91影视

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

Let x represent thepresident鈥檚 height.

Let y represent theopponent鈥檚 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.