
How do you find the y-intercept of the least square regression line for the data set $\left( 1,8 \right)\left( 2,7 \right)\left( 3,5 \right)$?
Answer
516.6k+ views
Hint: To obtain the y-intercept of the least squares regression line for the given data set use the formula $\hat{y}={{\hat{\beta }}_{1}}x+{{\hat{\beta }}_{0}}$ where ${{\hat{\beta }}_{1}}$ and ${{\hat{\beta }}_{0}}$ are the slope and y-intercept respectively. Find the value of the y-intercept by using the formula ${{\hat{\beta }}_{0}}=\dfrac{S{{S}_{xy}}}{S{{S}_{xx}}}$ and simplify it to get the desired answer.
Complete step by step answer:
The data set given is:
$\left( 1,8 \right)\left( 2,7 \right)\left( 3,5 \right)$
Firstly we will form a table containing x-value and y-values as:
Next we will find the value of ${{x}^{2}}$ and $xy$ for each value in the above set and form summation for each column as:
Now as we know y-intercept formula is:
${{\hat{\beta }}_{0}}=\bar{y}-{{\hat{\beta }}_{1}}\bar{x}$……$\left( 1 \right)$
Where,
${{\hat{\beta }}_{1}}=\dfrac{S{{S}_{xy}}}{S{{S}_{xx}}}$……$\left( 2 \right)$
$\bar{y}=\dfrac{\sum{y}}{n}$……$\left( 3 \right)$
$\bar{x}=\dfrac{\sum{x}}{n}$…..$\left( 4 \right)$
$S{{S}_{xx}}=\sum{{{x}^{2}}}-\dfrac{1}{n}{{\left( \sum{x} \right)}^{2}}$……$\left( 5 \right)$
$S{{S}_{xy}}=\sum{xy}-\dfrac{1}{n}\left( \sum{x} \right)\left( \sum{y} \right)$……$\left( 6 \right)$
Next we will find the value of equation (3), (4), (5), (6) by substituting the valued from the table as:
For equation (3)
$\begin{align}
& \bar{y}=\dfrac{20}{3} \\
& \therefore \bar{y}=6.67 \\
\end{align}$
For equation (4)
$\begin{align}
& \bar{x}=\dfrac{6}{3} \\
& \therefore \bar{x}=2 \\
\end{align}$
For equation (5)
$\begin{align}
& S{{S}_{xx}}=14-\dfrac{1}{3}{{\left( 6 \right)}^{2}} \\
& \Rightarrow S{{S}_{xx}}=14-12 \\
& \therefore S{{S}_{xx}}=2 \\
\end{align}$
For equation (6) we have:
$\begin{align}
& S{{S}_{xy}}=37-\dfrac{1}{3}\times 6\times 20 \\
& \Rightarrow S{{S}_{xy}}=37-40 \\
& \therefore S{{S}_{xy}}=-3 \\
\end{align}$
Put the above obtain value in equation (2) we get,
$\begin{align}
& {{{\hat{\beta }}}_{1}}=\dfrac{-3}{2} \\
& \therefore {{{\hat{\beta }}}_{1}}=-1.5 \\
\end{align}$
Substitute all value in equation (1)
$\begin{align}
& {{{\hat{\beta }}}_{0}}=\dfrac{20}{3}-\left( -1.5 \right)\times 2 \\
& \Rightarrow {{{\hat{\beta }}}_{0}}=6.67+3 \\
& \therefore {{{\hat{\beta }}}_{0}}=9.67 \\
\end{align}$
Hence the y-intercept of the least square regression line for the data set $\left( 1,8 \right)\left( 2,7 \right)\left( 3,5 \right)$ is $9.67$
Note: A least square regression line is the line that fits the best for the given data. Its y-intercept and slope is calculated in a very different way from other methods. It is the line that makes the vertical distance from the data points to their regression line as small as possible. The best line is the one that minimizes the value of the variance of a given data.
Complete step by step answer:
The data set given is:
$\left( 1,8 \right)\left( 2,7 \right)\left( 3,5 \right)$
Firstly we will form a table containing x-value and y-values as:
| $x$ | $y$ |
| $1$ | $8$ |
| $2$ | $7$ |
| $3$ | $5$ |
