictions made by the regression equation and, therefore, is a fundamental step in evaluating the fit of a regression model.

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Most popular questions from this chapter

Exercises 25 to 28 refer to the following setting. Does the color in which words are printed affect your ability to read them? Do the words themselves affect your ability to name the color in which they are printed? Mr. Starnes designed a study to investigate these questions using the 16 students in his AP \(^{\text {R }}\) Statistics class as subjects. Each student performed two tasks in a random order while a partner timed: ( 1 ) read 32 words aloud as quickly as possible, and ( 2 ) say the color in which each of 32 words is printed as quickly as possible. Try both tasks for yourself using the word list below $$ \begin{array}{llll} \text { YELLOW } & \text { RED } & \text { BLUE } & \text { GREEN } \\ \text { RED } & \text { GREEN } & \text { YELLOW } & \text { YELLOW } \\ \text { GREEN } & \text { RED } & \text { BLUE } & \text { BLUE } \\ \text { YELLOW } & \text { BLUE } & \text { GREEN } & \text { RED } \\ \text { BLUE } & \text { YELLOW } & \text { RED } & \text { RED } \\ \text { RED } & \text { BLUE } & \text { YELLOW } & \text { GREN } \\ \text { BLUE } & \text { GREEN } & \text { GREEN } & \text { BLUE } \\ \text { GREEN } & \text { YELLOW } & \text { RED } & \text { YELLOW } \end{array} $$ Color words (4.2) Let's review the design of the study. (a) Explain why this was an experiment and not an observational study. (b) Did Mr. Starnes use a completely randomized design or a randomized block design? Why do you think he chose this experimental design? (c) Explain the purpose of the random assignment in the context of the study. The data from Mr. Starnes's experiment are shown below. For each subject, the time to perform the two tasks is given to the nearest second. $$ \begin{array}{cccccc} \hline \text { Subject } & \text { Words } & \text { Colors } & \text { Subject } & \text { Words } & \text { Colors } \\ 1 & 13 & 20 & 9 & 10 & 16 \\ 2 & 10 & 21 & 10 & 9 & 13 \\ 3 & 15 & 22 & 11 & 11 & 11 \\ 4 & 12 & 25 & 12 & 17 & 26 \\ 5 & 13 & 17 & 13 & 15 & 20 \\ 6 & 11 & 13 & 14 & 15 & 15 \\ 7 & 14 & 32 & 15 & 12 & 18 \\ 8 & 16 & 21 & 16 & 10 & 18 \\ \hline \end{array} $$

Multiple choice: Select the best answer for Exercises, which are based on the following information. To determine property taxes, Florida reappraises real estate every year, and the county appraiser's Web site lists the current "fair market value" of each piece of property. Property usually sells for somewhat more than the appraised market value. We collected data on the appraised market values \(x\) and actual selling prices \(y\) (in thousands of dollars) of a random sample of 16 condominium units in Florida. We checked that the conditions for inference about the slope of the population regression line are met. Here is part of the Minitab output from a least-squares regression analysis using these data. \({ }^{13}\) $$ \begin{array}{lllll} \text { Predictor } & \text { Coef } & \text { SE Coef } & \text { T } & \text { P } \\ \text { Constant } & 127.27 & 79.49 & 1.60 & 0.132 \\ \text { Appraisal } & 1.0466 & 0.1126 & 9.29 & 0.000 \\ \mathrm{~S}=69.7299 & \mathrm{R}-\mathrm{Sq}=86.1 \% & \mathrm{R}-\mathrm{Sq}(\mathrm{adj}) & =85.1 \% \end{array} $$ Is there convincing evidence that selling price increases as appraised value increases? To answer this question, test the hypotheses (a) \(\quad H_{0}: \beta=0\) versus \(H_{a}: \beta > 0\). (b) \(H_{0}: \beta=0\) versus \(H_{a}: \beta < 0\). (c) \(\quad H_{0}: \beta=0\) versus \(H_{a}: \beta \neq 0\). (d) \(H_{0}: \beta > 0\) versus \(H_{a}: \beta=0\). (e) \(\quad H_{0}: \beta=1\) versus \(H_{a}: \beta>1\)

