Question

linear regression and correlation. x y 3 -1.75 4 -1.5 5 -6.55 6 -7.8 7 -6.65 8 -3.3 9 -16.65 10 -16.6 compute the equation of the linear regression line in the form y = mx + b, where m is the slope and b is the intercept. use at least 3 decimal places. (round if necessary) y = x + compute the correlation coefficient for this data set. use at least 3 decimal places. (round if necessary) r = > next question

Explanation:

Step1: Calculate sums

Let $n = 8$.
Calculate $\sum x=\sum_{i = 1}^{8}x_i=3 + 4+5+6+7+8+9+10 = 52$.
Calculate $\sum y=\sum_{i = 1}^{8}y_i=-1.75-1.5 - 6.55-7.8-6.65-3.3-16.65-16.6=-50.8$.
Calculate $\sum x^2=\sum_{i = 1}^{8}x_i^2=3^2 + 4^2+5^2+6^2+7^2+8^2+9^2+10^2=380$.
Calculate $\sum y^2=\sum_{i = 1}^{8}y_i^2=(-1.75)^2+(-1.5)^2+(-6.55)^2+(-7.8)^2+(-6.65)^2+(-3.3)^2+(-16.65)^2+(-16.6)^2 = 687.93$.
Calculate $\sum xy=\sum_{i = 1}^{8}x_iy_i=3\times(-1.75)+4\times(-1.5)+5\times(-6.55)+6\times(-7.8)+7\times(-6.65)+8\times(-3.3)+9\times(-16.65)+10\times(-16.6)=-439.2$.

Step2: Calculate slope $m$

The formula for the slope $m$ of the regression - line is $m=\frac{n\sum xy-\sum x\sum y}{n\sum x^2-(\sum x)^2}$.
Substitute the values:

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Step3: Calculate intercept $b$

The formula for the intercept $b$ is $b=\bar{y}-m\bar{x}$, where $\bar{x}=\frac{\sum x}{n}=\frac{52}{8}=6.5$ and $\bar{y}=\frac{\sum y}{n}=\frac{-50.8}{8}=-6.35$.

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Step4: Calculate correlation coefficient $r$

The formula for the correlation coefficient $r$ is $r=\frac{n\sum xy-\sum x\sum y}{\sqrt{(n\sum x^2 - (\sum x)^2)(n\sum y^2-(\sum y)^2)}}$.
We already know $n\sum xy-\sum x\sum y=-872$, $n\sum x^2-(\sum x)^2 = 336$, and

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Answer:

$y=-2.595x + 10.518$
$r=-0.880$