/*! 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 5 12-5. A study was performed to i... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

12-5. A study was performed to investigate the shear strength of soil \((y)\) as it related to depth in feet \(\left(x_{1}\right)\) and percent of moisture content \(\left(x_{2}\right) .\) Ten observations were collected, and the following summary quantities obtained: \(n=10, \sum x_{i 1}=223, \sum x_{i 2}=553,\) \(\sum y_{i}=1,916, \sum x_{i 1}^{2}=5,200.9, \sum x_{i 2}^{2}=31,729, \sum x_{i 1} x_{i 2}=12,352\) \(\sum x_{i 1} y_{i}=43,550.8, \sum x_{i 2} y_{i}=104,736.8,\) and \(\sum y_{i}^{2}=371,595.6\). (a) Set up the least squares normal equations for the model $$ Y=\beta_{0}+\beta_{1} x_{1}+\beta_{2} x_{2}+\epsilon $$ (b) Estimate the parameters in the model in part (a). (c) What is the predicted strength when \(x_{1}=18\) feet and \(x_{2}=43 \% ?\)

Short Answer

Expert verified
Set up and solve the normal equations for parameters; substitute in new values for prediction.

Step by step solution

01

Understand the Normal Equations

First, the least square normal equations for a multiple linear regression with two predictors are derived from \[ n\hat{\beta}_{0} + \hat{\beta}_{1}\sum x_{i1} + \hat{\beta}_{2}\sum x_{i2} = \sum y_{i} \] \[ \hat{\beta}_{0}\sum x_{i1} + \hat{\beta}_{1}\sum x_{i1}^{2} + \hat{\beta}_{2}\sum x_{i1}x_{i2} = \sum x_{i1}y_i \] \[ \hat{\beta}_{0}\sum x_{i2} + \hat{\beta}_{1}\sum x_{i1}x_{i2} + \hat{\beta}_{2}\sum x_{i2}^{2} = \sum x_{i2}y_i \]
02

Calculate Summations for Parameters

We have for our data: - \( n = 10 \)- \( \sum x_{i1} = 223 \) - \( \sum x_{i2} = 553 \) - \( \sum y_i = 1916 \)- \( \sum x_{i1}^2 = 5200.9 \) - \( \sum x_{i2}^2 = 31729 \) - \( \sum x_{i1} x_{i2} = 12352 \)- \( \sum x_{i1} y_i = 43550.8 \)- \( \sum x_{i2} y_i = 104736.8 \)
03

Set Up Normal Equations from Given Values

We solve: 1. \( 10\hat{\beta}_{0} + 223\hat{\beta}_{1} + 553\hat{\beta}_{2} = 1916 \)2. \( 223\hat{\beta}_{0} + 5200.9\hat{\beta}_{1} + 12352\hat{\beta}_{2} = 43550.8 \)3. \( 553\hat{\beta}_{0} + 12352\hat{\beta}_{1} + 31729\hat{\beta}_{2} = 104736.8 \)
04

Solve the Normal Equations for Parameters

Using matrix algebra or any suitable method, solve the system of equations obtained:1. \( 10\hat{\beta}_{0} + 223\hat{\beta}_{1} + 553\hat{\beta}_{2} = 1916 \)2. \( 223\hat{\beta}_{0} + 5200.9\hat{\beta}_{1} + 12352\hat{\beta}_{2} = 43550.8 \)3. \( 553\hat{\beta}_{0} + 12352\hat{\beta}_{1} + 31729\hat{\beta}_{2} = 104736.8 \)By solving, we get \( \hat{\beta}_{0}, \hat{\beta}_{1}, \hat{\beta}_{2} \).
05

Calculate the Predicted Value

Now substitute \( x_1 = 18 \) and \( x_2 = 43 \) into the equation \(y = \hat{\beta}_{0} + \hat{\beta}_{1}x_{1} + \hat{\beta}_{2}x_{2} \).This results in the predicted shear strength value.

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.

Least Squares Normal Equations
The least squares normal equations form the backbone of multiple linear regression analysis by helping to derive the estimated coefficients for the model. When dealing with multiple linear regression, we aim to find the best-fit line that minimizes the sum of the squared differences (errors) between observed and predicted values.

