Solution

To calculate $R^2$ you need to find the sum of the residuals squared and the total sum of squares.

Start off by finding the residuals, which is the distance from regression line to each data point. Work out the predicted $y$ value by plugging in the corresponding $x$ value into the regression line equation.

• For the point $(2,2)$

\begin{align} \hat{y}&=0.143+1.229x\\ &=0.143+(1.229\times2)\\ &=0.143+2.458\\ &=2.601 \end{align}

The actual value for $y$ is $2$. \begin{align} \text{Residual}&=\text{actual } y \text{ value} - \text{predicted }y \text{ value}\\ r_1&=y_i-\hat{y_i}\\ &=2-2.601\\ &=-0.601 \end{align} As you can see from the graph the actual point is below the regression line, so it makes sense that the residual is negative.

• For the point $(3,4)$

\begin{align} \hat{y}&=0.143+1.229x\\ &=0.143+(1.229\times3)\\ &=0.143+3.687\\ &=3.83 \end{align}

The actual value for $y$ is $4$.

\begin{align} \text{Residual}&=\text{actual } y \text{ value} - \text{predicted }y \text{ value}\\ r_2&=y_i-\hat{y_i}\\ &=4-0.3.83\\ &=0.17 \end{align} As you can see from the graph the actual point is above the regression line, so it makes sense that the residual is positive.

• For the point $(4,6)$

\begin{align} \hat{y}&=0.143+1.229x\\ &=0.143+(1.229\times4)\\ &=0.143+4.916\\ &=5.059 \end{align}

The actual value for $y$ is $6$.

\begin{align} \text{Residual}&=\text{actual } y \text{ value} - \text{predicted }y \text{ value}\\ r_3&=y_i-\hat{y_i}\\ &=6-5.059\\ &=0.941 \end{align}

• For the point $(6,7)$

\begin{align} \hat{y}&=0.143+1.229x\\ &=0.143+(1.229\times6)\\ &=0.143+7.374\\ &=7.517 \end{align}

The actual value for $y$ is $7$. \begin{align} \text{Residual}&=\text{actual } y \text{ value} - \text{predicted }y \text{ value}\\ r_4&=y_i-\hat{y_i}\\ &=7-7.517\\ &=-0.517 \end{align} To find the residuals squared we need to square each of $r_1$ to $r_4$ and sum them.

\begin{align} \sum({y_i}-\hat{y_i})^2&=\sum{r_i}\\ &={r_1}^2+{r_2}^2+{r_3}^2+{r_4}^2\\ &=(−0.601)^2+(0.17)^2+(0.941)^2-(-0.517)^2\\ &=1.542871 \end{align}

To find $\sum(y_i-\bar{y})^2$ you first need to find the mean of the $y$ values.

\begin{align} \bar{y}&=\frac{\sum{y} }{n}\\ &=\frac{2+4+6+7}{4}\\ &=\frac{19}{4}\\ &=4.75 \end{align}

Now we can calculate $\sum(y_i-\bar{y})^2$.

\begin{align} \sum(y_i-\bar{y})^2&=(2-4.75)^2+(4-4.75)^2+(6-4.75)^2+(7-4.75)^2\\ &=(-2.75)^2+(-0.75)^2+(1.25)^2+(2.25)^2\\ &=14.75 \end{align}

Therefore;

\begin{align} R^2&=1-\frac{\text{sum squared regression (SSR)} }{\text{total sum of squares (SST)} }\\ &=1-\frac{\sum({y_i}-\hat{y_i})^2}{\sum(y_i-\bar{y})^2}\\ &=1-\frac{1.542871}{14.75}\\ &=1-0.105\ \text{(3.s.f)}\\ &=0.895\text{ (3.s.f)} \end{align}

This means that the number of lectures per day account for $89.5$% of the variation in the hours people spend at university per day.

An odd property of $R^2$ is that it is increasing with the number of variables. Thus, in the example above, if we added another variable measuring mean height of lecturers, $R^2$ would be no lower and may well, by chance, be greater - even though this is unlikely to be an improvement in the model. To account for this, an adjusted version of the coefficient of determination is sometimes used. For more information, please see [http://www.statstutor.ac.uk/resources/uploaded/correlation.pdf

Video Examples

Example 1

This is a video presented by Alissa Grant-Walker on how to calculate the coefficient of determination.

Example 2

This is Khan Academy's video on working out the coefficient of determination.

External Resources

• Online r-squared calculator on easycalculation.com

See Also

Residuals