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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 \(A\) be the real symmetric matrix of a quadratic form \(Q\) in the observations of a random sample of size \(n\) from a distribution which is \(N\left(0, \sigma^{2}\right)\). Given that \(Q\) and the mean \(\bar{X}\) of the sample are independent, what can be said of the elements of each row (column) of \(\boldsymbol{A}\) ? Hint: Are \(Q\) and \(X^{2}\) independent?

Let \(\mu_{1}, \mu_{2}, \mu_{3}\) be, respectively, the means of three normal distributions with a common but unknown variance \(\sigma^{2}\). In order to test, at the \(\alpha=5\) percent significance level, the hypothesis \(H_{0}: \mu_{1}=\mu_{2}=\mu_{3}\) against all possible alternative hypotheses, we take an independent random sample of size 4 from each of these distributions. Determine whether we accept or reject \(H_{0}\) if the observed values from these three distributions are, respectively, $$ \begin{array}{lrrrr} X_{1}: & 5 & 9 & 6 & 8 \\ X_{2}: & 11 & 13 & 10 & 12 \\ X_{3}: & 10 & 6 & 9 & 9 \end{array} $$

Show that \(\sum_{i=1}^{n}\left[Y_{i}-\alpha-\beta\left(x_{i}-\bar{x}\right)\right]^{2}=n(\hat{\alpha}-\alpha)^{2}+(\hat{\beta}-\beta)^{2} \sum_{i=1}^{n}\left(x_{i}-\bar{x}\right)^{2}+\sum_{i=1}^{n}\left[Y_{i}-\hat{\alpha}-\hat{\beta}\left(x_{i}-\bar{x}\right)\right]^{2} .\)

Using the notation of this section, assume that the means satisfy the condition that \(\mu=\mu_{1}+(b-1) d=\mu_{2}-d=\mu_{3}-d=\cdots=\mu_{b}-d .\) That is, the last \(b-1\) means are equal but differ from the first mean \(\mu_{1}\), provided that \(d \neq 0\). Let independent random samples of size \(a\) be taken from the \(b\) normal distributions with common unknown variance \(\sigma^{2}\). (a) Show that the maximum likelihood estimators of \(\mu\) and \(d\) are \(\hat{\mu}=\bar{X} . .\) and $$ \hat{d}=\frac{\sum_{j=2}^{b} \bar{X}_{. j} /(b-1)-\bar{X}_{.1}}{b} $$ (b) Using Exercise \(9.1 .3\), find \(Q_{6}\) and \(Q_{7}=c \hat{d}^{2}\) so that, when \(d=0, Q_{7} / \sigma^{2}\) is \(\chi^{2}(1)\) and $$ \sum_{i=1}^{a} \sum_{j=1}^{b}\left(X_{i j}-\bar{X}_{n}\right)^{2}=Q_{3}+Q_{6}+Q_{7} $$ (c) Argue that the three terms in the right-hand member of Part (b), once divided by \(\sigma^{2}\), are independent random variables with chi-square distributions, provided that \(d=0\). (d) The ratio \(Q_{7} /\left(Q_{3}+Q_{6}\right)\) times what constant has an \(F\) -distribution, provided that \(d=0\) ? Note that this \(F\) is really the square of the two-sample \(T\) used to test the equality of the mean of the first distribution and the common mean of the other distributions, in which the last \(b-1\) samples are combined into one.

Let \(X_{i j k}, i=1, \ldots, a ; j=1, \ldots, b, k=1, \ldots, c\), be a random sample of size \(n=a b c\) from a normal distribution \(N\left(\mu, \sigma^{2}\right) .\) Let \(\bar{X}_{\ldots}=\sum_{k=1}^{c} \sum_{j=1}^{b} \sum_{i=1}^{a} X_{i j k} / n\) and $$ \begin{aligned} \bar{X}_{i_{r}} &=\sum_{k=1}^{c} \sum_{j=1}^{b} X_{i j k} / b c . \text { Prove that } \\ & \sum_{i=1}^{a} \sum_{j=1}^{b} \sum_{k=1}^{c}\left(X_{i j k}-\bar{X}_{\ldots}\right)^{2}=\sum_{i=1}^{a} \sum_{j=1}^{b} \sum_{k=1}^{c}\left(X_{i j k}-\bar{X}_{i . .}\right)^{2}+b c \sum_{i=1}^{a}\left(\bar{X}_{i .}-\bar{X}_{\cdots}\right)^{2} \end{aligned} $$ Show that \(\sum_{i=1}^{a} \sum_{j=1}^{b} \sum_{k=1}^{c}\left(X_{i j k}-\bar{X}_{i . .}\right)^{2} / \sigma^{2}\) has a chi-square distribution with \(a(b c-1)\) degrees of freedom. Prove that the two terms in the right-hand member are independent. What, then, is the distribution of \(b c \sum_{i=1}^{a}\left(\bar{X}_{i .}-\bar{X}_{\ldots}\right)^{2} / \sigma^{2} ?\) Furthermore, let \(X_{. j .}=\sum_{k=1}^{c} \sum_{i=1}^{a} X_{i j k} / a c\) and \(\bar{X}_{i j .}=\sum_{k=1}^{c} X_{i j k} / c .\) Show that $$ \begin{aligned} \sum_{i=1}^{a} \sum_{j=1}^{b} \sum_{k=1}^{c}\left(X_{i j k}-\bar{X}_{\cdots}\right)^{2}=& \sum_{i=1}^{a} \sum_{j=1}^{b} \sum_{k=1}^{c}\left(X_{i j k}-\bar{X}_{i j .}\right)^{2} \\ &+b c \sum_{i=1}^{a}\left(\bar{X}_{i_{n}}-\bar{X}_{\ldots}\right)^{2}+a c \sum_{j=1}^{b}\left(\bar{X}_{. j}-\bar{X}_{\ldots}\right)^{2} \\ &+c \sum_{i=1}^{a} \sum_{j=1}^{b}\left(\bar{X}_{i j .}-\bar{X}_{i .}-X_{. j .}+X_{\ldots}\right) \end{aligned} $$ Prove that the four terms in the right-hand member, when divided by \(\sigma^{2}\), are independent chi-square variables with \(a b(c-1), a-1, b-1\), and \((a-1)(b-1)\) degrees of freedom, respectively.

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