Paired tires Exercise 71 in Chapter 8 (page 529 ) compared two methods for estimating tire wear. The first method used the amount of weight lost by a tire. The second method used the amount of wear in the grooves of the tire. A random sample of 16 tires was obtained. Both methods were used to estimate the total distance traveled by each tire. The following scatterplot displays the two estimates (in thousands of miles) for each tire. \({ }^{12}\) Computer output from a least-squares regression analysis of these data is shown below. Assume that the conditions for regression inference are met. $$ \begin{array}{lllrl} \text { Predictor } & \text { Coef } & \text { SE Coef } & \text { T } & \text { P } \\ \text { Constant } & 1.351 & 2.105 & 0.64 & 0.531 \\ \text { Weight } & 0.79021 & 0.07104 & 11.12 & 0.000 \\ \mathrm{~S}=2.62078 & \mathrm{R}-\mathrm{Sq}=89.8 \% & \mathrm{R}-\mathrm{Sq}(\mathrm{adj}) & =89.1 \% \end{array} $$ (a) Verify that the \(99 \%\) confidence interval for the slope of the population regression line is (0.5787,1.0017) (b) Researchers want to test whether there is a difference in the two methods of estimating tire wear. Explain why the researchers might think that an appropriate pair of hypotheses for this test is \(H_{0}: \beta=1\) versus \(H_{a}: \beta \neq 1\) (c) Compute the test statistic and \(P\) -value for the test in part (b). What conclusion would you draw at the \(\alpha=0.01\) significance level? (d) Does the confidence interval in part (a) lead to the same conclusion as the test in part (c)? Explain.

Refer to the following setting. About 1100 high school teachers attended a weeklong summer institute for teaching \(\mathrm{AP}^{(\mathrm{(R)}}\) classes. After hearing about the survey in Exercise \(52,\) the teachers in the \(\mathrm{AP}^{(R)}\) Statistics class wondered whether the results of the tattoo survey would be similar for teachers. They designed a survey to find out. The class opted to take a random sample of 100 teachers at the institute. One of the questions on the survey was Do you have any tattoos on your body? (Circle one) YES \(\quad\) NO Tattoos (8.2,9.2) Of the 98 teachers who responded, \(23.5 \%\) said that they had one or more tattoos. (a) Construct and interpret a \(95 \%\) confidence interval for the actual proportion of teachers at the \(\mathrm{AP}^{\otimes}\) institute who would say they had tattoos. (b) Does the interval in part (a) provide convincing evidence that the proportion of teachers at the institute with tattoos is not 0.14 (the value cited in the Harris Poll report)? Justify your answer. (c) Two of the selected teachers refused to respond to the survey. If both of these teachers had responded, could your answer to part (b) have changed? Justify your answer.

Ideal proportions The students in Mr. Shenk's class measured the arm spans and heights (in inches) of a random sample of 18 students from their large high school. Some computer output from a least-squares regression analysis on these data is shown below. Construct and interpret a \(90 \%\) confidence interval for the slope of the population regression line. Assume that the conditions for performing inference are met. \(\begin{array}{lllrl}\text { Predictor } & \text { Coef } & \text { Stdev } & \text { t-ratio } & \text { p } \\ \text { Constant } & 11.547 & 5.600 & 2.06 & 0.056 \\ \text { Armspan } & 0.84042 & 0.08091 & 10.39 & 0.000 \\\ \mathrm{~S}=1.613 & \mathrm{R}-\mathrm{Sq}=87.1 \% & \mathrm{R}-\mathrm{Sq}(\mathrm{adj}) & =86.3 \%\end{array}\)
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