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 diameter/circumference data, find the best predicted circumference of a marble with a diameter of 1.50 cm. How does the result compare to the actual circumference of 4.7 cm?

Values are given on three variables namely, diameter, circumference, and volume.

Let x represent the diameter.

Let y represent thecircumference.

Themean value of xis given as,

\(\begin{array}{c}\bar x = \frac{{\sum\limits_{i = 1}^n {{x_i}} }}{n}\\ = \frac{{7.4 + 23.9 + .... + 9.7}}{8}\\ = 12.375\end{array}\)

Therefore, the mean value of x is 12.375.

Themean value of yis given as,

\(\begin{array}{c}\bar y = \frac{{\sum\limits_{i = 1}^n {{y_i}} }}{n}\\ = \frac{{23.2 + 75.1 + .... + 30.5}}{8}\\ = 38.888\end{array}\)

Therefore, the mean value of y is 38.888.

The standard deviation of x is given as,

\(\begin{array}{c}{s_x} = \sqrt {\frac{{\sum\limits_{i = 1}^n {{{({x_i} - \bar x)}^2}} }}{{n - 1}}} \\ = \sqrt {\frac{{{{\left( {7.4 - 12.375} \right)}^2} + {{\left( {23.9 - 12.375} \right)}^2} + ..... + {{\left( {9.7 - 12.375} \right)}^2}}}{{8 - 1}}} \\ = 8.371\end{array}\)

Therefore, the standard deviation of x is 8.371.

The standard deviation of y is given as,

\(\begin{array}{c}{s_y} = \sqrt {\frac{{\sum\limits_{i = 1}^n {{{({y_i} - \bar y)}^2}} }}{{n - 1}}} \\ = \sqrt {\frac{{{{\left( {23.2 - 38.888} \right)}^2} + {{\left( {75.1 - 38.888} \right)}^2} + ..... + {{\left( {30.5 - 38.888} \right)}^2}}}{{8 - 1}}} \\ = 26.307\end{array}\)

Therefore, the standard deviation of y is 26.307.

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,

\(\begin{array}{c}r = \frac{{n\left( {\sum {xy} } \right) - \left( {\sum x } \right)\left( {\sum y } \right)}}{{\sqrt {\left( {\left( {n\sum {{x^2}} } \right) - {{\left( {\sum x } \right)}^2}} \right)\left( {\left( {n\sum {{y^2}} } \right) - {{\left( {\sum y } \right)}^2}} \right)} }}\\ = \frac{{8\left( {5391.3} \right) - \left( {99} \right)\left( {311.1} \right)}}{{\sqrt {\left( {\left( {8 \times 1715.6} \right) - {{\left( {99} \right)}^2}} \right)\left( {\left( {8 \times 16942} \right) - {{\left( {311.1} \right)}^2}} \right)} }}\\ = 0.999999\end{array}\)

Therefore, the correlation coefficient is 0.999999.

The slopeof the regressionline is given as,

\(\begin{array}{c}{b_1} = r \times \frac{{{s_Y}}}{{{s_X}}}\\ = 0.999999 \times \frac{{26.307}}{{8.371}}\\ = 3.143\end{array}\)

Therefore, the value of slope is 3.14.

The interceptis computed as,

\(\begin{array}{c}{b_0} = \bar y - {b_1}\bar x\\ = 38.888 - \left( {3.143 \times 12.375} \right)\\ = - 0.00396\end{array}\)

Therefore, the value of intercept is -0.004.

Theregression equationis given as,

\(\begin{array}{c}\hat y = {b_0} + {b_1}x\\ = - 0.004 + 3.14x\end{array}\)

Thus, the regression equation is \(\hat y = - 0.00396 + 3.143x\).

Referring to exercise 27 of section 10-1,

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

2)The P-value is 0.000.

As the P-value is less than the level of significance (0.05), this implies the null hypothesis is rejected.

Therefore, the correlation is significant.

Referring to figure 10-5, the criteria for a good regression model are satisfied.

Therefore, the regression equation can be used to predict the value of y.

The best predicted circumference of marble with a diameter of 1.50 cm is computed as,

\(\begin{array}{c}\hat y = - 0.00396 + \left( {3.14 \times 1.50} \right)\\ = 4.70604\end{array}\)

Therefore, the best predicted circumference of marble with a diameter of 1.50 cm is 4.7 cm.

The predicted circumference of marble with a diameter of 1.50 cm is the same as the actual circumference.