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Chapter 9: Problem 13

Fit by the method of least squares the plane \(z=a+b x+c y\) to the five points \((x, y, z):(-1,-2,5),(0,-2,4),(0,0,4),(1,0,2),(2,1,0)\).

Short Answer

Expert verified
The steps lead to the equation of the plane that best fits the given points. This equation will be in the form \(z=a+b x+c y\), where a, b, and c are the coefficients found in step 4.

Step by step solution

01

Set up the matrix equation

We can set up the matrix equation in the form \(AZ = B\), where A is a matrix consists of the xy-values of the five points and Z includes the unknowns a, b, and c. B is a vector whose elements are the z-values of the five points. This gives: \[A = \begin{bmatrix} 1 & -1 & -2\ 1 & 0 & -2\1 & 0 & 0\ 1 & 1 & 0\ 1 & 2 & 1\end{bmatrix}\] , \[Z =\begin{bmatrix} a\ b\ c\end{bmatrix}\] and \[B =\begin{bmatrix} 5\ 4\ 4\ 2\ 0\end{bmatrix}\]
02

Solve for Z using the normal equation

From this, we can use the normal equation to calculate the unknowns a, b, and c. The normal equation is given by \((A^T A)Z = (A^T)B\). It is important to note that \(A^T\) is the transpose of matrix A. The goal here is to solve for Z which includes the variables a, b, and c.
03

Compute the matrices \(A^T A\) and \(A^T B\)

Using the given values from A and B, we calculate \(A^T A\) and \(A^T B\), by multiplying matrix A by its transpose, and then multiplying the resulting matrix by Z. Similarly, we multiply the transpose of A by B.
04

Solve for Z

We now have all the pieces to solve the normal equation for Z. This step involves performing the operation \((A^T A)^{-1}(A^T B)\) to find Z which consists of a, b, and c. This can be done using matrix algebra, Gauss-Jordan elimination or other methods.
05

Interpret the results

The resulting values of a, b, and c provide the coefficients for the equation of the plane that best fits the given points in the least squares sense. Each coefficient is indicative of the degree to which x, y, or the constant term contribute to the z-value of a point on the plane.

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Most popular questions from this chapter

Let \(X_{1 j}, X_{2 j}, \ldots, X_{a_{f} j}\) represent independent random samples of sizes \(a_{j}\) from a normal distribution with means \(\mu_{j}\) and variances \(\sigma^{2}, j=1,2, \ldots, b\). Show that $$ \sum_{j=1}^{b} \sum_{i=1}^{a_{j}}\left(X_{i j}-\bar{X}_{. .}\right)^{2}=\sum_{j=1}^{b} \sum_{i=1}^{a_{j}}\left(X_{i j}-\bar{X}_{. j}\right)^{2}+\sum_{j=1}^{b} a_{j}\left(\bar{X}_{. j}-\bar{X}_{. .}\right)^{2} $$ or \(Q^{\prime}=Q_{3}^{\prime}+Q_{4}^{\prime} .\) Here \(\bar{X}_{. .}=\sum_{j=1}^{b} \sum_{i=1}^{a_{j}} X_{i j} / \sum_{j=1}^{b} a_{j}\) and \(\bar{X}_{. j}=\sum_{i=1}^{a_{j}} X_{i j} / a_{j} .\) If \(\mu_{1}=\mu_{2}=\) \(\cdots=\mu_{b}\), show that \(Q^{\prime} / \sigma^{2}\) and \(Q_{3}^{\prime} / \sigma^{2}\) have chi-square distributions. Prove that \(Q_{3}^{\prime}\) and \(Q_{4}^{\prime}\) are independent, and hence \(Q_{4}^{\prime} / \sigma^{2}\) also has a chi-square distribution. If the likelihood ratio \(\Lambda\) is used to test \(H_{0}: \mu_{1}=\mu_{2}=\cdots=\mu_{b}=\mu, \mu\) unspecified and \(\sigma^{2}\) unknown against all possible alternatives, show that \(\Lambda \leq \lambda_{0}\) is equivalent to the computed \(F \geq c\), where $$ F=\frac{\left(\sum_{j=1}^{b} a_{j}-b\right) Q_{4}^{\prime}}{(b-1) Q_{3}^{\prime}} $$ What is the distribution of \(F\) when \(H_{0}\) is true?

Fit \(y=a+x\) to the data $$ \begin{array}{l|lll} \mathrm{x} & 0 & 1 & 2 \\ \hline \mathrm{y} & 1 & 3 & 4 \end{array} $$ by the method of least squares.

Let \(X_{1}, X_{2}, X_{3}\) be a random sample from the normal distribution \(N\left(0, \sigma^{2}\right)\). Are the quadratic forms \(X_{1}^{2}+3 X_{1} X_{2}+X_{2}^{2}+X_{1} X_{3}+X_{3}^{2}\) and \(X_{1}^{2}-2 X_{1} X_{2}+\frac{2}{3} X_{2}^{2}-\) \(2 X_{1} X_{2}-X_{3}^{2}\) independent or dependent?

Given the following observations associated with a two-way classification with \(a=3\) and \(b=4\), compute the \(F\) -statistic used to test the equality of the column means \(\left(\beta_{1}=\beta_{2}=\beta_{3}=\beta_{4}=0\right)\) and the equality of the row means \(\left(\alpha_{1}=\alpha_{2}=\alpha_{3}=0\right)\), respectively. $$ \begin{array}{ccccc} \hline \text { Row/Column } & 1 & 2 & 3 & 4 \\ \hline 1 & 3.1 & 4.2 & 2.7 & 4.9 \\ 2 & 2.7 & 2.9 & 1.8 & 3.0 \\ 3 & 4.0 & 4.6 & 3.0 & 3.9 \\ \hline \end{array} $$

Using the notation of Section 9.2, assume that the means \(\mu_{j}\) satisfy a linear function of \(j\), nanely \(\mu_{j}=c+d[j-(b+1) / 2] .\) Let independent random samples of size \(a\) be taken from the \(b\) normal distributions having means \(\mu_{1}, \mu_{2}, \ldots, \mu_{b}\), respectively, and common unknown variance \(\sigma^{2}\). (a) Show that the maximum likelihood estimators of \(c\) and \(d\) are, respectively, \(\hat{c}=\bar{X}_{. .}\) and $$ \hat{d}=\frac{\sum_{j=1}^{b}[j-(b-1) / 2]\left(\bar{X}_{. j}-\bar{X}_{. .}\right)}{\sum_{j=1}^{b}[j-(b+1) / 2]^{2}} $$ (b) Show that $$ \begin{aligned} \sum_{i=1}^{a} \sum_{j=1}^{b}\left(X_{i j}-\bar{X}_{. .}\right)^{2}=\sum_{i=1}^{a} \sum_{j=1}^{b}\left[X_{i j}-\bar{X}_{. .}-\hat{d}\left(j-\frac{b+1}{2}\right)\right]^{2} \\ &+\hat{d}^{2} \sum_{j=1}^{b} a\left(j-\frac{b+1}{2}\right)^{2} \end{aligned} $$ (c) Argue that the two terms in the right-hand member of Part (b), once divided by \(\sigma^{2}\), are independent random variables with \(\chi^{2}\) distributions provided that \(d=0\) (d) What \(F\) -statistic would be used to test the equality of the means, that is, \(H_{0}: d=0 ?\)
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