Effects of an Outlier Refer to the Minitab-generated scatterplot given in Exercise 11 of

Section 10-1 on page 485.

a. Using the pairs of values for all 10 points, find the equation of the regression line.

b. After removing the point with coordinates (10, 10), use the pairs of values for the remaining 9 points and find the equation of the regression line.

c. Compare the results from parts (a) and (b).

A set of 10 pairs of values is considered.

a.

The regression equation of y on x has the following notation:

\(\hat y = {b_0} + {b_1}x\),where

\({b_0}\)is the intercept term, and

\({b_1}\)is the slope coefficient.

The following data points are considered:

The following table shows the necessary calculations:

The value of the y-intercept is computed below.

\(\begin{array}{c}{b_0} = \frac{{\left( {\sum y } \right)\left( {\sum {{x^2}} } \right) - \left( {\sum x } \right)\left( {\sum {xy} } \right)}}{{n\left( {\sum {{x^2}} } \right) - {{\left( {\sum x } \right)}^2}}}\\ = \frac{{\left( {28} \right)\left( {142} \right) - \left( {28} \right)\left( {136} \right)}}{{10\left( {142} \right) - {{\left( {28} \right)}^2}}}\\ = 0.264\end{array}\).

The value of the slope coefficient is computed below.

\(\begin{array}{c}{b_1} = \frac{{n\left( {\sum {xy} } \right) - \left( {\sum x } \right)\left( {\sum y } \right)}}{{n\left( {\sum {{x^2}} } \right) - {{\left( {\sum x } \right)}^2}}}\\ = \frac{{\left( {10} \right)\left( {136} \right) - \left( {28} \right)\left( {28} \right)}}{{10\left( {142} \right) - {{\left( {28} \right)}^2}}}\\ = 0.906\end{array}\).

Thus, the regression equation becomes

\(\hat y = 0.264 - 0.906x\).

b.

The following 9 pairs of data points are considered:

The following table shows the necessary calculations:

The value of the y-intercept is computed below.

\(\begin{array}{c}{b_0} = \frac{{\left( {\sum y } \right)\left( {\sum {{x^2}} } \right) - \left( {\sum x } \right)\left( {\sum {xy} } \right)}}{{n\left( {\sum {{x^2}} } \right) - {{\left( {\sum x } \right)}^2}}}\\ = \frac{{\left( {18} \right)\left( {42} \right) - \left( {18} \right)\left( {36} \right)}}{{9\left( {42} \right) - {{\left( {18} \right)}^2}}}\\ = 2.000\end{array}\).

The value of the slope coefficient is computed below.

\(\begin{array}{c}{b_1} = \frac{{n\left( {\sum {xy} } \right) - \left( {\sum x } \right)\left( {\sum y } \right)}}{{n\left( {\sum {{x^2}} } \right) - {{\left( {\sum x } \right)}^2}}}\\ = \frac{{\left( 9 \right)\left( {36} \right) - \left( {18} \right)\left( {18} \right)}}{{9\left( {42} \right) - {{\left( {18} \right)}^2}}}\\ = 0.000\end{array}\).

Thus, the regression equation becomes

\(\hat y = 2.00 - 0.00x\).

c.

The regression equations obtained in parts (a) and (b) are completely different from one another.

Thus, the presence of an extreme data pair (10,10) can greatly influence the regression equation.