Question

linear regression
use linear regression to find the equation for the linear function that best fits this data. round both numbers to two decimal places. write your final answer in a form of an equation y = mx + b

x	1	2	3	4	5	6
y	105	124	136	157	179	204

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Explanation:

Step1: Calculate the means of x and y

$n = 6$
$\bar{x}=\frac{1 + 2+3+4+5+6}{6}=\frac{21}{6}=3.5$
$\bar{y}=\frac{105 + 124+136+157+179+204}{6}=\frac{905}{6}\approx150.83$

Step2: Calculate the numerator and denominator for slope m

$\sum_{i = 1}^{n}(x_i-\bar{x})(y_i - \bar{y})=(1 - 3.5)(105-150.83)+(2 - 3.5)(124 - 150.83)+(3 - 3.5)(136-150.83)+(4 - 3.5)(157-150.83)+(5 - 3.5)(179-150.83)+(6 - 3.5)(204-150.83)$
$=(- 2.5)(-45.83)+(-1.5)(-26.83)+(-0.5)(-14.83)+(0.5)(6.17)+(1.5)(28.17)+(2.5)(53.17)$
$=114.575 + 40.245+7.415 + 3.085+42.255+132.925$
$=340.5$
$\sum_{i = 1}^{n}(x_i-\bar{x})^2=(1 - 3.5)^2+(2 - 3.5)^2+(3 - 3.5)^2+(4 - 3.5)^2+(5 - 3.5)^2+(6 - 3.5)^2$
$=(-2.5)^2+(-1.5)^2+(-0.5)^2+(0.5)^2+(1.5)^2+(2.5)^2$
$=6.25+2.25 + 0.25+0.25+2.25+6.25$
$=17.5$
$m=\frac{\sum_{i = 1}^{n}(x_i-\bar{x})(y_i - \bar{y})}{\sum_{i = 1}^{n}(x_i-\bar{x})^2}=\frac{340.5}{17.5}\approx19.46$

Step3: Calculate the y - intercept b

$b=\bar{y}-m\bar{x}$
$b = 150.83-19.46\times3.5$
$b=150.83 - 68.11$
$b = 82.72$

Answer:

$y = 19.46x+82.72$