Question

what is the linear regression of the data?
what is the linear regression of the data?
y = square
(use integers or decimals for any numbers in the expression. round to the nearest tenth as needed.)

Explanation:

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

The \(x\) - values are \(x_1 = 1,x_2 = 3,x_3 = 5,x_4 = 7,x_5 = 9\).
The mean of \(x\), \(\bar{x}=\frac{1 + 3+5 + 7+9}{5}=\frac{25}{5}=5\)
The \(y\) - values are \(y_1 = 2,y_2 = 2,y_3 = 3,y_4 = 5,y_5 = 6\).
The mean of \(y\), \(\bar{y}=\frac{2 + 2+3 + 5+6}{5}=\frac{18}{5}=3.6\)

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

The formula for the slope \(b=\frac{\sum_{i = 1}^{n}(x_i-\bar{x})(y_i - \bar{y})}{\sum_{i = 1}^{n}(x_i-\bar{x})^2}\)
\((x_1-\bar{x})(y_1 - \bar{y})=(1 - 5)(2 - 3.6)=(-4)\times(-1.6)=6.4\)
\((x_2-\bar{x})(y_2 - \bar{y})=(3 - 5)(2 - 3.6)=(-2)\times(-1.6)=3.2\)
\((x_3-\bar{x})(y_3 - \bar{y})=(5 - 5)(3 - 3.6)=0\times(-0.6)=0\)
\((x_4-\bar{x})(y_4 - \bar{y})=(7 - 5)(5 - 3.6)=2\times1.4 = 2.8\)
\((x_5-\bar{x})(y_5 - \bar{y})=(9 - 5)(6 - 3.6)=4\times2.4 = 9.6\)
\(\sum_{i = 1}^{5}(x_i-\bar{x})(y_i - \bar{y})=6.4+3.2 + 0+2.8+9.6=22\)

\((x_1-\bar{x})^2=(1 - 5)^2=16\)
\((x_2-\bar{x})^2=(3 - 5)^2 = 4\)
\((x_3-\bar{x})^2=(5 - 5)^2=0\)
\((x_4-\bar{x})^2=(7 - 5)^2=4\)
\((x_5-\bar{x})^2=(9 - 5)^2=16\)
\(\sum_{i = 1}^{5}(x_i-\bar{x})^2=16 + 4+0+4+16=40\)

The slope \(b=\frac{22}{40}=0.55\)

Step3: Calculate the intercept \(a\)

The formula for the intercept \(a=\bar{y}-b\bar{x}\)
\(a = 3.6-0.55\times5=3.6 - 2.75=0.85\)

Answer:

\(y = 0.9x+0.9\) (rounded to the nearest tenth)