Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

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.

Short Answer

Expert verified
The line that best fits the given data by the method of least squares is \(y = 1 + 1.67\).

Step by step solution

01

Calculate Summations

First, calculate the summations needed in the following calculations, which are \(\sum x\) and \(\sum y\). Here, \(\sum x = 0 + 1 + 2 = 3\) and \(\sum y = 1 + 3 + 4 = 8\)
02

Find the Mean Values

Find the mean values of x and y with the formulas \(\overline{x} =\frac{\sum x}{n}\) and \(\overline{y} =\frac{\sum y}{n}\). In this case, \(\overline{x} =\frac{3}{3} = 1\) and \(\overline{y} = \frac{8}{3} \approx 2.67\)
03

Determine the Value of 'a'

As the given line is \(y = a + x\), this can be re-written in slope-intercept form as \(y = mx + b\), with \(m=1\) and \(b=a\). By comparing to the general line equation, determine 'a' as the y-intercept which is the mean value of y minus the mean value of x multiplied by the slope. Thus, \(a =\overline{y} - m\cdot\overline{x}\), i.e., \(a = 2.67 - 1\cdot 1 = 1.67\)

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Suppose \(\boldsymbol{Y}\) is an \(n \times 1\) random vector, \(\boldsymbol{X}\) is an \(n \times p\) matrix of known constants of rank \(p\), and \(\beta\) is a \(p \times 1\) vector of regression coefficients. Let \(\boldsymbol{Y}\) have a \(N\left(\boldsymbol{X} \boldsymbol{\beta}, \sigma^{2} \boldsymbol{I}\right)\) distribution. Discuss the joint pdf of \(\hat{\boldsymbol{\beta}}=\left(\boldsymbol{X}^{\prime} \boldsymbol{X}\right)^{-1} \boldsymbol{X}^{\prime} \boldsymbol{Y}\) and \(\boldsymbol{Y}^{\prime}\left[\boldsymbol{I}-\boldsymbol{X}\left(\boldsymbol{X}^{\prime} \boldsymbol{X}\right)^{-1} \boldsymbol{X}^{\prime}\right] \boldsymbol{Y} / \sigma^{2}\)

Let \(Q_{1}\) and \(Q_{2}\) be two nonnegative quadratic forms in the observations of a random sample from a distribution which is \(N\left(0, \sigma^{2}\right) .\) Show that another quadratic form \(Q\) is independent of \(Q_{1}+Q_{2}\) if and only if \(Q\) is independent of each of \(Q_{1}\) and \(Q_{2}\) Hint: \(\quad\) Consider the orthogonal transformation that diagonalizes the matrix of \(Q_{1}+Q_{2}\). After this transformation, what are the forms of the matrices \(Q, Q_{1}\) and \(Q_{2}\) if \(Q\) and \(Q_{1}+Q_{2}\) are independent?

In Exercise 9.2.1, show that the linear functions \(X_{i j}-X_{. j}\) and \(X_{. j}-X\).. are uncorrelated. Hint: Recall the definition of \(\bar{X}_{. j}\) and \(\bar{X}_{. .}\) and, without loss of generality, we can let \(E\left(X_{i j}\right)=0\) for all \(i, j\)

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

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?

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free