
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
517.2k+ 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.
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