Identify lines of best fit - HiSet: High School Equivalency Test: Math

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Question

Determine the linear regression line for the following paired data:

$\begin{matrix} \underline{x }&\underline{y} \ 1 & 3.5 \ 2 & 4.7\ 3 & 6.0\ 4 & 7.8 \end{matrix}$

Answer

The formula for a regression line for a set of paired data is the line of the equation

$y =b_{1}x+ b_{0}$ ,

where

$b_{1} = \frac{n (\sum xy)-(\sum x)(\sum y)}{n(\sum x^{2})-(\sum x)^{2}}$

and

$b_{0} = \overline{y} - b_{1}\overline{x}$ .

, the number of pairs.

$\sum x$ is the sum of the values:

$\sum x = 1 + 2 + 3 + 4 = 10$

$\sum y$ is the sum of the values:

$\sum y = 3.5 + 4.7 + 6.0 + 7.8 = 22.0$

$\sum x^{2}$ is the sum of the squares of the values:

$\sum x^{2} = 1^{2} + 2^{2} + 3 ^{2} + 4 ^{2} = 1 + 4 + 9 + 16 = 30$

$\sum xy$ is the sum of the products:

$\sum xy = 1(3.5 )+ 2(4.7)+ 3(6.0)+ 4 (7.8) = 3.5 + 9.4 + 18.0 + 31.2 = 62.1$

Substituting:

$b_{1} = \frac{4 (62.1)-(10)( 22.0)}{4(30)-(10)^{2}}$

$= \frac{248.4-220}{120-100}$

$= \frac{28.4}{20}$

$\overline{x}$ and $\overline{y}$ are the arithmetic means of the and values, respectively - the sums divided by the number of values:

$\overline{x} = \frac{\sum x}{n} = \frac{10}{4} = 2.5$

$\overline{y} = \frac{\sum y}{n} = \frac{22}{4} = 5.5$

$b_{0} = \overline{y} - b_{1}\overline{x}$

$b_{0} =5.5 - 1.42 (2.5)$

The equation of the linear regression line is .

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Question

Determine the linear regression line for the following paired data:

$\begin{matrix} \underline{x }&\underline{y} \ 1 &10.0 \ 2 &8.7\ 3 & 7.1 \ 4 & 5.0 \end{matrix}$

Answer

The formula for a regression line for a set of paired data is the line of the equation

$y =b_{1}x+ b_{0}$ ,

where

$b_{1} = \frac{n (\sum xy)-(\sum x)(\sum y)}{n(\sum x^{2})-(\sum x)^{2}}$

and

$b_{0} = \overline{y} - b_{1}\overline{x}$ .

, the number of pairs.

$\sum x$ is the sum of the values:

$\sum x = 1 + 2 + 3 + 4 = 10$

$\sum y$ is the sum of the values:

$\sum y= 10.0+8.7+ 7.1 +5.0 = 30.8$

$\sum x^{2}$ is the sum of the squares of the values:

$\sum x^{2} = 1^{2} + 2^{2} + 3 ^{2} + 4 ^{2} = 1 + 4 + 9 + 16 = 30$

$\sum xy$ is the sum of the products:

$\sum xy = 1(10.0 )+ 2(8.7)+ 3(7.1)+ 4 (5.0) = 10.0+ 17.4+21.3 + 20.0= 68.7$

Substituting:

$b_{1} = \frac{4 (68.7)-(10)( 30.8)}{4(30)-(10)^{2}}$

$= \frac{274.8 - 308}{120-100 }$

$= \frac{-33.2}{20 }$

$\overline{x}$ and $\overline{y}$ are the arithmetic means of the and values, respectively - the sums divided by the number of values:

$\overline{x} = \frac{\sum x}{n} = \frac{10}{4} = 2.5$

$\overline{y} = \frac{\sum y}{n} = \frac{30.8}{4} = 7.7$

$b_{0} = \overline{y} - b_{1}\overline{x}$

The linear regression line is the line of the equation .

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