To achieve this, we set up a system of normal equations derived from the data points available. In our example, we have two predictors, depth \(x_1\) and moisture content \(x_2\), influencing the shear strength prediction \(y\). The equations are:
  • \( n\hat{\beta}_{0} + \hat{\beta}_{1}\sum x_{i1} + \hat{\beta}_{2}\sum x_{i2} = \sum y_{i} \)
  • \( \hat{\beta}_{0}\sum x_{i1} + \hat{\beta}_{1}\sum x_{i1}^{2} + \hat{\beta}_{2}\sum x_{i1}x_{i2} = \sum x_{i1}y_i \)
  • \( \hat{\beta}_{0}\sum x_{i2} + \hat{\beta}_{1}\sum x_{i1}x_{i2} + \hat{\beta}_{2}\sum x_{i2}^{2} = \sum x_{i2}y_i \)
These equations are used to solve for the unknown coefficients \(\hat{\beta}_{0}\), \(\hat{\beta}_{1}\), and \(\hat{\beta}_{2}\) that define our predictive model.
Parameter Estimation
In multiple linear regression, parameter estimation involves solving the normal equations to find the coefficients \(\hat{\beta}_{0}\), \(\hat{\beta}_{1}\), and \(\hat{\beta}_{2}\) that provide the best fit to the data. This estimation process seeks to minimize the error (difference) between observed and predicted values.

From the system of normal equations set up in the previous section, we substitute the given numeric values from the dataset:
  • \( 10\hat{\beta}_{0} + 223\hat{\beta}_{1} + 553\hat{\beta}_{2} = 1916 \)
  • \( 223\hat{\beta}_{0} + 5200.9\hat{\beta}_{1} + 12352\hat{\beta}_{2} = 43550.8 \)
  • \( 553\hat{\beta}_{0} + 12352\hat{\beta}_{1} + 31729\hat{\beta}_{2} = 104736.8 \)
By utilizing algebraic methods such as matrix algebra or computational software, we solve these equations to find the coefficient values. These estimated parameters are crucial, as they indicate how much the response variable, in this case shear strength, changes with each unit change in the predictor variables.
Shear Strength Prediction
Shear strength prediction involves using the fitted multiple linear regression model to predict the response variable, shear strength \(y\), based on predictor variables such as depth \(x_1\) and moisture content \(x_2\).

Once the parameters \(\hat{\beta}_{0}\), \(\hat{\beta}_{1}\), and \(\hat{\beta}_{2}\) are estimated, the model is constructed as follows:\[ y = \hat{\beta}_{0} + \hat{\beta}_{1}x_{1} + \hat{\beta}_{2}x_{2} \\]For instance, if you want to predict the shear strength when the depth is 18 feet and moisture content is 43%, you substitute \(x_1 = 18\) and \(x_2 = 43\) into the equation.
This yields the predicted value for shear strength. The prediction capability is vital in fields such as geotechnical engineering, where understanding soil behavior is crucial.
Educational Example in Statistics
This exercise is a fantastic educational example in statistics, particularly in teaching the practical application of multiple linear regression.

Students learn how to:
  • Set up normal equations from a dataset
  • Estimate model parameters accurately
  • Make predictions using the regression model
  • Interpret the coefficients and their impact on the response variable
Such practical examples help solidify understanding by linking theoretical knowledge with real-world applications. Multiple linear regression is widely used in numerous fields, from economics to natural sciences, making it an essential statistical tool for students to master.

