Question

Variation and Prediction Intervals. In Exercises 17–20, find the (a) explained variation, (b) unexplained variation, and (c) indicated prediction interval. In each case, there is sufficient evidence to support a claim of a linear correlation, so it is reasonable to use the regression equation when making predictions.

Weighing Seals with a Camera The table below lists overhead widths (cm) of seals measured from photographs and the weights (kg) of the seals (based on “Mass Estimation of Weddell Seals Using Techniques of Photogrammetry,” by R. Garrott of Montana State University). For the prediction interval, use a 99% confidence level with an overhead width of 9.0 cm.

Overhead Width	7.2	7.4	9.8	9.4	8.8	8.4
Weight	116	154	245	202	200	191

Short answer

(a) Explained Variation:8880.1818

(b) Unexplained Variation:991.1515

(c) 99% Prediction Interval: (124.97 cm, 284.55 cm)

Step-by-step solution

Given information

Data are given on two variables, “Overhead Width” and “Weight”.

Regression equation

Let x denote the variable “Overhead Width.”

Let y denote the variable “Weight”.

The regression equation of y on x has the following notation:

\(\hat y = {b_0} + {b_1}x\)where

\({b_0}\)is the intercept term

\({b_1}\)is the slope coefficient

The following calculations are done to compute the intercept and the slope coefficient:

The value of the y-intercept is computed below:

\(\begin{aligned}{c}{b_0} &= \frac{{\left( {\sum y } \right)\left( {\sum {{x^2}} } \right) - \left( {\sum x } \right)\left( {\sum {xy} } \right)}}{{n\left( {\sum {{x^2}} } \right) - {{\left( {\sum x } \right)}^2}}}\\ &= \frac{{\left( {1108} \right)\left( {439} \right) - \left( {81} \right)\left( {9639} \right)}}{{8\left( {439} \right) - {{\left( {81} \right)}^2}}}\\ &= - 156.878788\\ \approx - 156.88\end{aligned}\)

The value of the slope coefficient is computed below:

\(\begin{aligned}{c}{b_1} &= \frac{{n\left( {\sum {xy} } \right) - \left( {\sum x } \right)\left( {\sum y } \right)}}{{n\left( {\sum {{x^2}} } \right) - {{\left( {\sum x } \right)}^2}}}\\ &= \frac{{\left( 6 \right)\left( {9639} \right) - \left( {51} \right)\left( {1108} \right)}}{{6\left( {439} \right) - {{\left( {51} \right)}^2}}}\\ &= 40.181818\\ &\approx 40.18\end{aligned}\)

Thus, the regression equation becomes:

\(\begin{aligned}{c}\hat y &= - 156.878788 + 40.181818x\\ \approx - 156.88 + 40.18x\end{aligned}\)

Predicted values

The mean value of observed y is computed below:

\(\begin{aligned}{c}\bar y &= \frac{{\sum y }}{n}\\ &= \frac{{1108}}{6}\\ &= 184.6667\end{aligned}\)

The following table shows the predicted values (upon substituting the values of x in the regression equation), and other important calculations are done below:

The value of the explained variation is shown below:

\(\sum {{{\left( {\hat y - \bar y} \right)}^2}} = 8880.1818\)

Thus, the explained variation is equal to 8880.182.

The value of the unexplained variation is shown below:

\(\sum {{{\left( {y - \hat y} \right)}^2}} = 991.1515\)

Thus, the unexplained variation is equal to 991.1515.

Predicted value at \(\left( {{x_0}} \right)\)

Substituting the value of\({x_0} = 9\)in the regression equation, the predicted value is obtained as follows:

\(\begin{aligned}{c}\hat y &= - 156.878788 + 40.181818x\\ &= - 156.878788 + 40.181818\left( 9 \right)\\ &= 204.7576\end{aligned}\)

Level of significance and degrees of freedom

The following formula is used to compute the level of significance

\(\begin{aligned}{c}{\rm{Confidence}}\;{\rm{Level}} &= 99\% \\100\left( {1 - \alpha } \right) &= 99\\1 - \alpha &= 0.99\\ &= 0.01\end{aligned}\)

The degrees of freedom for computing the value of the t-multiplier are shown below:

\(\begin{aligned}{c}df &= n - 2\\ &= 6 - 2\\ &= 4\end{aligned}\)

Value of t-multiplier, \({t_{\frac{\alpha }{2}}}\)

The value of the t-multiplier for a level of significance equal to 0.01and degrees of freedom equal to 4 is equal to 4.6041.

Standard error of the estimate

The value of the standard error of the estimate is computed below:

\(\begin{array}{c}{s_e} = \sqrt {\frac{{\sum {{{\left( {y - \hat y} \right)}^2}} }}{{n - 2}}} \\ = \sqrt {\frac{{991.1515}}{{6 - 2}}} \\ = 15.74128\end{array}\)

Value of \(\bar x\)

The value of\(\bar x\)is computed as follows:

\(\begin{array}{c}\bar x = \frac{{\sum x }}{n}\\ = \frac{{51}}{6}\\ = 8.5\end{array}\)

Prediction interval

Substitute the values obtained above to calculate the value of margin of error (E) as shown:

\(\begin{aligned}{c}E &= {t_{\frac{\alpha }{2}}}{s_e}\sqrt {1 + \frac{1}{n} + \frac{{n{{\left( {{x_0} - \bar x} \right)}^2}}}{{n\left( {\sum {{x^2}} } \right) - {{\left( {\sum x } \right)}^2}}}} \\ &= \left( {4.6041} \right)\left( {15.74128} \right)\sqrt {1 + \frac{1}{6} + \frac{{6{{\left( {9 - 8.5} \right)}^2}}}{{6\left( {439} \right) - {{\left( {51} \right)}^2}}}} \\ &= 79.791718\end{aligned}\)

Thus, the prediction interval becomes:

\(\begin{aligned}{c}PI &= \left( {\hat y - E,\hat y + E} \right)\\ &= \left( {204.7576 - 79.791718,204.7576 + 79.791718} \right)\\ &= \left( {124.966,284.549} \right)\\ &\approx \left( {124.97,284.55} \right)\end{aligned}\)

Therefore, the 99% prediction interval for the overhead width for the given value of weight equal to 9.0 cm is (124.97 cm, 284.55 cm).