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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

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.

The driver of a diesel-powered automobile decided to test the quality of three types of diesel fuel sold in the area based on mpg. Test the null hypothesis that the three means are equal using the following data. Make the usual assumptions and take \(\alpha=0.05\). $$ \begin{array}{llllll} \text { Brand A: } & 38.7 & 39.2 & 40.1 & 38.9 & \\ \text { Brand B: } & 41.9 & 42.3 & 41.3 & & \\ \text { Brand C: } & 40.8 & 41.2 & 39.5 & 38.9 & 40.3 \end{array} $$

Here \(Q_{1}\) and \(Q_{2}\) are quadratic forms in observations of a random sample from \(N(0,1) .\) If \(Q_{1}\) and \(Q_{2}\) are independent and if \(Q_{1}+Q_{2}\) has a chi-square distribution, prove that \(Q_{1}\) and \(Q_{2}\) are chi-square variables.

Let the independent normal random variables \(Y_{1}, Y_{2}, \ldots, Y_{n}\) have, respectively, the probability density functions \(N\left(\mu, \gamma^{2} x_{i}^{2}\right), i=1,2, \ldots, n\), where the given \(x_{1}, x_{2}, \ldots, x_{n}\) are not all equal and no one of which is zero. Discuss the test of the hypothesis \(H_{0}: \gamma=1, \mu\) unspecified, against all alternatives \(H_{1}: \gamma \neq 1, \mu\) unspecified.

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?

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