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Consider the following inverse model matrix. \(\left(\mathbf{X}^{\prime} \mathbf{X}\right)^{-1}=\left[\begin{array}{cccc}0.125 & 0 & 0 & 0 \\ 0 & 0.125 & 0 & 0 \\ 0 & 0 & 0.125 & 0 \\ 0 & 0 & 0 & 0.125\end{array}\right]\) (a) How many regressors are in this model? (b) What was the sample size? (c) Notice the special diagonal structure of the matrix. What does that tell you about the columns in the original X matrix?

Short Answer

Expert verified
(a) 4 regressors, (b) Sample size is 12, (c) Columns in \(\mathbf{X}\) are orthogonal and possibly orthonormal.

Step by step solution

01

Identify the Diagonal Matrix Structure

The given inverse model matrix \(\left(\mathbf{X}^{\prime}\mathbf{X}\right)^{-1} = \begin{bmatrix}0.125 & 0 & 0 & 0 \ 0 & 0.125 & 0 & 0 \ 0 & 0 & 0.125 & 0 \ 0 & 0 & 0 & 0.125\end{bmatrix} \) is a diagonal matrix. This indicates that the columns of the original matrix \(\mathbf{X}\) are orthogonal to each other.
02

Calculate Number of Regressors

Since the inverse matrix is a 4x4 diagonal matrix, this implies there are 4 regressors in the model (one per row/column of the matrix). Each diagonal element corresponds to the variance of a parameter estimate.
03

Determine the Sample Size

The elements on the diagonal of the inverse matrix are all the same, each equal to 0.125. The relationship between these elements and the sample size \(n\) is given as 0.125 = 1/(n-4). Solving for \(n\), we see that \(0.125 = 1/(n-4)\) or \(n-4 = 8\), hence \(n = 12\).
04

Interpretation of Special Diagonal Structure

The diagonal structure, with equal variances across the diagonal, suggests that the original \(\mathbf{X}\) columns were not only orthogonal, but were also standardized such that each regressor has equal variance. This often occurs when regressors are orthonormal.

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Key Concepts

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

Regressors in Statistical Models
In statistical models, regressors, also called predictors or independent variables, are crucial components used to predict or explain the variability in a dependent variable. Each regressor represents a specific piece of information or characteristic used in forming a model.
For instance, in a linear regression model, the typical formulation would be:
\[ y = \beta_0 + \beta_1 x_1 + \beta_2 x_2 + ... + \beta_p x_p + \epsilon \]
Here, \( x_1, x_2, ..., x_p \) are the regressors, while \( \beta_1, \beta_2, ..., \beta_p \) are coefficients indicating the strength and direction of the relationship between each regressor and the dependent variable \( y \). Regressors don't just introduce new data points to analyze; they shape the interpretation of results, influencing decisions and conclusions.
By analyzing the number of rows and columns in a model's matrix, we can determine the number of regressors used, which in the case of a 4x4 inverse matrix, signifies 4 regressors in the model.
Orthogonal Columns
When we talk about orthogonal columns in the context of matrix algebra, we're referring to columns of a matrix that are mutually independent. Simply put, the dot product of any pair of different columns equals zero, showing no correlation between them. This has significant implications in regression analysis.
In an orthogonal configuration, each predictor's effect on the dependent variable is distinct. This separation means that changes in one predictor do not affect the estimates of others. When the matrix \( \mathbf{X} \) is orthogonal, the matrix \( \mathbf{X}' \mathbf{X} \) becomes a diagonal matrix. Because of this, the inverse, \( (\mathbf{X}' \mathbf{X})^{-1} \), also takes a special diagonal form, as seen in the exercise.
The diagonal structure not only indicates orthogonality but often suggests ortho-normality if further transformed. In practice, orthogonality simplifies calculations and interpretations, especially when each regressor is standardized to equal variance.
Sample Size in Regression Analysis
Sample size is a critical factor in regression analysis. It impacts the precision of parameter estimates and the power of statistical tests. To estimate the sample size from an inverse model matrix, specific relationships between matrix elements are exploited.
In our given problem, the diagonal elements of \( (\mathbf{X}' \mathbf{X})^{-1} \) were equal to 0.125. This consistent diagonal value provides insight into how large the sample was. By understanding the formula \( \text{Element} = 1/(n-p) \) where \( n \) is the sample size and \( p \) is the number of parameters, we can derive the sample size.
Here, since there were 4 regressors, setting \( 0.125 = 1/(n-4) \) and solving for \( n \), we establish \( n = 12 \). In broader terms, greater sample sizes tend to yield more reliable estimates, as they minimize the standard error. Small sample sizes might lead to significant variability in estimates, thus impacting the overall analysis efficiency.

