A Linear regression line has an equation in the form,
y= ax+b
where Y= Dependent variable
X= Independent variable
a= The intercept
b= The slope of the line.
So we are to get the summation of all the tables formulated.
[tex]\begin{gathered} \sum ^{}_{}x=\text{ 15} \\ \sum ^{}_{}y=797 \\ \sum ^{}_{}xy=2464 \\ \sum ^{}_{}x^2=\text{ }55 \\ \sum ^{}_{}y^2=\text{ 119875} \end{gathered}[/tex]
We are to solve for the mean of x, mean of y, a and b.
[tex]\begin{gathered} \bar{x}=\text{ }\frac{\sum ^{}_{}x}{n} \\ \bar{y}=\text{ }\frac{\sum ^{}_{}y}{n} \\ a=\text{ }\frac{n\sum ^{}_{}xy-(\sum ^{}_{}x)(\sum ^{}_{}y)}{n\sum ^{}_{}x^2-(\sum ^{}_{}x)^2} \\ b=\text{ }\bar{y}-a\bar{x} \end{gathered}[/tex][tex]\begin{gathered} where\text{ n=6} \\ \bar{x}=\text{ }\frac{15}{6}=2.5 \\ \bar{y}=\text{ }\frac{797}{6}=\text{ 132.8} \end{gathered}[/tex][tex]\begin{gathered} a=\frac{6(2464)-(15)(797)}{6(55)-(15)^2} \\ a=\text{ }\frac{14784-11955}{330-225} \\ a=\text{ }\frac{2829}{105} \\ a=\text{ 26.9} \end{gathered}[/tex][tex]\begin{gathered} b=\bar{y}-\bar{ax} \\ b=\text{ }132.8-(26.9)(2.5) \\ b=\text{ 132.8- 67.25} \\ b=\text{ 65.55}\approx65.6 \end{gathered}[/tex]
Hence the regression line equation is:
y=ax+b
y=26.9x+65.6
To find the projected profit of the year 2008, we would make use of the derived equation, y=26.9x+65.6
where x= 2008-1997= 11years,
y=?
Solving for y,
y= 26.9×11+65.6
y= 295.9+65.6
y= 361.15(thousand dollars) to the nearest thousand dollars
y= $361,000.
Hence the value for y= $361,000.