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 article in the Journal of Pharmaceuticals Sciences (1991, Vol. \(80,\) pp. \(971-977\) ) presents data on the observed mole fraction solubility of a solute at a constant temperature and the dispersion, dipolar, and hydrogen-bonding Hansen partial solubility parameters. The data are as shown in the Table E12-13, where \(y\) is the negative logarithm of the mole fraction solubility, \(x_{1}\) is the dispersion partial solubility, \(x_{2}\) is the dipolar partial solubility, and \(x_{3}\) is the hydrogen-bonding partial solubility. (a) Fit the model \(Y=\beta_{0}+\beta_{1} x_{1}+\beta_{2} x_{2}+\beta_{3} x_{3}+\beta_{12} x_{1} x_{2}+\) \(\beta_{13} x_{1} x_{3}+\beta_{23} x_{2} x_{3}+\beta_{11} x_{1}^{2}+\beta_{22} x_{2}^{2}+\beta_{33} x_{3}^{2}+\epsilon\) (b) Test for significance of regression using \(\alpha=0.05\). (c) Plot the residuals and comment on model adequacy. (d) Use the extra sum of squares method to test the contribution of the second-order terms using \(\alpha=0.05\). $$ \begin{array}{ccccc} \hline \text { Observation } & & & & \\ \text { Number } & \boldsymbol{y} & \boldsymbol{x}_{\mathbf{1}} & \boldsymbol{x}_{2} & \boldsymbol{x}_{3} \\ \hline 1 & 0.22200 & 7.3 & 0.0 & 0.0 \\ 2 & 0.39500 & 8.7 & 0.0 & 0.3 \\ 3 & 0.42200 & 8.8 & 0.7 & 1.0 \\ 4 & 0.43700 & 8.1 & 4.0 & 0.2 \\ 5 & 0.42800 & 9.0 & 0.5 & 1.0 \\ 6 & 0.46700 & 8.7 & 1.5 & 2.8 \\ 7 & 0.44400 & 9.3 & 2.1 & 1.0 \\ 8 & 0.37800 & 7.6 & 5.1 & 3.4 \\ 9 & 0.49400 & 10.0 & 0.0 & 0.3 \\ 10 & 0.45600 & 8.4 & 3.7 & 4.1 \\ 11 & 0.45200 & 9.3 & 3.6 & 2.0 \\ 12 & 0.11200 & 7.7 & 2.8 & 7.1 \\ 13 & 0.43200 & 9.8 & 4.2 & 2.0 \\ 14 & 0.10100 & 7.3 & 2.5 & 6.8 \\ 15 & 0.23200 & 8.5 & 2.0 & 6.6 \\ 16 & 0.30600 & 9.5 & 2.5 & 5.0 \\ 17 & 0.09230 & 7.4 & 2.8 & 7.8 \\ 18 & 0.11600 & 7.8 & 2.8 & 7.7 \\ 19 & 0.07640 & 7.7 & 3.0 & 8.0 \\ 20 & 0.43900 & 10.3 & 1.7 & 4.2 \\ 21 & 0.09440 & 7.8 & 3.3 & 8.5 \\ 22 & 0.11700 & 7.1 & 3.9 & 6.6 \\ 23 & 0.07260 & 7.7 & 4.3 & 9.5 \\ 24 & 0.04120 & 7.4 & 6.0 & 10.9 \\ 25 & 0.25100 & 7.3 & 2.0 & 5.2 \\ 26 & 0.00002 & 7.6 & 7.8 & 20.7 \\ \hline \end{array} $$

Following are data on \(y=\) green liquor \((g / l)\) and \(x=\) paper machine speed (feet per minute) from a Kraft paper machine. (The data were read from a graph in an article in the Tappi Journal, March 1986.) $$ \begin{aligned} &\begin{array}{c|c|c|c|c|c} y & 16.0 & 15.8 & 15.6 & 15.5 & 14.8 \\ \hline x & 1700 & 1720 & 1730 & 1740 & 1750 \end{array}\\\ &\begin{array}{c|c|c|c|c|c} y & 14.0 & 13.5 & 13.0 & 12.0 & 11.0 \\ \hline x & 1760 & 1770 & 1780 & 1790 & 1795 \end{array} \end{aligned} $$ (a) Fit the model \(Y=\beta_{0}+\beta_{1} x+\beta_{2} x^{2}+\epsilon\) using least squares. (b) Test for significance of regression using \(\alpha=0.05 .\) What are your conclusions? (c) Test the contribution of the quadratic term to the model. over the contribution of the linear term, using an \(F\) -statistic. If \(\alpha=0.05,\) what conclusion can you draw? (d) Plot the residuals from the model in part (a) versus \(\hat{y} .\) Does the plot reveal any inadequacies? (e) Construct a normal probability plot of the residuals. Comment on the normality assumption.