Next we will find the value of ${{x}^{2}}$ and $xy$ for each value in the above set and form summation for each column as:
| $x$ | $y$ | ${{x}^{2}}$ | $xy$ |
| $1$ | $8$ | $1$ | $8$ |
| $2$ | $7$ | $4$ | $14$ |
| $3$ | $5$ | $9$ | $15$ |
| $\sum{x}=6$ | $\sum{y}=20$ | $\sum{{{x}^{2}}}=14$ | $\sum{xy}=37$ |
Now as we know y-intercept formula is:
${{\hat{\beta }}_{0}}=\bar{y}-{{\hat{\beta }}_{1}}\bar{x}$……$\left( 1 \right)$
Where,
${{\hat{\beta }}_{1}}=\dfrac{S{{S}_{xy}}}{S{{S}_{xx}}}$……$\left( 2 \right)$
$\bar{y}=\dfrac{\sum{y}}{n}$……$\left( 3 \right)$
$\bar{x}=\dfrac{\sum{x}}{n}$…..$\left( 4 \right)$
$S{{S}_{xx}}=\sum{{{x}^{2}}}-\dfrac{1}{n}{{\left( \sum{x} \right)}^{2}}$……$\left( 5 \right)$
$S{{S}_{xy}}=\sum{xy}-\dfrac{1}{n}\left( \sum{x} \right)\left( \sum{y} \right)$……$\left( 6 \right)$
Next we will find the value of equation (3), (4), (5), (6) by substituting the valued from the table as:
For equation (3)
$\begin{align}
& \bar{y}=\dfrac{20}{3} \\
& \therefore \bar{y}=6.67 \\
\end{align}$
For equation (4)
$\begin{align}
& \bar{x}=\dfrac{6}{3} \\
& \therefore \bar{x}=2 \\
\end{align}$
For equation (5)
$\begin{align}
& S{{S}_{xx}}=14-\dfrac{1}{3}{{\left( 6 \right)}^{2}} \\
& \Rightarrow S{{S}_{xx}}=14-12 \\
& \therefore S{{S}_{xx}}=2 \\
\end{align}$
For equation (6) we have:
$\begin{align}
& S{{S}_{xy}}=37-\dfrac{1}{3}\times 6\times 20 \\
& \Rightarrow S{{S}_{xy}}=37-40 \\
& \therefore S{{S}_{xy}}=-3 \\
\end{align}$
Put the above obtain value in equation (2) we get,
$\begin{align}
& {{{\hat{\beta }}}_{1}}=\dfrac{-3}{2} \\
& \therefore {{{\hat{\beta }}}_{1}}=-1.5 \\
\end{align}$
Substitute all value in equation (1)
$\begin{align}
& {{{\hat{\beta }}}_{0}}=\dfrac{20}{3}-\left( -1.5 \right)\times 2 \\
& \Rightarrow {{{\hat{\beta }}}_{0}}=6.67+3 \\
& \therefore {{{\hat{\beta }}}_{0}}=9.67 \\
\end{align}$
Hence the y-intercept of the least square regression line for the data set $\left( 1,8 \right)\left( 2,7 \right)\left( 3,5 \right)$ is $9.67$
Note: A least square regression line is the line that fits the best for the given data. Its y-intercept and slope is calculated in a very different way from other methods. It is the line that makes the vertical distance from the data points to their regression line as small as possible. The best line is the one that minimizes the value of the variance of a given data.
Recently Updated Pages
Master Class 12 Economics: Engaging Questions & Answers for Success

Master Class 12 Maths: Engaging Questions & Answers for Success

Master Class 12 Biology: Engaging Questions & Answers for Success

Master Class 12 Physics: Engaging Questions & Answers for Success

Master Class 8 Maths: Engaging Questions & Answers for Success

Class 8 Question and Answer - Your Ultimate Solutions Guide

Trending doubts
Which animal has three hearts class 11 biology CBSE

1 Quintal is equal to a 110 kg b 10 kg c 100kg d 1000 class 11 physics CBSE

Explain zero factorial class 11 maths CBSE

Mention the basic forces in nature class 11 physics CBSE

What is centripetal acceleration Derive the expression class 11 physics CBSE

Bond order ofO2 O2+ O2 and O22 is in order A O2 langle class 11 chemistry CBSE

