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
1	22.6
2	23.2
3	24.9
4	25.9
5	27.2
6	29

\hat{y}= + x

Explanation:

Step1: Calculate the means of X and Y

$n = 6$
$\bar{x}=\frac{1 + 2+3+4+5+6}{6}=\frac{21}{6}=3.5$
$\bar{y}=\frac{22.6 + 23.2+24.9+25.9+27.2+29}{6}=\frac{152.8}{6}\approx25.4667$

Step2: Calculate the numerator and denominator for the slope

$\sum_{i = 1}^{n}(x_i-\bar{x})(y_i - \bar{y})=(1 - 3.5)(22.6-25.4667)+(2 - 3.5)(23.2 - 25.4667)+(3 - 3.5)(24.9-25.4667)+(4 - 3.5)(25.9-25.4667)+(5 - 3.5)(27.2-25.4667)+(6 - 3.5)(29-25.4667)$
$=(- 2.5)(-2.8667)+(-1.5)(-2.2667)+(-0.5)(-0.5667)+(0.5)(0.4333)+(1.5)(1.7333)+(2.5)(3.5333)$
$=7.16675 + 3.40005+0.28335 + 0.21665+2.60005+8.83325=22.5$
$\sum_{i = 1}^{n}(x_i-\bar{x})^2=(1 - 3.5)^2+(2 - 3.5)^2+(3 - 3.5)^2+(4 - 3.5)^2+(5 - 3.5)^2+(6 - 3.5)^2$
$=(-2.5)^2+(-1.5)^2+(-0.5)^2+(0.5)^2+(1.5)^2+(2.5)^2$
$=6.25+2.25 + 0.25+0.25+2.25+6.25 = 17.5$
The slope $b=\frac{\sum_{i = 1}^{n}(x_i-\bar{x})(y_i - \bar{y})}{\sum_{i = 1}^{n}(x_i-\bar{x})^2}=\frac{22.5}{17.5}\approx1.2857$

Step3: Calculate the y - intercept

$a=\bar{y}-b\bar{x}=25.4667-1.2857\times3.5$
$=25.4667 - 4.50995=20.95675\approx20.9568$

Answer:

$20.9568+1.2857x$