Question

based on the data shown below, calculate the regression line (each value to at least two decimal places) y = x +

x	y
4	9.48
5	9.85
6	13.22
7	10.39
8	11.96
9	13.33

Explanation:

Step1: Calculate means of x and y

Let $x_i$ and $y_i$ be the data - points.
$n = 6$
$\bar{x}=\frac{4 + 5+6+7+8+9}{6}=\frac{39}{6}=6.5$
$\bar{y}=\frac{9.48 + 9.85+13.22+10.39+11.96+13.33}{6}=\frac{68.23}{6}\approx11.37$

Step2: Calculate numerator and denominator for slope (b)

$\sum_{i = 1}^{n}(x_i-\bar{x})(y_i - \bar{y})=(4 - 6.5)(9.48-11.37)+(5 - 6.5)(9.85 - 11.37)+(6 - 6.5)(13.22-11.37)+(7 - 6.5)(10.39-11.37)+(8 - 6.5)(11.96-11.37)+(9 - 6.5)(13.33-11.37)$
$=(- 2.5)(-1.89)+(-1.5)(-1.52)+(-0.5)(1.85)+(0.5)(-0.98)+(1.5)(0.59)+(2.5)(1.96)$
$=4.725 + 2.28-0.925-0.49+0.885 + 4.9$
$=11.475$
$\sum_{i = 1}^{n}(x_i-\bar{x})^2=(4 - 6.5)^2+(5 - 6.5)^2+(6 - 6.5)^2+(7 - 6.5)^2+(8 - 6.5)^2+(9 - 6.5)^2$
$=(-2.5)^2+(-1.5)^2+(-0.5)^2+(0.5)^2+(1.5)^2+(2.5)^2$
$=6.25+2.25 + 0.25+0.25+2.25+6.25$
$=17.5$
$b=\frac{\sum_{i = 1}^{n}(x_i-\bar{x})(y_i - \bar{y})}{\sum_{i = 1}^{n}(x_i-\bar{x})^2}=\frac{11.475}{17.5}\approx0.66$

Step3: Calculate intercept (a)

$a=\bar{y}-b\bar{x}$
$a = 11.37-0.66\times6.5$
$a = 11.37 - 4.29$
$a = 7.08$

Answer:

$y = 0.66x+7.08$