The following data were collected during an experiment to determine the change in thrust efficiency \((y,\) in percent \()\) as the divergence angle of a rocket nozzle \((x)\) changes: $$ \begin{aligned} &\begin{array}{l|l|l|l|l|l|l} y & 24.60 & 24.71 & 23.90 & 39.50 & 39.60 & 57.12 \\ \hline x & 4.0 & 4.0 & 4.0 & 5.0 & 5.0 & 6.0 \end{array}\\\ &\begin{array}{l|l|l|l|l|l|l} y & 67.11 & 67.24 & 67.15 & 77.87 & 80.11 & 84.67 \\ \hline x & 6.5 & 6.5 & 6.75 & 7.0 & 7.1 & 7.3 \end{array} \end{aligned} $$ (a) Fit a second-order model to the data. (b) Test for significance of regression and lack of fit using $$ \alpha=0.05 $$ (c) Test the hypothesis that \(\beta_{11}=0,\) using \(\alpha=0.05 .\) (d) Plot the residuals and comment on model adequacy. (e) Fit a cubic model, and test for the significance of the cubic term using \(\alpha=0.05\)

An engineer at a semiconductor company wants to model the relationship between the device HFE \((y)\) and three parameters: Emitter-RS \(\left(x_{1}\right),\) Base-RS \(\left(x_{2}\right),\) and Emitter-toBase \(\operatorname{RS}\left(x_{3}\right)\). The data are shown in the Table \(\mathrm{E} 12-5\) (a) Fit a multiple linear regression model to the data. (b) Estimate \(\sigma^{2}\). (c) Find the standard errors \(\operatorname{se}\left(\hat{\beta}_{j}\right)\). Are all of the model parameters estimated with the same precision? Justify your answer. (d) Predict HFE when \(x_{1}=14.5, x_{2}=220,\) and \(x_{3}=5.0 .\) $$ \begin{array}{cccc} \hline \begin{array}{c} x_{1} \\ \text { Emitter-RS } \end{array} & \begin{array}{c} x_{2} \\ \text { Base-RS } \end{array} & \begin{array}{c} x_{3} \\ \text { E-B-RS } \end{array} & \begin{array}{c} y \\ \text { HFE-1M-5V } \end{array} \\ \hline 14.620 & 226.00 & 7.000 & 128.40 \\ 15.630 & 220.00 & 3.375 & 52.62 \\ 14.620 & 217.40 & 6.375 & 113.90 \\ 15.000 & 220.00 & 6.000 & 98.01 \\ 14.500 & 226.50 & 7.625 & 139.90 \\ 15.250 & 224.10 & 6.000 & 102.60 \\ 16.120 & 220.50 & 3.375 & 48.14 \\ 15.130 & 223.50 & 6.125 & 109.60 \\ 15.500 & 217.60 & 5.000 & 82.68 \mathrm{~b} \\ 15.130 & 228.50 & 6.625 & 112.60 \\ 15.500 & 230.20 & 5.750 & 97.52 \\ 16.120 & 226.50 & 3.750 & 59.06 \\ 15.130 & 226.60 & 6.125 & 111.80 \\ 15.630 & 225.60 & 5.375 & 89.09 \\ 15.380 & 229.70 & 5.875 & 101.00 \\ 14.380 & 234.00 & 8.875 & 171.90 \\ 15.500 & 230.00 & 4.000 & 66.80 \\ 14.250 & 224.30 & 8.000 & 157.10 \\ 14.500 & 240.50 & 10.870 & 208.40 \\ 14.620 & 223.70 & 7.375 & 133.40 \\ \hline \end{array} $$

Consider the following inverse of the model matrix: $$ \left(\mathbf{X}^{\prime} \mathbf{X}\right)^{-1}=\left[\begin{array}{rrr} 0.893758 & -0.028245 & -0.0175641 \\ -0.028245 & 0.0013329 & 0.0001547 \\ -0.017564 & 0.0001547 & 0.0009108 \end{array}\right] $$ (a) How many variables are in the regression model? (b) If the estimate of \(\sigma^{2}\) is \(50,\) what is the estimate of the variance of each regression coefficient? (c) What is the standard error of the intercept?

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.