Question

annual low 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).
\\(\hat{y}=\\)

x	y
5	10.8
6	11.62
7	13.04
8	15.46
9	14.78
10	16.1
11	15.92
12	19.14

question help: video message instructor

Explanation:

Step1: Calculate the means of \(x\) and \(y\)

\(\bar{x}=\frac{4 + 5+6+7+8+9+10+11+12}{9}=\frac{72}{9} = 8\)
\(\bar{y}=\frac{10.28+10.8+11.62+13.04+15.46+14.78+16.1+15.92+19.14}{9}=\frac{127.14}{9}\approx14.1267\)

Step2: Calculate the numerator and denominator for the slope \(b_1\)

The numerator \(S_{xy}=\sum_{i = 1}^{n}(x_i-\bar{x})(y_i - \bar{y})\)
\((4 - 8)(10.28-14.1267)+(5 - 8)(10.8 - 14.1267)+(6 - 8)(11.62-14.1267)+(7 - 8)(13.04-14.1267)+(8 - 8)(15.46-14.1267)+(9 - 8)(14.78-14.1267)+(10 - 8)(16.1-14.1267)+(11 - 8)(15.92-14.1267)+(12 - 8)(19.14-14.1267)\)
\(=(- 4)(-3.8467)+(-3)(-3.3267)+(-2)(-2.5067)+(-1)(-1.0867)+0\times1.3333 + 1\times0.6533+2\times1.9733+3\times1.7933+4\times5.0133\)
\(=15.3868+9.9801+5.0134 + 1.0867+0+0.6533+3.9466+5.3799+20.0532\)
\(=61.499\)
The denominator \(S_{xx}=\sum_{i=1}^{n}(x_i-\bar{x})^2\)
\((4 - 8)^2+(5 - 8)^2+(6 - 8)^2+(7 - 8)^2+(8 - 8)^2+(9 - 8)^2+(10 - 8)^2+(11 - 8)^2+(12 - 8)^2\)
\(=16 + 9+4 + 1+0+1+4+9+16\)
\(=60\)
\(b_1=\frac{S_{xy}}{S_{xx}}=\frac{61.499}{60}\approx1.0250\)

Step3: Calculate the intercept \(b_0\)

\(b_0=\bar{y}-b_1\bar{x}\)
\(b_0 = 14.1267-1.0250\times8\)
\(b_0=14.1267 - 8.2\)
\(b_0 = 5.9267\)

Answer:

\(5.9267+1.0250x\)