/*! 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 12 a. Could a linear regression res... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

a. Could a linear regression result in residuals \(23,-27,5\), \(17,-8,9\), and 15 ? Why or why not? b. Could a linear regression result in residuals \(23,-27,5\), \(17,-8,-12\), and 2 corresponding to \(x\) values \(3,-4,8\), \(12,-14,-20\), and 25 ? Why or why not?

Short Answer

Expert verified
a. No, because the sum of residuals is not zero. b. Yes, because the sum of residuals is zero.

Step by step solution

01

Understanding Residuals in Linear Regression

In linear regression, a residual is the difference between the observed value and the value predicted by the model. Mathematically, it is expressed as \( e_i = y_i - \hat{y}_i \), where \( y_i \) is the observed value and \( \hat{y}_i \) is the predicted value for each point \( i \). The sum of these residuals always equals zero in a properly fitting linear regression.
02

Calculating Sum of Residuals for Part a

We are given residuals of \(23, -27, 5, 17, -8, 9,\) and \(15\). To determine if these could be residuals in linear regression, calculate their sum: \(23 - 27 + 5 + 17 - 8 + 9 + 15 = 34\). The sum is not zero, indicating these cannot be residuals of a linear regression.
03

Analyzing Residuals in Part b

Before proceeding to check, we note that for linear regression with residuals corresponding to specific \(x\) values, it is not necessary for residuals to exactly sum to zero due to potential floating point inaccuracies. Instead, we check the necessary condition over different points such that the average sum approaches zero. However, let’s calculate the sum of residuals \(23, -27, 5, 17, -8, -12,\) and \(2\), which results in \(0\). This satisfies the linear regression condition.
04

Checking Linear Relationship for Part b

For the given \(x\) values \(3, -4, 8, 12, -14, -20,\) and \(25\), we consider if a linear model fits. In practice, this means calculating coefficients for best-fit line, but the theoretical check suffices with residual sums zero, suggesting a potential fit for a linear model.

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.

Residuals
In the context of linear regression, residuals are vital to understanding how well a model fits the data. A residual is the difference between an observed value and its corresponding predicted value, which is expressed mathematically as follows: \( e_i = y_i - \hat{y}_i \) where \( e_i \) is the residual for the \( i^{th} \) observation, \( y_i \) is the actual observed value, and \( \hat{y}_i \) is the value predicted by the linear regression model. Residuals provide insight into the accuracy of the predictions the model makes.
  • If the residuals add up to zero, it suggests a well-fitted model.
  • Positive residuals indicate where actual values exceed the predicted ones.
  • Negative residuals show where predictions are higher than the actual values.
The sum of residuals in any linear regression model should ideally be zero or very close to zero. This ensures that the overestimations and underestimations by the model balance each other out.
Linear Model Fit
A linear model fit aims to establish a 'best-fit' line through a dataset that minimizes the sum of the squared residuals. This method, known as least squares, helps ensure that the overall prediction error is minimized. The fit of the linear model involves two main factors:
  • Coefficient Calculation: These are the slope and the y-intercept of the best-fit line, calculated to minimize prediction errors.
  • Model Simplicity: Linear models assume a linear relationship between the variables.
When assessing a linear model fit: - Look at the residual plots, which should appear scattered without any particular pattern if the fit is appropriate. - Verify that the residuals have a sum close to zero.
Remember, the fit is only valid if the assumption of linearity holds true in the context of your data.
Regression Analysis
Regression analysis is a cornerstone of statistical methods used to determine the relationships between variables. In the simplest form, such as linear regression, it models the relationship between a dependent variable and one or more independent variables.Through regression analysis, predictions can be made based on the data trends uncovered. In the case of linear regression, the relationship is described by a straight line, whose formula is expressed as:\[ y = mx + b \] where: - \( y \) is the dependent variable,- \( m \) is the slope of the line,- \( x \) is the independent variable, and- \( b \) is the y-intercept.Key aspects of regression analysis include:
  • Assumptions: Linearity, independence, homoscedasticity, and normality of residuals.
  • Model Evaluation: This often involves looking at measures like R-squared which indicates the fit of the model.
  • Prediction: Regression allows for the forecasting of data trends based on existing data points.
Effective regression analysis provides insights into the significance and strength of the relationships within the data, laying the foundation for both theoretical research and practical decision-making.

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

An investigation of a die casting process resulted in the accompanying data on \(x_{1}=\) furnace temperature, \(x_{2}=\) die close time, and \(y=\) temperature difference on the die surface ("A Multiple-Objective Decision-Making Approach for Assessing Simultaneous Improvement in Die Life and Casting Quality in a Die Casting Process," Quality Engineering, 1994: 371-383). a. Carry out the model utility test. b. Calculate and interpret a \(95 \%\) confidence interval for \(\beta_{2}\), the population regression coefficient of \(x_{2}\). c. When \(x_{1}=1300\) and \(x_{2}=7\), the estimated standard deviation of \(\hat{y}\) is \(s_{\bar{Y}}=.353\). Calculate a \(95 \%\) confidence interval for true average temperature difference when furnace temperature is 1300 and die close time is \(7 .\) d. Calculate a \(95 \%\) prediction interval for the temperature difference resulting from a single experimental run with a furnace temperature of 1300 and a die close time of 7 .

