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 listed actress/actor ages, find the best predicted age of the Best Actor given that the age of the Best Actress is 54 years. Is the result reasonably close to the Best Actor’s (Eddie Redmayne) actual age of 33 years, which happened in 2015, when the Best Actress was Julianne Moore, who was 54 years of age?

Values are given on two variables namely, actress age and actor age.

Let x represent the age of Best Actress.

Let y represent the age of Best Actor.

Themean value of xis given as,

\(\begin{array}{c}\bar x = \frac{{\sum\limits_{i = 1}^n {{x_i}} }}{n}\\ = \frac{{28 + 30 + .... + 54}}{{12}}\\ = 39.083\end{array}\)

Therefore, the mean value of x is 39 years.

Themean value of yis given as,

\(\begin{array}{c}\bar y = \frac{{\sum\limits_{i = 1}^n {{y_i}} }}{n}\\ = \frac{{43 + 37 + .... + 33}}{{12}}\\ = 45.167\end{array}\)

Therefore, the mean value of y is 45 years.

The standard deviation of x is given as,

\(\begin{array}{c}{s_x} = \sqrt {\frac{{\sum\limits_{i = 1}^n {{{({x_i} - \bar x)}^2}} }}{{n - 1}}} \\ = \sqrt {\frac{{{{\left( {28 - 39.083} \right)}^2} + {{\left( {30 - 39.083} \right)}^2} + ..... + {{\left( {54 - 39.083} \right)}^2}}}{{12 - 1}}} \\ = 13.734\end{array}\)

Therefore, the standard deviation of x is 13.734.

The standard deviation of yis given as,

\(\begin{array}{c}{s_y} = \sqrt {\frac{{\sum\limits_{i = 1}^n {{{({y_i} - \bar y)}^2}} }}{{n - 1}}} \\ = \sqrt {\frac{{{{\left( {43 - 45.167} \right)}^2} + {{\left( {37 - 45.167} \right)}^2} + ..... + {{\left( {33 - 45.167} \right)}^2}}}{{12 - 1}}} \\ = 7.872\end{array}\)

Therefore, the standard deviation of y is 7.872.

Thecorrelation coefficientis 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,

\(\begin{array}{l}r = \frac{{n\left( {\sum {xy} } \right) - \left( {\sum x } \right)\left( {\sum y } \right)}}{{\sqrt {\left( {\left( {n\sum {{x^2}} } \right) - {{\left( {\sum x } \right)}^2}} \right)\left( {\left( {n\sum {{y^2}} } \right) - {{\left( {\sum y } \right)}^2}} \right)} }}\\ = \frac{{12\left( {20841} \right) - \left( {469} \right)\left( {542} \right)}}{{\sqrt {\left( {\left( {12 \times 20405} \right) - {{\left( {469} \right)}^2}} \right)\left( {\left( {12 \times 25162} \right) - {{\left( {542} \right)}^2}} \right)} }}\\ = - 0.288\end{array}\)

Therefore, the correlation coefficient is -0.288.

The slopeof the regression line is given as,

\(\begin{array}{c}{b_1} = r\frac{{{s_Y}}}{{{s_X}}}\\ = - 0.288 \times \frac{{7.872}}{{13.734}}\\ = - 0.165\end{array}\)

Therefore, the value of slope is - 0.165.

The interceptis computed as,

\(\begin{array}{c}{b_0} = \bar y - {b_1}\bar x\\ = 45.167 - \left( { - 0.165 \times 39.083} \right)\\ = 51.6\end{array}\)

Therefore, the value of intercept is 51.6.

Theregression equationis given as,

\(\begin{array}{c}\hat y = {b_0} + {b_1}x\\ = 51.6 - 0.165x\end{array}\)

Thus, the regression equation is \(\hat y = 51.6 - 0.165x\).

Referring to exercise 21 of section 10-1 for the following result

1)The scatter plot shows approximately a linear relationship between the variables.

2)The P-value is 0.364.

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, thecriteria for a good regression model are not satisfied.

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