/*! 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 115 A multiple regression model was ... [FREE SOLUTION] | 91Ó°ÊÓ

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A multiple regression model was used to relate \(y=\) viscosity of a chemical product to \(x_{1}=\) temperature and \(x_{2}=\) reaction time. The data set consisted of \(n=15\) observations. (a) The estimated regression coefficients were \(\hat{\beta}_{0}=300.00\) \(\hat{\beta}_{1}=0.85,\) and \(\hat{\beta}_{2}=10.40 .\) Calculate an estimate of mean viscosity when \(x_{1}=100^{\circ} \mathrm{F}\) and \(x_{2}=2\) hours. (b) The sums of squares were \(S S_{T}=1230.50\) and \(S S_{E}=120.30\). Test for significance of regression using \(\alpha=0.05 .\) What conclusion can you draw? (c) What proportion of total variability in viscosity is accounted for by the variables in this model? (d) Suppose that another regressor, \(x_{3}=\) stirring rate, is added to the model. The new value of the error sum of squares is \(S S_{E}=117.20 .\) Has adding the new variable resulted in a smaller value of \(M S_{E} ?\) Discuss the significance of this result. (e) Calculate an \(F\) -statistic to assess the contribution of \(x_{3}\) to the model. Using \(\alpha=0.05,\) what conclusions do you reach?

Short Answer

Expert verified
(a) 405.80. (b) Regression is significant. (c) 90.21%. (d) Higher MSE; no benefit. (e) No significant contribution.

Step by step solution

01

Calculate Mean Viscosity

To estimate the mean viscosity, we use the regression equation: \( \hat{y} = \hat{\beta}_{0} + \hat{\beta}_{1}x_{1} + \hat{\beta}_{2}x_{2} \).Substituting the given values: \( \hat{y} = 300.00 + 0.85 \times 100 + 10.40 \times 2 \).Calculate: \( \hat{y} = 300.00 + 85 + 20.80 = 405.80 \).Thus, the estimated mean viscosity is 405.80.
02

Test for Significance of Regression

To test for the significance of the regression, we use the formula for the F-statistic:\[ F = \frac{(SS_{T} - SS_{E})/p}{SS_{E}/(n-p-1)} \]Where \( SS_{T} = 1230.50 \), \( SS_{E} = 120.30 \), \( p = 2 \), and \( n = 15 \).Calculate sums:\( SS_{R} = SS_{T} - SS_{E} = 1230.50 - 120.30 = 1110.20 \).Then, calculate the F-statistic:\[ F = \frac{1110.20/2}{120.30/12} = \frac{555.10}{10.025} = 55.37 \]Using \( \alpha = 0.05 \) and \( df_1 = 2 \), \( df_2 = 12 \), compare with \( F_{critical} \approx 3.89 \).Since 55.37 > 3.89, the regression is significant.
03

Calculate Proportion of Variability Explained

The proportion of total variability explained by the model is given by the coefficient of determination, \( R^2 \):\[ R^2 = \frac{SS_{R}}{SS_{T}} = \frac{1110.20}{1230.50} \approx 0.9021 \]Thus, approximately 90.21% of the variability is explained by the model.
04

Assess MS Error Reduction with New Variable

The original mean square error \( MS_{E} \) is:\[ MS_{E} = \frac{SS_{E}}{n - p - 1} = \frac{120.30}{12} = 10.025 \]With the new variable, the \( MSE \) with \( p+1 = 3 \) regressors is:\[ MS_{E_{new}} = \frac{117.20}{11} = 10.655 \]The new \( MS_{E} \) has increased, indicating the new variable did not decrease error variance.
05

Calculate F-statistic for New Variable Contribution

For the new variable \( x_3 \), calculate the F-statistic using:\[ F = \frac{(MS_{E} - MS_{E_{new}})/1}{MS_{E_{new}}} = \frac{(10.025 - 10.655)}{10.655/1} \approx -0.630 \]Since this F-value is negative and not greater than \( F_{critical} \approx 4.84 \) for \( df_1 = 1 \) and \( df_2 = 11 \), the new variable does not make a significant contribution at \( \alpha = 0.05 \).