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Most popular questions from this chapter

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 \(x_{2}=43 \% ?\)

A chemical engineer is investigating how the amount of conversion of a product from a raw material \((y)\) depends on reaction temperature \(\left(x_{1}\right)\) and the reaction time \(\left(x_{2}\right) .\) He has developed the following regression models: 1\. \(\hat{y}=100+2 x_{1}+4 x_{2}\) 2\. \(\hat{y}=95+1.5 x_{1}+3 x_{2}+2 x_{1} x_{2}\) Both models have been built over the range \(0.5 \leq x_{2} \leq 10 .\) (a) What is the predicted value of conversion when \(x_{2}=2 ?\) Repeat this calculation for \(x_{2}=8 .\) Draw a graph of the predicted values for both conversion models. Comment on the effect of the interaction term in model 2 . (b) Find the expected change in the mean conversion for a unit change in temperature \(x_{1}\) for model 1 when \(x_{2}=5 .\) Does this quantity depend on the specific value of reaction time selected? Why? (c) Find the expected change in the mean conversion for a unit change in temperature \(x_{1}\) for model 2 when \(x_{2}=5 .\) Repeat this calculation for \(x_{2}=2\) and \(x_{2}=8\). Does the result depend on the value selected for \(x_{2}\) ? Why?

You have fit a regression model with two regressors to a data set that has 20 observations. The total sum of squares is 1000 and the model sum of squares is 750 (a) What is the value of \(R^{2}\) for this model? (b) What is the adjusted \(R^{2}\) for this model? (c) What is the value of the \(F\) -statistic for testing the significance of regression? What conclusions would you draw about this model if \(\alpha=0.05 ?\) What if \(\alpha=0.01 ?\) (d) Suppose that you add a third regressor to the model and as a result, the model sum of squares is now \(785 .\) Does it seem to you that adding this factor has improved the model?

