Given the ordered pairs:
(2500, 2000), (2650, 2001), (3000, 2003), (3500, 2006), (4200, 2010)
Let's find the linear regression line.
Apply the formula:
y = mx + b
Where m si the slope and b is the y-intercept.
To find the slope, apply the formula:
[tex]m=\frac{n(\sum^{}_{}xy)-\sum^{}_{}x\sum^{}_{}y}{n(\sum^{}_{}x^2)-(\sum^{}_{}x)^2}[/tex]Where:
∑x = 2500 + 2650 + 3000 + 3500 + 4200 = 15850
∑y = 2000 + 2001 + 2003 + 2006 + 2010 = 10020
∑xy = (2500 * 2000) + (2650 * 2001) + (3000 * 2003) + (3500 * 2006) +(4200 * 2010) = 31774650
∑x² = 2500² + 2650² + 3000² + 3500² + 4200² = 52162500
∑y² = 2000² + 2001² + 2003² + 2006² + 2010² = 20080146
n = number of ordered pairs = 5
Thus, we have:
[tex]\begin{gathered} m=\frac{5(31774650)-15850\ast10020}{5(52162500)-(15850)^2} \\ \\ m=0.00587 \end{gathered}[/tex]The slope of the regression line is 0.00587
To find the y-intercept, apply the formula:
[tex]b=\frac{(\sum ^{}_{}y)(\sum ^{}_{}x^2)-\sum ^{}_{}x\sum ^{}_{}xy}{n(\sum ^{}_{}x^2)-(\sum ^{}_{}x)^2}[/tex]Thus, we have:
[tex]\begin{gathered} b=\frac{(10020)(52162500)-15850\ast10020}{5(52162500)-(15850)^2} \\ \\ b=1985.41 \end{gathered}[/tex]Therefore, the regression line is:
[tex]y=0.00587x+1985.41[/tex]To determine the year the population will hit 8000, substitute 8000 for x and solve for y:
[tex]\begin{gathered} y=0.00587(8000)+1985.41 \\ \\ y=46.92+1985.41 \\ \\ x=2032.33 \end{gathered}[/tex]Therefore, the population will hit 8000 in the year 2032.
To find the R-value, apply the formula:
[tex]R=\frac{n(\sum ^{}_{}xy)-\sum ^{}_{}x\sum ^{}_{}y}{(n(\sum ^{}_{}x^2)-(\sum ^{}_{}x)^2)-(n(\sum ^{}_{}y^2^{})-(\sum ^{}_{}y)^2}[/tex]Thus, we have:
[tex]\begin{gathered} R=\frac{5(31774650)-15850\ast10020}{\sqrt{5(52162500)-(15850)^2)-(5(20080146)-(10020)^2}} \\ \\ R=18.16 \end{gathered}[/tex]The R value is
