/*! 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 24 The output shown was obtained fr... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

The output shown was obtained from Minitab. (a) The least-squares regression equation is \(\hat{y}=1.3962 x+\) 12.396. What is the predicted value of \(y\) at \(x=10 ?\) (b) What is the mean of \(y\) at \(x=10 ?\) (c) The standard error, \(s_{e}\), is \(2.167 .\) What is an estimate of the standard deviation of \(y\) at \(x=10 ?\) (d) If the requirements for inference on the least-squares regression model are satisfied, what is the distribution of \(y\) at \(x=10 ?\)

Short Answer

Expert verified
a) 26.358. b) 26.358. c) 2.167. d) Normal distribution with mean 26.358 and standard deviation 2.167.

Step by step solution

01

Predicted Value of y at x=10

To find the predicted value of y when x = 10, substitute x = 10 into the least-squares regression equation. The equation given is \(\hat{y} = 1.3962 x + 12.396 \). By substituting x = 10 into the equation, \(\hat{y} = 1.3962 \times 10 + 12.396 \). Calculate the value: \(\hat{y} = 13.962 + 12.396 = 26.358 \). Therefore, the predicted value of y at x = 10 is 26.358.
02

Mean of y at x=10

The least-squares regression equation predicts the mean value of y for a given x. Hence, the predicted value \(\hat{y}\) is also the mean of y at x = 10. Therefore, the mean of y at x = 10 is 26.358.
03

Estimate of Standard Deviation of y at x=10

The standard error of the estimate \(s_e\) is given as 2.167. This standard error is an estimate of the standard deviation of y-values about the regression line. Therefore, the estimated standard deviation of y at x = 10 is 2.167.
04

Distribution of y at x=10

For inference on the least-squares regression model, it is assumed that for a given value of x, y is normally distributed with mean \(\hat{y}\) and standard error \(s_e\). Therefore, the distribution of y at x = 10 is approximately normal with mean 26.358 and standard deviation 2.167.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Predicted Value
The predicted value is a key concept in regression analysis. It represents the expected value of the dependent variable (y) for a given value of the independent variable (x) based on the regression equation. The least-squares regression equation given is \(\backslashhat\backslash{y} = 1.3962 x + 12.396\). To find the predicted value of y at x = 10, we substitute 10 for x in the equation: \(\backslashhat\backslash{y} = 1.3962 \times 10 + 12.396 = 26.358\). Hence, the predicted value of y at x = 10 is 26.358. This figure tells us what we can expect y to be when x is 10, based on our regression model.
Mean
In statistics, the mean is the average value of a set of numbers. For our regression equation, the predicted value \(\backslashhat\backslash{y}\) at a given x is also the mean value of y at that x. Therefore, the mean of y at x = 10 is the same as the predicted value we calculated earlier, 26.358. This is because the regression line represents the average relationship between x and y across all the data points. Knowing the mean helps us understand the central tendency of y at a specific x value.
Standard Error
The standard error, often denoted as \(\backslashsigma\_e\), provides a measure of the typical size of the residuals, or the differences between the observed y-values and the predicted y-values from the regression model. In this case, the standard error is given as 2.167. This value estimates the standard deviation of y-values about the regression line. It helps us understand how much the y-values scatter around the regression line. Essentially, a smaller standard error indicates a more accurate regression model, while a larger standard error suggests more variability around the regression line.
Normal Distribution
Normal distribution is a fundamental concept in statistics, often described as a bell-shaped curve. When it comes to regression, we assume that for any given x, the y-values follow a normal distribution with a mean represented by the predicted value \(\backslashhat\backslash{y}\) and a standard deviation equal to the standard error \(\backslashsigma\_e\). In our example, for x = 10, y is normally distributed with a mean of 26.358 and a standard deviation of 2.167. This assumption enables us to make probabilistic inferences about y. Understanding that y follows a normal distribution around its mean helps us predict the likelihood of different y-values occurring.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Wanting to know if there is a linear relation among wind chill temperature, air temperature (in degrees Fahrenheit), and wind speed (in miles per hour), a researcher collected the following data for various days. $$ \begin{array}{ccc} \text { Air Temp. } & \text { Wind Speed } & \text { Wind Chill } \\ \hline 15 & 10 & 3 \\ \hline 15 & 15 & 0 \\ \hline 15 & 25 & -4 \\ \hline 0 & 5 & -11 \\ \hline 0 & 20 & -22 \\ \hline-5 & 10 & -22 \\\ \hline-5 & 25 & -31 \\ \hline-10 & 15 & -32 \\ \hline-10 & 20 & -35 \\\ \hline-15 & 25 & -44 \\ \hline-15 & 35 & -48 \\ \hline-15 & 50 & -52 \\\ \hline 5 & 40 & -22 \\ \hline 10 & 45 & -16 \\ \hline \end{array} $$ (a) Find the least-squares regression equation \(\hat{y}=b_{0}+b_{1} x_{1}+b_{2} x_{2},\) where \(x_{1}\) is air temperature, \(x_{2}\) is wind speed, and \(y\) is the response variable "wind chill." (b) Draw residual plots to assess the adequacy of the model. What might you conclude based on the plot of residuals against wind speed?

