/*! 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 1 Given the linear regression equa... [FREE SOLUTION] | 91影视

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Chapter 10: Problem 1

Given the linear regression equation $$ x_{1}=1.6+3.5 x_{2}-7.9 x_{3}+2.0 x_{4} $$ (a) Which variable is the response variable? Which variables are the explanatory variables? (b) Which number is the constant term? List the coefficients with their corresponding explanatory variables. (c) If \(x_{2}=2, x_{3}=1\), and \(x_{4}=5\), what is the predicted value for \(x_{1}\) ? (d) Explain how each coefficient can be thought of as a "slope" under certain conditions. Suppose \(x_{3}\) and \(x_{4}\) were held at fixed but arbitrary values and \(x_{2}\) increased by 1 unit. What would be the corresponding change in \(x_{1} ?\) Suppose \(x_{2}\) increased by 2 units. What would be the expected change in \(x_{1}\) ? Suppose \(x_{2}\) decreased by 4 units. What would be the expected change in \(x_{1} ?\) (e) Suppose that \(n=12\) data points were used to construct the given regression equation and that the standard error for the coefficient of \(x_{2}\) is \(0.419\). Construct a \(90 \%\) confidence interval for the coefficient of \(x_{2}\). (f) Using the information of part (e) and level of significance \(5 \%\), test the claim that the coefficient of \(x_{2}\) is different from zero. Explain how the conclusion of this test would affect the regression equation.

Short Answer

Expert verified
The response variable is \( x_{1} \); coefficients are \( 3.5 \), \(-7.9 \), \( 2.0 \). Predicted \( x_{1} \) is 10.7. Change in \( x_{1} \) per unit \( x_{2} \) is \( 3.5 \). Confidence interval for \( x_{2} \) is [2.732, 4.268]. Coefficient is significant.

Step by step solution

01

Identify Variables

For the equation \( x_{1} = 1.6 + 3.5 x_{2} - 7.9 x_{3} + 2.0 x_{4} \), \( x_{1} \) is the response variable because it is the dependent variable being predicted. The explanatory variables are \( x_{2} \), \( x_{3} \), and \( x_{4} \) because they are the independent variables used to make the prediction.
02

Identify the Constant & Coefficients

The constant term in the equation is \( 1.6 \). The coefficients for the explanatory variables are: \( 3.5 \) for \( x_{2} \), \( -7.9 \) for \( x_{3} \), and \( 2.0 \) for \( x_{4} \).
03

Calculate Predicted Value

To find the predicted value of \( x_{1} \) when \( x_{2}=2 \), \( x_{3}=1 \), and \( x_{4}=5 \), substitute these values into the equation: \[ x_{1} = 1.6 + 3.5(2) - 7.9(1) + 2.0(5) \]. Calculate: \( 1.6 + 7.0 - 7.9 + 10.0 = 10.7 \). Thus, the predicted value is \( 10.7 \).
04

Coefficient Interpretation & Change Effect

Each coefficient represents the expected change in the response variable \( x_{1} \) for a one-unit increase in that explanatory variable, holding others constant. If \( x_{3} \) and \( x_{4} \) are fixed, an increase of 1 unit in \( x_{2} \) will increase \( x_{1} \) by \( 3.5 \). Thus, increasing \( x_{2} \) by 2 units changes \( x_{1} \) by \( 7.0 \), and decreasing \( x_{2} \) by 4 units changes \( x_{1} \) by \( -14.0 \).
05

Confidence Interval for Coefficient

To construct the \( 90\% \) confidence interval for \( x_{2} \)'s coefficient, use \( \text{margin of error} = t_{\alpha/2} \times \text{standard error of } x_{2} \). With \( n = 12 \), degrees of freedom \( df = 9 \). Use \( t_{0.05} \approx 1.833 \) (from t-distribution table). Calculate \( 1.833 \times 0.419 = 0.768 \). Confidence interval is \( (3.5 - 0.768, 3.5 + 0.768) = (2.732, 4.268) \).
06

Hypothesis Testing

To test \( H_0: \beta_2 = 0 \) with a \( 5\% \) significance, calculate \( t = \frac{3.5}{0.419} \approx 8.35 \). With \( df = 9 \), look up \( t_{0.025} \approx 2.262 \). Since \( 8.35 > 2.262 \), reject \( H_0 \); the coefficient of \( x_{2} \) is significantly different from zero, confirming it affects the regression outcome.