Continuous recording of heart rate can be used to obtain information about the level of exercise intensity or physical strain during sports participation, work, or other daily activities. The article "The Relationship between Heart Rate and Oxygen Uptake During Non-Steady State Exercise" (Ergonomics, 2000: 1578-1592) reported on a study to investigate using heart rate response \((x\), as a percentage of the maximum rate) to predict oxygen uptake ( \(y\), as a percentage of maximum uptake) during exercise. The accompanying data was read from a graph in the paper. \begin{tabular}{l|cccccccc} \(\mathrm{HR}\) & \(43.5\) & \(44.0\) & \(44.0\) & \(44.5\) & \(44.0\) & \(45.0\) & \(48.0\) & \(49.0\) \\ \hline \(\mathrm{VO}_{2}\) & \(22.0\) & \(21.0\) & \(22.0\) & \(21.5\) & \(25.5\) & \(24.5\) & \(30.0\) & \(28.0\) \end{tabular} \begin{tabular}{l|llllllll} \(\mathrm{HR}\) & \(49.5\) & \(51.0\) & \(54.5\) & \(57.5\) & \(57.7\) & \(61.0\) & \(63.0\) & \(72.0\) \\ \hline \(\mathrm{VO}_{2}\) & \(32.0\) & \(29.0\) & \(38.5\) & \(30.5\) & \(57.0\) & \(40.0\) & \(58.0\) & \(72.0\) \end{tabular} Use a statistical software package to perform a simple linear regression analysis, paying particular attention to the presence of any unusual or influential observations.

The following data resulted from an experiment to assess the potential of unburnt colliery spoil as a medium for plant growth. The variables are \(x=\) acid extractable cations and \(y=\) exchangeable acidity/total cation exchange capacity ("Exchangeable Acidity in Unburnt Colliery Spoil," Nature, 1969: 161): \begin{tabular}{r|rrrrrrr} \(x\) & \(-23\) & \(-5\) & 16 & 26 & 30 & 38 & 52 \\ \hline\(y\) & \(1.50\) & \(1.46\) & \(1.32\) & \(1.17\) & \(.96\) & \(.78\) & \(.77\) \\ \(x\) & 58 & 67 & 81 & 96 & 100 & 113 & \\ \hline\(y\) & \(.91\) & \(.78\) & \(.69\) & \(.52\) & \(.48\) & \(.55\) & \end{tabular} Standardizing the independent variable \(x\) to obtain \(x^{\prime}=\) \((x-\bar{x}) / s_{x}\) and fitting the regression function \(y=\beta_{0}^{*}+\) \(\beta_{1}^{+} x^{\prime}+\beta_{2}^{*}\left(x^{\prime}\right)^{2}\) yielded the accompanying computer output. \begin{tabular}{ccc} Parameter & Estimate & Estimated SD \\ \hline\(\beta_{0}^{*}\) & \(.8733\) & \(.0421\) \\ \(\beta_{1}^{*}\) & \(-.3255\) & \(.0316\) \\ \(\beta_{2}^{*}\) & \(.0448\) & \(.0319\) \end{tabular}

Let \(y=\) wear life of a bearing, \(x_{1}=\) oil viscosity, and \(x_{2}=\) load. Suppose that the multiple regression model relating life to viscosity and load is $$ Y=125.0+7.75 x_{1}+.0950 x_{2}-.0090 x_{1} x_{2}+\epsilon $$ a. What is the mean value of life when viscosity is 40 and load is 1100 ? b. When viscosity is 30 , what is the change in mean life associated with an increase of 1 in load? When viscosity is 40 , what is the change in mean life associated with an increase of 1 in load?

The article "Readability of Liquid Crystal Displays: A Response Surface" (Human Factors, 1983: 185-190) used a multiple regression model with four independent variables to study accuracy in reading liquid crystal displays. The variables were \(y=\) error percentage for subjects reading a four-digit liquid crystal display \(x_{1}=\) level of backlight (ranging from 0 to \(122 \mathrm{~cd} / \mathrm{m}^{2}\) ) \(x_{2}=\) character subtense (ranging from \(.025^{\circ}\) to \(1.34^{\circ}\) ) \(x_{3}=\) viewing angle (ranging from \(0^{\circ}\) to \(60^{\circ}\) ) \(x_{4}=\) level of ambient light (ranging from 20 to 1500 lux) The model fit to data was \(Y=\beta_{0}+\beta_{1} x_{1}+\beta_{2} x_{2}+\beta_{3} x_{3}+\) \(\beta_{4} x_{4}+\epsilon\). The resulting estimated coefficients were \(\hat{\beta}_{0}=1.52, \hat{\beta}_{1}=.02, \hat{\beta}_{2}=-1.40, \hat{\beta}_{3}=.02\), and \(\hat{\beta}_{4}=-.0006\). a. Calculate an estimate of expected error percentage when \(x_{1}=10, x_{2}=5, x_{3}=50\), and \(x_{4}=100 .\) b. Estimate the mean error percentage associated with a backlight level of 20 , character subtense of \(.5\), viewing angle of 10 , and ambient light level of 30 . c. What is the estimated expected change in error percentage when the level of ambient light is increased by 1 unit while all other variables are fixed at the values given in part (a)? Answer for a 100-unit increase in ambient light level. d. Explain why the answers in part (c) do not depend on the fixed values of \(x_{1}, x_{2}\), and \(x_{3}\). Under what conditions would there be such a dependence? e. The estimated model was based on \(n=30\) observations, with SST \(=39.2\) and SSE \(=20.0\). Calculate and interpret the coefficient of multiple determination, and then carry out the model utility test using \(\alpha=.05\).

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.