Answer
a) Least squares line is $Y=0.978x+0.066$ and $r=0.978$
b) Least squares line is $Y=1.5$ and $r=0$
c) In part a) data points are linearly correlated while there is no linear correlation between the data points in part b).
Work Step by Step
a) The least squares line for the data points $(x_1,y_1), (x_2,y_2),...,(x_n,y_n)$ is given by the quation $ Y=mx+b$ where,
$m=\frac{n(\sum{xy})-(\sum{x})(\sum{y})}{n(\sum{x^2})-(\sum{x})^2}$ and $b=\frac{\sum{y}-m(\sum{x})}{n}$
$x \hspace{0.4cm}1 \hspace{0.2cm}1 \hspace{0.2cm}2 \hspace{0.2cm} 2 \hspace{0.2cm} 9$
$y \hspace{0.4cm}1\hspace{0.2cm}2\hspace{0.2cm}1\hspace{0.2cm}2\hspace{0.2cm}9$
$n=5$
$\sum{x}=15$
$\sum{y}=15$
$\sum{x^2}=91$
$\sum{y^2}=91$
$\sum{xy}=90$
$m=\frac{(5\times90)-(15\times15)}{(5\times91)-15^2}=0.978$
$b=\frac{15-(0.978\times15)}{5}=0.066$
The least squares line is given by $Y=0.978x+0.066$.
The correlation coefficient is given by:
$r=\frac{n(\sum{xy})-(\sum{x})(\sum{y})}{\sqrt{n(\sum{x^2})-(\sum{x})^2}\sqrt{n(\sum{y^2})-(\sum{y})^2}}$
$=\frac{(5\times90)-(15\times15)}{\sqrt{(5\times91)-15^2}\sqrt{(5\times91)-15^2}}=0.978$
b) Deleting the last point we have $4$ data points.
$x \hspace{0.4cm}1 \hspace{0.2cm}1 \hspace{0.2cm}2 \hspace{0.2cm} 2 $
$y \hspace{0.4cm}1\hspace{0.2cm}2\hspace{0.2cm}1\hspace{0.2cm}2$
$n=4$
$\sum{x}=6$
$\sum{y}=6$
$\sum{x^2}=10$
$\sum{y^2}=10$
$\sum{xy}=9$
$m=\frac{(4\times9)-(6\times6)}{(4\times10)-6^2}=0$
$b=\frac{6-(0\times6)}{4}=1.5$
The least squares line is now given by the equation: $Y=1.5$.
The correlation coefficient $r=\frac{(4\times6)-(6\times6)}{\sqrt{(4\times10)-6^2}\sqrt{(4\times10)-6^2}}=0$
c) The graph of the data points along with the least squares line is given below:
When the data point $(9,9)$ marked as E is removed, the strong linear correlation $(r=0.978)$ that the data points had vanishes and the correlation becomes $0$. This indicates that the data is no longer linearly correlated.