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Key Concepts

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

Response Variable
In the realm of linear regression, identifying the response variable is the first step to understanding the model. The response variable, also known as the dependent variable, is what we are trying to predict or explain through our regression model. In the given equation \( x_{1} = 1.6 + 3.5 x_{2} - 7.9 x_{3} + 2.0 x_{4} \), \( x_{1} \) is the response variable. This variable's value is influenced by the input values of other variables, termed as explanatory variables. The goal of a linear regression model is to establish a mathematical relationship between the response variable and one or more explanatory variables.
Explanatory Variables
Explanatory variables, often referred to as independent variables, are the variables that we manipulate to see how they influence the response variable. In the equation \( x_{1} = 1.6 + 3.5 x_{2} - 7.9 x_{3} + 2.0 x_{4} \), the explanatory variables include \( x_{2} \), \( x_{3} \), and \( x_{4} \). They are considered the predictors or input variables and are crucial in determining the outcome or prediction of \( x_{1} \). Each of these explanatory variables has a corresponding coefficient that indicates the strength and direction of its impact on the response variable. For example, \( x_{2} \) has a coefficient of 3.5, meaning that for every one-unit increase in \( x_{2} \), the response variable \( x_{1} \) is expected to increase by 3.5 units, assuming all other factors are held constant.
Confidence Interval
Confidence intervals are statistical tools used to estimate a range of values within which we expect an unknown parameter to fall, with a certain level of confidence. In linear regression, confidence intervals can provide insights into the reliability of the coefficients of explanatory variables. For instance, with a 90% confidence interval calculated for the coefficient of \( x_{2} \), as \( (2.732, 4.268) \), we can be 90% confident that the true coefficient lies within this range. This interval is constructed by taking into account the standard error of the coefficient and a corresponding t-value derived from the t-distribution table based on the degrees of freedom in the dataset. A narrower confidence interval reflects more precise estimates of the coefficient.
Hypothesis Testing
Hypothesis testing in regression analysis enables us to make inferences about the population parameters based on sample data. It's used to assess whether the observed relationships in the data are statistically significant. In the context of this regression model, a hypothesis test was conducted to determine if the coefficient of \( x_{2} \) is different from zero. By setting up the null hypothesis \( H_0: \beta_2 = 0 \) against the alternative hypothesis \( H_a: \beta_2 eq 0 \), and calculating a t-statistic, we found it to be 8.35. This value was significantly greater than the critical t-value of 2.262 at a \( 5\% \) significance level. Consequently, we rejected the null hypothesis, leading to the conclusion that \( x_{2} \) has a statistically significant effect on \( x_{1} \). This result reinforces the importance of \( x_{2} \) as a contributor to changes in the response variable.

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Most popular questions from this chapter

Can a low barometer reading be used to predict maximum wind speed of an approaching tropical cyclone? Data for this problem are based on information taken from Weatherwise \((\) Vol. 46, No. 1\()\), a publication of the American Meteorological Society. For a random sample of tropical cyclones, let \(x\) be the lowest pressure (in millibars) as a cyclone approaches, and let \(y\) be the maximum wind speed (in miles per hour) of the cyclone. $$ \begin{array}{l|rrrrrr} \hline x & 1004 & 975 & 992 & 935 & 985 & 932 \\ \hline y & 40 & 100 & 65 & 145 & 80 & 150 \\ \hline \end{array} $$ (a) Make a scatter diagram and draw the line you think best fits the data. (b) Would you say the correlation is low, moderate, or strong? positive or negative? (c) Use a calculator to verify that \(\Sigma x=5823, \Sigma x^{2}=5,655,779, \Sigma y=580\), \(\Sigma y^{2}=65,750\), and \(\Sigma x y=556,315 .\) Compute \(r .\) As \(x\) increases, does the value of \(r\) imply that \(y\) should tend to increase or decrease? Explain.

Over the past 50 years, there has been a strong negative correlation between average annual income and the record time to run 1 mile. In other words, average annual incomes have been rising while the record time to run 1 mile has been decreasing. (a) Do you think increasing incomes cause decreasing times to run the mile? Explain. (b) What lurking variables might be causing the increase in one or both of the variables? Explain.

The following data are based on information from the Harvard Business Review (Vol. 72, No. 1). Let \(x\) be the number of different research programs, and let \(y\) be the mean number of patents per program. As in any business, a company can spread itself too thin. For example, too many research programs might lead to a decline in overall research productivity. The following data are for a collection of pharmaceutical companies and their research programs: $$ \begin{array}{c|rrrrrr} \hline x & 10 & 12 & 14 & 16 & 18 & 20 \\ \hline y & 1.8 & 1.7 & 1.5 & 1.4 & 1.0 & 0.7 \\ \hline \end{array} $$ Complete parts (a) through (e), given \(\Sigma x=90, \Sigma y=8.1, \Sigma x^{2}=1420\), \(\Sigma y^{2}=11.83, \Sigma x y=113.8\), and \(r \approx-0.973 .\) (f) Suppose a pharmaceutical company has 15 different research programs. What does the least-squares equation forecast for \(y=\) mean number of patents per program?

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 \(r\).

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