Chapter 13.3, Problem 18E

To determine

To find: the correlation coefficient r for the given data and also how well does the linear model fit the data.

Answer to Problem 18E

Correlation coefficient r $\approx 0.9845$ and the least square regression line $y=1.05x-1.4$ fit the data very well.

Explanation of Solution

Given information:

The points are

  $\left(0,-1\right),\left(2,0\right),\left(4,3\right),\left(6,5\right)$

The least square egression line $y=1.05x-1.4$

Formula used:

For n number of set of observations correlation coefficient r is given by

  $r=\frac{n{\displaystyle \sum _{i=1}^n{x}_i{y}_i}-\left(\left({\displaystyle \sum _{i=1}^n{x}_i}\right)\left({\displaystyle \sum _{i=1}^n{y}_i}\right)\right)}{\sqrt{n{\displaystyle \sum _{i=1}^n{\left(x_i\right)}^2}-{\left({\displaystyle \sum _{i=1}^n{x}_i}\right)}^2}\sqrt{n{\displaystyle \sum _{i=1}^n{\left(y_i\right)}^2}-{\left({\displaystyle \sum _{i=1}^n{y}_i}\right)}^2}}\cdots \left(i\right)$

Calculation:

Here number of observations (n) is 4

Construct a table as shown below:

  $\begin{array}{ccccc}x& y& xy& x^2& y^2\\ 0& -1& 0& 0& 1\\ 2& 0& 0& 4& 0\\ 4& 3& 12& 16& 9\\ 6& 5& 30& 36& 25\\ \begin{array}{l}{\displaystyle \sum _{i=1}^4{x}_i}=\left(0+2+4+6\right)\\ =12\end{array}& \begin{array}{l}{\displaystyle \sum _{i=1}^4{y}_i}=\left(-1+0+3+5\right)\\ =7\end{array}& \begin{array}{l}{\displaystyle \sum _{i=1}^4{x}_i{y}_i}=\left(0+0+12+30\right)\\ =42\end{array}& \begin{array}{l}{\displaystyle \sum _{i=1}^4{x}_i{}^2}=\left(0+4+16+36\right)\\ =56\end{array}& \begin{array}{l}{\displaystyle \sum _{i=1}^4{y}_i{}^2}=\left(1+0+9+25\right)\\ =35\end{array}\end{array}$

Put these values in $\left(i\right)$

  $\begin{array}{c}r=\frac{4\left(42\right)-\left(12\times 7\right)}{\sqrt{4\left(56\right)-{\left(12\right)}^2}\sqrt{4\left(35\right)-{\left(7\right)}^2}}\\ =\frac{84}{\sqrt{80}\times \sqrt{91}}\\ =0.98449518497\\ \approx 0.9845\end{array}$

Since the value of r is closer to 1 there exists a strong positive linear correlation. Hence the least square regression line $y=1.05x-1.4$ fits the data very well.