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Key Concepts

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

Regression Coefficients
In the context of multiple regression analysis, regression coefficients represent the influence that each independent variable has on the dependent variable. For the given exercise, we are dealing with temperature (\( x_1 \)) and reaction time (\( x_2 \)) impacting the viscosity of a chemical product (\( y \)). The regression coefficients (\( \hat{\beta}_0 = 300.00 \), \( \hat{\beta}_1 = 0.85 \), \( \hat{\beta}_2 = 10.40 \)) indicate the estimated changes in viscosity with changes in temperature and time.
- \( \hat{\beta}_0 = 300.00 \) is the intercept, indicating the expected viscosity when both temperature and reaction time are zero. However, in practical terms, settings like zero temperature and zero time may be nonsensical, but this intercept helps improve the model's fit.
- \( \hat{\beta}_1 = 0.85 \) shows that with each 1-degree increase in temperature, viscosity is expected to increase by 0.85 units, assuming reaction time remains constant.
- \( \hat{\beta}_2 = 10.40 \) demonstrates the expected increase of 10.40 units in viscosity for every additional hour of reaction time, assuming temperature remains constant.
By plugging in values such as \( x_1 = 100 \) and \( x_2 = 2 \) into the regression equation, we estimate the mean viscosity, allowing predictions based on different experimental conditions.
F-statistic
The F-statistic is an integral part of regression analysis, as it measures the overall significance of the regression model. It evaluates whether there is a meaningful relationship between the dependent and independent variables. To arrive at the F-statistic, one can use the formula:
\[ F = \frac{(SS_T - SS_E)/p}{SS_E/(n-p-1)} \]
In this problem, given the sums of squares (\( SS_T = 1230.50 \), \( SS_E = 120.30 \)), number of predictors (\( p = 2 \)), and number of observations (\( n = 15 \)), we calculate:
- \( SS_R \), the regression sum of squares, is \( 1110.20 \).
- The F-statistic comes out to be 55.37, which strongly exceeds the critical value of 3.89 for \( \alpha = 0.05 \).
This indicates a significant relationship, meaning the combination of temperature and reaction time provides valuable predictive information for viscosity. Thus, the regression model as a whole is deemed to be statistically significant.
Coefficient of Determination
The coefficient of determination, denoted as \( R^2 \), is a statistical measure that provides insight into how well the independent variables explain the variability of the dependent variable. In this specific exercise, \( R^2 \) is calculated as:
\[ R^2 = \frac{SS_R}{SS_T} = \frac{1110.20}{1230.50} \approx 0.9021 \]
Here, approximately 90.21% of the variability in viscosity is explained by the model, a very high proportion which suggests a strong predictive power.
The closer \( R^2 \) is to 1, the better the data points fit the regression line. This high value indicates that temperature and reaction time substantially account for changes in viscosity, leaving very little unexplained variance. Such insight helps validate the model's robustness.
Error Sum of Squares
The error sum of squares (\( SS_E \)) is crucial in assessing model fit. It represents the unexplained variance after fitting the model. Lower \( SS_E \) implies a better fitting model. Initially, \( SS_E \) was 120.30 for two predictors.
Upon adding a third predictor (\( x_3 \), stirring rate), the new \( SS_E \) becomes 117.20. Despite this decrease, one must consider the \( MS_E \) or mean square error:
- With two predictors, \( MS_E = 10.025 \).
- With three, it increases slightly to \( 10.655 \).
The rise in \( MS_E \) indicates the new variable didn’t contribute meaningfully to reducing error, emphasizing that just because a variable appears in a model, it doesn't automatically ensure beneficial results. Sometimes, superfluous predictors can increase complexity without offering predictive improvement, a vital consideration during model refinement.

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Most popular questions from this chapter

Constrained Least Squares. Suppose we wish to find the least squares estimator of \(\beta\) in the model \(\mathbf{y}=\mathbf{X} \boldsymbol{\beta}+\boldsymbol{\epsilon}\) subject to a set of equality constraints, say, \(\mathbf{T} \boldsymbol{\beta}=\mathbf{c} .\) (a) Show that the estimator is $$ \begin{array}{l} \hat{\boldsymbol{\beta}}_{c}=\hat{\boldsymbol{\beta}}+\left(\mathbf{X}^{\prime} \mathbf{X}\right)^{-1} \times \mathbf{T}^{\prime}\left[\mathbf{T}\left(\left(\mathbf{X}^{\prime} \mathbf{X}\right)^{-1}\right) \mathbf{T}^{\prime}\right]^{-1}(\mathbf{c}-\mathbf{T} \hat{\boldsymbol{\beta}}) \\ \text { where } \hat{\boldsymbol{\beta}}=\left(\mathbf{X}^{\prime} \mathbf{X}\right)^{-1} \mathbf{X}^{\prime} \mathbf{y} \end{array} $$ (b) Discuss situations where this model might be appropriate.

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.

Hsuie, Ma, and Tsai ("Separation and Characterizations of Thermotropic Copolyesters of p-Hydroxybenzoic Acid, Sebacic Acid, and Hydroquinone," (1995, Vol. 56) studied the effect of the molar ratio of sebacic acid (the regressor) on the intrinsic viscosity of copolyesters (the response). The following display presents the data. $$ \begin{array}{cc} \hline \text { Ratio } & \text { Viscosity } \\ \hline 1.0 & 0.45 \\ 0.9 & 0.20 \\ 0.8 & 0.34 \\ 0.7 & 0.58 \\ 0.6 & 0.70 \\ 0.5 & 0.57 \\ 0.4 & 0.55 \\ 0.3 & 0.44 \end{array} $$ (a) Construct a scatterplot of the data. (b) Fit a second-order prediction equation.

A sample of 25 observations is used to fit a regression model in seven variables. The estimate of \(\sigma^{2}\) for this full model is \(M S_{E}=10\). (a) A forward selection algorithm has put three of the original seven regressors in the model. The error sum of squares for the three-variable model is \(S S_{E}=300 .\) Based on \(C_{p}\), would you conclude that the three- variable model has any remaining bias? (b) After looking at the forward selection model in part (a), suppose you could add one more regressor to the model. This regressor will reduce the error sum of squares to \(275 .\) Will the addition of this variable improve the model? Why?

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 \(\alpha=0.05\)

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