The pull strength of a wire bond is an important characteristic. Table \(\mathrm{E} 12-4\) gives information on pull strength \((y)\) die height \(\left(x_{1}\right),\) post height \(\left(x_{2}\right),\) loop height \(\left(x_{3}\right),\) wire length \(\left(x_{4}\right),\) bond width on the die \(\left(x_{5}\right),\) and bond width on the post \(\left(x_{6}\right)\) (a) Fit a multiple linear regression model using \(x_{2}, x_{3}, x_{4},\) and \(x_{5}\) as the regressors (b) Estimate \(\sigma^{2}\). (c) Find the \(\operatorname{se}\left(\hat{\boldsymbol{\beta}}_{j}\right) .\) How precisely are the regression coefficients estimated in your opinion? (d) Use the model from part (a) to predict pull strength when \(x_{2}=20, x_{3}=30, x_{4}=90,\) and \(x_{5}=2.0 .\) $$ \begin{array}{rcccccc} \hline y & x_{1} & x_{2} & x_{3} & x_{4} & x_{5} & x_{6} \\ \hline 8.0 & 5.2 & 19.6 & 29.6 & 94.9 & 2.1 & 2.3 \\ 8.3 & 5.2 & 19.8 & 32.4 & 89.7 & 2.1 & 1.8 \\ 8.5 & 5.8 & 19.6 & 31.0 & 96.2 & 2.0 & 2.0 \\ 8.8 & 6.4 & 19.4 & 32.4 & 95.6 & 2.2 & 2.1 \\ 9.0 & 5.8 & 18.6 & 28.6 & 86.5 & 2.0 & 1.8 \\ 9.3 & 5.2 & 18.8 & 30.6 & 84.5 & 2.1 & 2.1 \\ 9.3 & 5.6 & 20.4 & 32.4 & 88.8 & 2.2 & 1.9 \\ 9.5 & 6.0 & 19.0 & 32.6 & 85.7 & 2.1 & 1.9 \\ 9.8 & 5.2 & 20.8 & 32.2 & 93.6 & 2.3 & 2.1 \\ 10.0 & 5.8 & 19.9 & 31.8 & 86.0 & 2.1 & 1.8 \\ 10.3 & 6.4 & 18.0 & 32.6 & 87.1 & 2.0 & 1.6 \\ 10.5 & 6.0 & 20.6 & 33.4 & 93.1 & 2.1 & 2.1 \\ 10.8 & 6.2 & 20.2 & 31.8 & 83.4 & 2.2 & 2.1 \\ 11.0 & 6.2 & 20.2 & 32.4 & 94.5 & 2.1 & 1.9 \\ 11.3 & 6.2 & 19.2 & 31.4 & 83.4 & 1.9 & 1.8 \\ 11.5 & 5.6 & 17.0 & 33.2 & 85.2 & 2.1 & 2.1 \\ 11.8 & 6.0 & 19.8 & 35.4 & 84.1 & 2.0 & 1.8 \\ 12.3 & 5.8 & 18.8 & 34.0 & 86.9 & 2.1 & 1.8 \\ 12.5 & 5.6 & 18.6 & 34.2 & 83.0 & 1.9 & 2.0 \\ \hline \end{array} $$

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 \(\alpha=0.05\). $$ \begin{array}{ccccc} \hline \text { Observation } & & & & \\ \text { Number } & \boldsymbol{y} & \boldsymbol{x}_{\mathbf{1}} & \boldsymbol{x}_{2} & \boldsymbol{x}_{3} \\ \hline 1 & 0.22200 & 7.3 & 0.0 & 0.0 \\ 2 & 0.39500 & 8.7 & 0.0 & 0.3 \\ 3 & 0.42200 & 8.8 & 0.7 & 1.0 \\ 4 & 0.43700 & 8.1 & 4.0 & 0.2 \\ 5 & 0.42800 & 9.0 & 0.5 & 1.0 \\ 6 & 0.46700 & 8.7 & 1.5 & 2.8 \\ 7 & 0.44400 & 9.3 & 2.1 & 1.0 \\ 8 & 0.37800 & 7.6 & 5.1 & 3.4 \\ 9 & 0.49400 & 10.0 & 0.0 & 0.3 \\ 10 & 0.45600 & 8.4 & 3.7 & 4.1 \\ 11 & 0.45200 & 9.3 & 3.6 & 2.0 \\ 12 & 0.11200 & 7.7 & 2.8 & 7.1 \\ 13 & 0.43200 & 9.8 & 4.2 & 2.0 \\ 14 & 0.10100 & 7.3 & 2.5 & 6.8 \\ 15 & 0.23200 & 8.5 & 2.0 & 6.6 \\ 16 & 0.30600 & 9.5 & 2.5 & 5.0 \\ 17 & 0.09230 & 7.4 & 2.8 & 7.8 \\ 18 & 0.11600 & 7.8 & 2.8 & 7.7 \\ 19 & 0.07640 & 7.7 & 3.0 & 8.0 \\ 20 & 0.43900 & 10.3 & 1.7 & 4.2 \\ 21 & 0.09440 & 7.8 & 3.3 & 8.5 \\ 22 & 0.11700 & 7.1 & 3.9 & 6.6 \\ 23 & 0.07260 & 7.7 & 4.3 & 9.5 \\ 24 & 0.04120 & 7.4 & 6.0 & 10.9 \\ 25 & 0.25100 & 7.3 & 2.0 & 5.2 \\ 26 & 0.00002 & 7.6 & 7.8 & 20.7 \\ \hline \end{array} $$

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