Using the diameter/volume data from the preceding exercise, find the best predicted volume of a marble with a diameter of 1.50 cm. How does the result compare to the actual volume of 1.8 cm3?

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

Let x represent thediameter.

Let y represent thevolume.

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,

\(\begin{array}{c}\bar y = \frac{{\sum\limits_{i = 1}^n {{y_i}} }}{n}\\ = \frac{{212.2 + 7148.1 + .... + 477.9}}{8}\\ = 2287.2\end{array}\)

Therefore, the mean value of y is 2287.2.

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 yis given as,

\(\begin{array}{c}{s_y} = \sqrt {\frac{{\sum\limits_{i = 1}^n {{{({y_i} - \bar y)}^2}} }}{{n - 1}}} \\ = \sqrt {\frac{{{{\left( {212.2 - 2287.2} \right)}^2} + {{\left( {7148.1 - 2287.2} \right)}^2} + ..... + {{\left( {477.9 - 2287.2} \right)}^2}}}{{8 - 1}}} \\ = 2971.994\end{array}\)

Therefore, the standard deviation of y is 2971.994.

The correlation 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,

\(\begin{array}{c}r = \frac{{n\left( {\sum {xy} } \right) - \left( {\sum x } \right)\left( {\sum y } \right)}}{{\sqrt {\left( {\left( {n\sum {{x^2}} } \right) - {{\left( {\sum x } \right)}^2}} \right)\left( {\left( {n\sum {{y^2}} } \right) - {{\left( {\sum y } \right)}^2}} \right)} }}\\ = \frac{{8\left( {396776} \right) - \left( {99} \right)\left( {18297.6} \right)}}{{\sqrt {\left( {\left( {8 \times 1715.6} \right) - {{\left( {99} \right)}^2}} \right)\left( {\left( {8 \times 103679501} \right) - {{\left( {18297.6} \right)}^2}} \right)} }}\\ = 0.9782\end{array}\)

Therefore, the correlation coefficient is 0.9782.

The slopeof the regression line is given as,

\(\begin{array}{c}{b_1} = r\frac{{{s_Y}}}{{{s_X}}}\\ = 0.9782 \times \frac{{2971.994}}{{8.371}}\\ = 347.30\end{array}\)

Therefore, the value of slope is 347.30.

The interceptis computed as,

\(\begin{array}{c}{b_0} = \bar y - {b_1}\bar x\\ = 2287.2 - \left( {347.30 \times 12.375} \right)\\ = - 2010.67\end{array}\)

Therefore, the value of intercept is -2010.67.

Theregression equationis given as,

\(\begin{array}{c}\hat y = {b_0} + {b_1}x\\ = - 2010.67 + 347.30x\end{array}\)

Thus, the regression equation is \(\hat y = - 2010.67 + 347.67x\).

Referring to exercise 28 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 good regression model are satisfied.

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

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

\(\begin{array}{c}\hat y = - 2010.67 + \left( {347.30 \times 1.50} \right)\\ = - 1489.71\end{array}\)

Therefore, thebest predicted volume of marble with a diameter of 1.50 cm is

\( - 1489.71\;{\rm{c}}{{\rm{m}}^3}\).

Since the predicted volume of marble witha diameter of 1.50 cm is negative, this makes no sense.

Therefore, there is no comparison between the predicted and actual volume.