Why is it desirable to have the explanatory variables spread out to test a hypothesis regarding \(\beta_{1}\) or to construct confidence intervals about \(\beta_{1} ?\)

For the data set below, use a partial \(F\) -test to determine whether the variables \(x_{4}\) and \(x_{5}\) do not significantly help to predict the response variable, \(y .\) Use the \(\alpha=0.05\) level of significance. $$ \begin{array}{llllll} x_{1} & x_{2} & x_{3} & x_{4} & x_{5} & y \\ \hline 0.8 & 2.8 & 2.5 & 10.6 & 15.7 & 11.0 \\ \hline 3.9 & 2.6 & 5.7 & 9.2 & 4.2 & 10.8 \\ \hline 1.8 & 2.4 & 7.8 & 10.1 & 1.5 & 10.6 \\ \hline 5.1 & 2.3 & 7.1 & 9.2 & 1.9 & 10.3 \\ \hline 4.9 & 2.5 & 5.9 & 11.2 & 5.6 & 10.3 \\ \hline 8.4 & 2.1 & 8.6 & 10.4 & 4.9 & 10.3 \\ \hline 12.9 & 2.3 & 9.2 & 11.1 & 1.9 & 10.0 \\ \hline 6.0 & 2.0 & 1.2 & 8.6 & 22.3 & 9.4 \\ \hline 14.6 & 2.2 & 3.7 & 10.5 & 11.5 & 8.7 \\ \hline 9.3 & 1.1 & 5.5 & 8.8 & 6.1 & 8.7 \\ \hline \end{array} $$

What is multicollinearity? How can we check for it? What are the consequences of multicollinearity?

An economist was interested in modeling the relation among annual income, level of education, and work experience. The level of education is the number of years of education beyond eighth grade, so 1 represents completing I year of high school, 8 means completing 4 years of college, and so on. Work experience is the number of years employed in the current profession. From a random sample of 12 individuals, he obtained the following data: $$ \begin{array}{ccc} \begin{array}{l} \text { Work Experience } \\ \text { (years) } \end{array} & \begin{array}{l} \text { Level of } \\ \text { Education } \end{array} & \begin{array}{l} \text { Annual Income } \\ \text { (\$ thousands) } \end{array} \\ \hline 21 & 6 & 34.7 \\ \hline 14 & 3 & 17.9 \\ \hline 4 & 8 & 22.7 \\\ \hline 16 & 8 & 63.1 \\ \hline 12 & 4 & 33.0 \\ \hline 20 & 4 & 41.4 \\\ \hline 25 & 1 & 20.7 \\ \hline 8 & 3 & 14.6 \\ \hline 24 & 12 & 97.3 \\\ \hline 28 & 9 & 72.1 \\ \hline 4 & 11 & 49.1 \\ \hline 15 & 4 & 52.0 \\\ \hline \end{array} $$ (a) Construct a correlation matrix between work experience, level of education, and annual income. Is there any reason to be concerned with multicollinearity based on the correlation matrix? (b) Find the least-squares regression equation \(\hat{y}=b_{0}+\) \(b_{1} x_{1}+b_{2} x_{2},\) where \(x_{1}\) is work experience, \(x_{2}\) is level of education, and \(y\) is the response variable, annual income. (c) Draw residual plots and a boxplot of the residuals to assess the adequacy of the model. (d) Interpret the regression coefficients for the least-squares regression equation. (e) Determine and interpret \(R^{2}\) and the adjusted \(R^{2}\) (f) Test \(H_{0}: \beta_{1}=\beta_{2}=0\) versus \(H_{1} ;\) at least one of the \(\beta_{i} \neq 0\) at the \(\alpha=0.05\) level of significance. (g) Test the hypotheses \(H_{0}: \beta_{1}=0\) versus \(H_{1}: \beta_{1} \neq 0\) and \(H_{0}: \beta_{2}=0\) versus \(H_{1}: \beta_{2} \neq 0\) at the \(\alpha=0.05\) level of significance. (h) Predict the mean income of all individuals whose experience is 12 years and level of education is 4 (i) Predict the income of a single individual whose experience is 12 years and level of education is 4 (j) Construct \(95 \%\) confidence and prediction intervals for income when experience is 12 years and level of education is 4

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.