The linear regression for a given data set has the form
[tex]y=a+bx[/tex]
where the values a and b can be solved using the equation
[tex]\begin{gathered} a=\frac{(\sum y)(\sum x^2)-(\sum x)(\sum xy)}{n(\sum x^2)-(\sum x)^2} \\ b=\frac{n(\sum xy)-(\sum x)(\sum y)}{n(\sum x^2)-(\sum x)^2} \end{gathered}[/tex]
Based on the given data set, we have n equals 5. We will solve for the values of the summation first. We have the following
[tex]\begin{gathered} \sum y=4+4+6+6+8=28 \\ \sum x=1+3+5+7+9=25 \\ \sum xy=(1\cdot4)+(3\cdot4)+(5\cdot6)+(7\cdot6)+(8\cdot9)=160 \\ \sum x^2=1^2+3^2+5^2+7^2+9^2=165^{} \\ (\sum x)^2=25^2=625 \end{gathered}[/tex]
Using these values to compute for the values of a and b, we get
[tex]\begin{gathered} a=\frac{(28\cdot165)-(25\cdot160)}{5(165)-625}=\frac{31}{10}=3.1 \\ b=\frac{5(160)-(28\cdot25)}{5(165)-625}=\frac{1}{2}=0.5 \end{gathered}[/tex]
Take note that the problem wants us to reduce the numbers to the nearest tenth. Hence, the linear regression for the given data set is written as
[tex]y=3.1+0.5x[/tex]
