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

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?

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.

Let the independent random variables \(Y_{1}, Y_{2}, \ldots, Y_{n}\) have, respectively, the probability density functions \(N\left(\beta x_{i}, \gamma^{2} x_{i}^{2}\right), i=1,2, \ldots, n\), where the given numbers \(x_{1}, x_{2}, \ldots, x_{n}\) are not all equal and no one is zero. Find the maximum likelihood estimators of \(\beta\) and \(\gamma^{2}\).

Let \(X_{1}, X_{2}, X_{3}, X_{4}\) be a random sample of size \(n=4\) from the normal distribution \(N(0,1) .\) Show that \(\sum_{i=1}^{4}\left(X_{i}-\bar{X}\right)^{2}\) equals $$ \frac{\left(X_{1}-X_{2}\right)^{2}}{2}+\frac{\left[X_{3}-\left(X_{1}+X_{2}\right) / 2\right]^{2}}{3 / 2}+\frac{\left[X_{4}-\left(X_{1}+X_{2}+X_{3}\right) / 3\right]^{2}}{4 / 3} $$ and argue that these three terms are independent, each with a chi-square distribution with 1 degree of freedom.

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 \(\alpha=0.05\). $$ \begin{array}{llllll} \text { Brand A: } & 38.7 & 39.2 & 40.1 & 38.9 & \\ \text { Brand B: } & 41.9 & 42.3 & 41.3 & & \\ \text { Brand C: } & 40.8 & 41.2 & 39.5 & 38.9 & 40.3 \end{array} $$

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