According to the least-squares property, the regression line minimizes the sum of the squares of the residuals. Refer to the data in table 10-1 on page 469.

a. Find the sum of squares of the residuals.

b. Show that the regression equation\(\hat y = - 3 + 2.5x\)results in a larger sum of squares of residuals.

Values are given for two variables, namely, Chocolate and Nobel.

Let x representChocolate.

Let y representNobel.

Themean value of xis given below:

\(\begin{array}{c}\bar x = \frac{{\sum\limits_{i = 1}^n {{x_i}} }}{n}\\ = \frac{{4.5 + 10.2 + .... + 5.3}}{{23}}\\ = 5.80435\end{array}\)

Therefore, the mean value of x is 5.80435.

Themean value of yis given below:

\(\begin{array}{c}\bar y = \frac{{\sum\limits_{i = 1}^n {{y_i}} }}{n}\\ = \frac{{5.5 + 24.3 + .... + 10.8}}{{23}}\\ = 11.10435\end{array}\)

Therefore, the mean value of y is 11.10435.

The standard deviation of x is given below:

\(\begin{array}{c}{s_x} = \sqrt {\frac{{\sum\limits_{i = 1}^n {{{({x_i} - \bar x)}^2}} }}{{n - 1}}} \\ = \sqrt {\frac{{{{\left( {4.5 - 5.80435} \right)}^2} + {{\left( {10.2 - 5.80435} \right)}^2} + ... + {{\left( {5.3 - 5.80435} \right)}^2}}}{{23 - 1}}} \\ = 3.27920\end{array}\)

Therefore, the standard deviation of x is 3.27920.

The standard deviation of yis given below:

\(\begin{array}{c}{s_y} = \sqrt {\frac{{\sum\limits_{i = 1}^n {{{({y_i} - \bar y)}^2}} }}{{n - 1}}} \\ = \sqrt {\frac{{{{\left( {5.5 - 11.10435} \right)}^2} + {{\left( {24.3 - 11.10435} \right)}^2} + ..... + {{\left( {10.8 - 11.10435} \right)}^2}}}{{23 - 1}}} \\ = 10.2116\end{array}\)

Therefore, the standard deviation of y is 10.2116.

The correlation coefficient is given below:

\(r = \frac{{n\left( {\su

m {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 below:

\(\begin{array}{c}r = \frac{{n\left( {\sum {xy} } \right) - \left( {\sum x } \right)\left( {\sum y } \right)}}{{\sqrt {\left( {\left( {n\sum {{x^2}} } \right) - {{\left( {\sum x } \right)}^2}} \right)\left( {\left( {n\sum {{y^2}} } \right) - {{\left( {\sum y } \right)}^2}} \right)} }}\\ = \frac{{23\left( {2072.23} \right) - \left( {133.5} \right)\left( {255.4} \right)}}{{\sqrt {\left( {\left( {23 \times 1011.45} \right) - {{\left( {133.5} \right)}^2}} \right)\left( {\left( {23 \times 5130.14} \right) - {{\left( {255.4} \right)}^2}} \right)} }}\\ = 0.80061\end{array}\)

Therefore, the correlation coefficient is 0.80061.

The slope of the regression line is given below:

\(\begin{array}{c}{b_1} = r\frac{{{s_Y}}}{{{s_X}}}\\ = 0.80061 \times \frac{{10.2116}}{{3.27920}}\\ = 2.49313\\ \approx 2.50\end{array}\)

Therefore, the value of the slope is 2.50.

The intercept is computed below:

\(\begin{array}{c}{b_0} = \bar y - {b_1}\bar x\\ = 11.10435 - \left( {2.50 \times 5.80435} \right)\\ = - 3.37\end{array}\)

Therefore, the value of the intercept is –3.37.

Theregression equationis givenbelow:

\(\begin{array}{c}\hat y = {b_0} + {b_1}x\\ = - 3.37 + 2.50x\end{array}\)

Thus,the best-fit regression equation is\(\hat y = - 3.37 + 2.50x\).

a.

The residual is computedbelow:

\(\begin{array}{c}{\mathop{\rm Residual}\nolimits} = {\rm{observed}}\;{\rm{y}} - \;{\rm{predicted}}\;{\rm{y}}\\ = y - \hat y\end{array}\)

The calculations are as follows:

b.

The calculations are as follows:

The sum of squares of residuals is computed below:

\(7.56 + 3.24 + 0.36 + ... + 0.30 = 827.45\)

Therefore, the sum of squares of residual is 827.45.

The sum of squares of residuals from the given regression line is 827.45 which is larger than the sum of squares of the residuals obtained from best-fit regression line 823.64.