Regression and Predictions. Exercises 13–28 use the same data sets as Exercises 13–28 in Section 10-1. In each case, find the regression equation, letting the first variable be the predictor (x) variable. Find the indicated predicted value by following the prediction procedure summarized in Figure 10-5 on page 493.

Using the president/opponent heights, find the best predicted height of an opponent of a president who is 190 cm tall. Does it appear that heights of opponents can be predicted from the heights of the presidents?

President	178	182	188	175	179	183	192	182	177	185	188	188	183
Opponent	180	180	182	173	178	182	180	180	183	177	173	188	185

Values are given on two variables namely, the president’s height and the opponent’s height.

Let x represent thepresident’s height.

Let y represent theopponent’s height.

Themean value of xis given as,

\(\begin{array}{c}\bar x = \frac{{\sum\limits_{i = 1}^n {{x_i}} }}{n}\\ = \frac{{178 + 182 + .... + 188}}{{14}}\\ = 183.429\end{array}\)

Therefore, the mean value of x is 183.429.

Themean value of yis given as,

\(\begin{array}{c}\bar y = \frac{{\sum\limits_{i = 1}^n {{y_i}} }}{n}\\ = \frac{{180 + 180 + .... + 175}}{{14}}\\ = 179.714\end{array}\)

Therefore, the mean value of y is 179.714.

The standard deviation of x is given as,

\(\begin{array}{c}{s_x} = \sqrt {\frac{{\sum\limits_{i = 1}^n {{{({x_i} - \bar x)}^2}} }}{{n - 1}}} \\ = \sqrt {\frac{{{{\left( {178 - 183.429} \right)}^2} + {{\left( {182 - 183.429} \right)}^2} + ..... + {{\left( {188 - 183.429} \right)}^2}}}{{14 - 1}}} \\ = 5.003\end{array}\)

Therefore, the standard deviation of x is 5.003.

The standard deviation of y is given as,

\(\begin{array}{c}{s_y} = \sqrt {\frac{{\sum\limits_{i = 1}^n {{{({y_i} - \bar y)}^2}} }}{{n - 1}}} \\ = \sqrt {\frac{{{{\left( {180 - 179.714} \right)}^2} + {{\left( {180 - 179.714} \right)}^2} + ..... + {{\left( {175 - 179.714} \right)}^2}}}{{14 - 1}}} \\ = 4.304\end{array}\)

Therefore, the standard deviation of y is 4.304.

Thecorrelation 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,

\(\begin{array}{c}r = \frac{{n\left( {\sum {xy} } \right) - \left( {\sum x } \right)\left( {\sum y } \right)}}{{\sqrt {\left( {\left( {n\sum {{x^2}} } \right) - {{\left( {\sum x } \right)}^2}} \right)\left( {\left( {n\sum {{y^2}} } \right) - {{\left( {\sum y } \right)}^2}} \right)} }}\\ = \frac{{14\left( {461538} \right) - \left( {2568} \right)\left( {2516} \right)}}{{\sqrt {\left( {\left( {14 \times 471370} \right) - {{\left( {2568} \right)}^2}} \right)\left( {\left( {14 \times 452402} \right) - {{\left( {2516} \right)}^2}} \right)} }}\\ = 0.1133\end{array}\)

Therefore, the correlation coefficient is 0.1133.

The slopeof the regression line is given as,

\(\begin{array}{c}{b_1} = r\frac{{{s_Y}}}{{{s_X}}}\\ = 0.1133 \times \frac{{4.304}}{{5.003}}\\ = 0.097\end{array}\)

Therefore, the value of slope is 0.097.

The interceptis computed as,

\(\begin{array}{c}{b_0} = \bar y - {b_1}\bar x\\ = 179.714 - \left( {0.1133 \times 183.429} \right)\\ = 161.838\end{array}\)

Therefore, the value of intercept is 161.838.

Theregression equationis given as,

\(\begin{array}{c}\hat y = {b_0} + {b_1}x\\ = 161.838 + 0.097x\end{array}\)

Thus, the regression equation is \(\hat y = 161.838 + 0.097x\).

Referring to exercise 26 of section 10-1,

1)The scatter plot does not show a linear relationship between the variables.

2)The P-value is 0.700.

As the P-value is greater than the level of significance (0.05), this implies the null hypothesis fails to reject.

Therefore, the correlation is not significant.

Referring to figure 10-5,the criteria for a good regression model are not satisfied.

Therefore, the regression equation cannot be used to predict the value of y.

The best predicted height of an opponent of a president who is 190 cm tall is computed as,

\(\begin{array}{c}\hat y = \bar y\\ = 179.71\end{array}\)

Therefore, the best predicted height of an opponent of a president who is 190 cm tall is 180 cm.

No, the heights of opponents cannot be predicted from the heights of the presidents using the regression equation as the regression model is not good due to insignificant correlation.