Question

annual high temperatures in a certain location have been tracked for several years. let x represent the year and y the high temperature. based on the data shown below, calculate the regression line (each value to four decimal places - be careful on where the slope/y - intercept is located in the equation below).
x y
3 11.4
4 13.6
5 13.5
6 12.9
7 14.5
8 13.3
\hat{y}=\square+\square x

Explanation:

Step1: Calculate means of X and Y

Let $n = 6$.
$\bar{x}=\frac{3 + 4+5+6+7+8}{6}=\frac{33}{6}=5.5$
$\bar{y}=\frac{11.4 + 13.6+13.5+12.9+14.5+13.3}{6}=\frac{79.2}{6}=13.2$

Step2: Calculate numerator and denominator for slope

$S_{xy}=\sum_{i = 1}^{n}(x_i-\bar{x})(y_i - \bar{y})$
$(3 - 5.5)(11.4-13.2)+(4 - 5.5)(13.6 - 13.2)+(5 - 5.5)(13.5 - 13.2)+(6 - 5.5)(12.9 - 13.2)+(7 - 5.5)(14.5 - 13.2)+(8 - 5.5)(13.3 - 13.2)$
$=(- 2.5)(-1.8)+(-1.5)(0.4)+(-0.5)(0.3)+(0.5)(-0.3)+(1.5)(1.3)+(2.5)(0.1)$
$=4.5-0.6 - 0.15-0.15 + 1.95+0.25=5.8$

$S_{xx}=\sum_{i = 1}^{n}(x_i-\bar{x})^2$
$(3 - 5.5)^2+(4 - 5.5)^2+(5 - 5.5)^2+(6 - 5.5)^2+(7 - 5.5)^2+(8 - 5.5)^2$
$=(-2.5)^2+(-1.5)^2+(-0.5)^2+(0.5)^2+(1.5)^2+(2.5)^2$
$=6.25 + 2.25+0.25+0.25+2.25+6.25 = 17.5$

The slope $b_1=\frac{S_{xy}}{S_{xx}}=\frac{5.8}{17.5}\approx0.3314$

Step3: Calculate y - intercept

The y - intercept $b_0=\bar{y}-b_1\bar{x}$
$b_0 = 13.2-0.3314\times5.5$
$b_0=13.2 - 1.8227=11.3773$

Answer:

$y = 11.3773+0.3314x$