In Exercises 9 and 10, use the given data to find the equation of the regression line. Examine the scatterplot and identify a characteristic of the data that is ignored by the regression line.

Values are given on two variables namely, x and y.

Themean value of xis given as,

\(\begin{array}{c}\bar x = \frac{{\sum\limits_{i = 1}^n {{x_i}} }}{n}\\ = \frac{{10 + 8 + .... + 5}}{{11}}\\ = 9\end{array}\)

Therefore, the mean value of x is 9.

Themean value of yis given as,

\(\begin{array}{c}\bar y = \frac{{\sum\limits_{i = 1}^n {{y_i}} }}{n}\\ = \frac{{7.46 + 6.77 + .... + 5.73}}{{11}}\\ = 7.5\end{array}\)

Therefore, the mean value of y is 7.5

The standard deviation of x is given as,

\(\begin{array}{c}{s_x} = \sqrt {\frac{{\sum\limits_{i = 1}^n {{{({x_i} - \bar x)}^2}} }}{{n - 1}}} \\ = \sqrt {\frac{{{{\left( {10 - 9} \right)}^2} + {{\left( {8 - 9} \right)}^2} + ... + {{\left( {5 - 9} \right)}^2}}}{{11 - 1}}} \\ = 3.3166\end{array}\)

Therefore, the standard deviation of x is 3.3166.

The standard deviation of y is given as,

\(\begin{array}{c}{s_y} = \sqrt {\frac{{\sum\limits_{i = 1}^n {{{({y_i} - \bar y)}^2}} }}{{n - 1}}} \\ = \sqrt {\frac{{{{\left( {7.46 - 7.5} \right)}^2} + {{\left( {6.77 - 7.5} \right)}^2} + ..... + {{\left( {5.73 - 7.5} \right)}^2}}}{{11 - 1}}} \\ = 2.0304\end{array}\)

Therefore, the standard deviation of y is 2.0304.

The correlation coefficient is given as,

\(r = \frac{{n\left( {\sum {xy} } \right) - \left( {\sum x } \right)\left( {\sum y } \right)}}{{\sqrt {\left( {\left( {n\sum {{x^2}} } \right) - {{\left( {\sum x } \right)}^2}} \right)\left( {\left( {n\sum {{y^2}} } \right) - {{\left( {\sum y } \right)}^2}} \right)} }}\)

The calculations required to compute the correlation coefficient are as follows:

The correlation coefficient is given as,

\(\begin{array}{l}r = \frac{{n\left( {\sum {xy} } \right) - \left( {\sum x } \right)\left( {\sum y } \right)}}{{\sqrt {\left( {\left( {n\sum {{x^2}} } \right) - {{\left( {\sum x } \right)}^2}} \right)\left( {\left( {n\sum {{y^2}} } \right) - {{\left( {\sum y } \right)}^2}} \right)} }}\\ = \frac{{11\left( {797.47} \right) - \left( {99} \right)\left( {82.5} \right)}}{{\sqrt {\left( {\left( {11 \times 1001} \right) - {{\left( {99} \right)}^2}} \right)\left( {\left( {11 \times 659.9762} \right) - {{\left( {82.5} \right)}^2}} \right)} }}\\ = 0.8163\end{array}\)

Therefore, the correlation coefficient is 0.8163.

The slopeof the regression line is given as,

\(\begin{array}{c}{b_1} = r\frac{{{s_Y}}}{{{s_X}}}\\ = 0.8163 \times \frac{{2.030}}{{3.317}}\\ = 0.4996\\ \approx 0.500\end{array}\)

Therefore, the value of slope is 0.500.

The interceptis computed as,

\(\begin{array}{c}{b_0} = \bar y - {b_1}\bar x\\ = 7.5 - \left( {0.500 \times 9} \right)\\ = 3.002\end{array}\)

Therefore, the value of intercept is 3.00.

Theregression equationis given as,

\(\begin{array}{c}\hat y = {b_0} + {b_1}x\\ = 3.002 + 0.500x\end{array}\)

Thus, the regression equation is \(\hat y = 3.002 + 0.500x\).

Use the following steps to plot a scatter plot between x and y:

• Consider x and y.
• Mark the values 0, 2, and so on until 14 on the vertical axis.
• Mark the values 0, 5, and so on until 15 on the horizontal axis.
• Plot the points on the graph corresponding to the pairs of values for the two variables.
• Label the horizontal axis as “y” and the vertical axis as “x”.

The following scatterplot is generated:

It can be observed from the above scatter plot that an observation (13,12.74) is extreme and deviates largely from a straight line pattern. Thus the characteristic that is been ignored is thatthere exists an outlier at (13,12.74).