Question

Finding a Prediction Interval. In Exercises 13–16, use the paired data consisting of registered Florida boats (tens of thousands) and manatee fatalities from boat encounters listed in Data Set 10 “Manatee Deaths” in Appendix B. Let x represent number of registered boats and let y represent the corresponding number of manatee deaths. Use the given number of registered boats and the given confidence level to construct a prediction interval estimate of manatee deaths.

Boats Use x = 96 (for 960,000 registered boats) with a 95% confidence level.

Short answer

The 95% prediction interval for the number of manatee deaths when the number of registered boats is equal to 960,000 is (65.1 manatees,106.8 manatees).

Step-by-step solution

Given information

The paired data for the variables ‘number of registered boats’ and ‘number of manatee deaths’ are provided.

Some important values inferred from the question are as follows.

\(\begin{array}{c}Confidence\;Level = 95\% \\{x_0} = 96\\n = 24\end{array}\).

Regression equation

Let x denote the variable ‘registered boats’.

Let y denote the variable ‘number of manatee deaths’

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, and

\({b_1}\)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.

\(\begin{array}{c}{b_0} = \frac{{\left( {\sum y } \right)\left( {\sum {{x^2}} } \right) - \left( {\sum x } \right)\left( {\sum {xy} } \right)}}{{n\left( {\sum {{x^2}} } \right) - {{\left( {\sum x } \right)}^2}}}\\ = \frac{{\left( {1700} \right)\left( {177128} \right) - \left( {2046} \right)\left( {148731} \right)}}{{24\left( {177128} \right) - {{\left( {2046} \right)}^2}}}\\ = - 49.048987\end{array}\).

The value of the slope coefficient is computed below.

\(\begin{array}{c}{b_1} = \frac{{n\left( {\sum {xy} } \right) - \left( {\sum x } \right)\left( {\sum y } \right)}}{{n\left( {\sum {{x^2}} } \right) - {{\left( {\sum x } \right)}^2}}}\\ = \frac{{\left( {24} \right)\left( {148731} \right) - \left( {2046} \right)\left( {1700} \right)}}{{24\left( {177128} \right) - {{\left( {2046} \right)}^2}}}\\ = 1.4062442\end{array}\).

Thus, the regression equation becomes

\(\hat y = - 49.048987 + 1.4062442x\).

Predicted value \(\left( {\hat y} \right)\)

The regression equation of y on x is

\(\hat y = - 49.048987 + 1.4062442x\).

Substituting the value of\({x_0} = 96\), the following value of\(\hat y\)is obtained:

\(\begin{array}{c}\hat y = - 49.048987 + 1.4062442\left( {96} \right)\\ = 85.9504562\end{array}\).

Level of significance and degrees of freedom

The following formula is used to compute the level of significance:

\(\begin{array}{c}Confidence\;Level = 95\% \\100\left( {1 - \alpha } \right) = 95\\1 - \alpha = 0.95\\\alpha = 1 - 0.95\\ = 0.05\end{array}\).

Therefore,

\(\begin{array}{c}\frac{\alpha }{2} = \frac{{0.05}}{2}\\ = 0.025\end{array}\).

The degree of freedom for computing the value of the t-multiplier isshown below.

\(\begin{array}{c}df = n - 2\\ = 24 - 2\\ = 22\end{array}\).

Value of \({t_{\frac{\alpha }{2}}}\)

The value of the t-multiplier for a level of significance equal to 0.025 and a degree of freedom equal to 22 is 2.0739.

Value of \({s_e}\)

The given table shows all the important values to compute the standard error of the estimate.

The value of the standard error of the estimate is computed, as shown below.

\(\begin{array}{c}{s_e} = \sqrt {\frac{{\sum {{{\left( {y - \hat y} \right)}^2}} }}{{n - 2}}} \\ = \sqrt {\frac{{2053.167806}}{{24 - 2}}} \\ = 9.6605284\end{array}\).

Thus, \({s_e} = 9.6605284\)

Value of \(\bar x\)

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

\(\begin{array}{c}\bar x = \frac{{68 + 68 + .... + 90}}{{24}}\\ = 85.25\end{array}\).

Value of \({\left( {\sum x } \right)^2}\)

The value of the term\({\left( {\sum x } \right)^2}\)is computed, as shown below.

\(\begin{array}{c}{\left( {\sum x } \right)^2} = {\left( {68 + 68 + ..... + 90} \right)^2}\\ = 4186116\end{array}\),

Value of \(\left( {\sum {{x^2}} } \right)\)

The value of the term\(\left( {\sum {{x^2}} } \right)\)is computed, as shown below.

\(\begin{array}{c}\left( {\sum {{x^2}} } \right) = {68^2} + {68^2} + ...... + {90^2}\\ = 177128\end{array}\)

Prediction interval

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

\(\begin{array}{c}E = {t_{\frac{\alpha }{2}}}{s_e}\sqrt {1 + \frac{1}{n} + \frac{{n{{\left( {{x_0} - \bar x} \right)}^2}}}{{n\left( {\sum {{x^2}} } \right) - {{\left( {\sum x } \right)}^2}}}} \\ = \left( {2.0739} \right)\left( {9.6605284} \right)\sqrt {1 + \frac{1}{{24}} + \frac{{24{{\left( {96 - 85.25} \right)}^2}}}{{24\left( {177128} \right) - \left( {4186116} \right)}}} \\ = 20.8629828\end{array}\)

Thus, the prediction interval becomes

\(\begin{array}{c}PI = \left( {\hat y - E,\hat y + E} \right)\\ = \left( {85.9504562 - 20.8629828,85.9504562 + 20.8629828} \right)\\ \approx \left( {65.1,106.8} \right)\end{array}\)

Therefore, the 95% prediction interval for the number of manatee deaths when the number of registered boats is equal to 960,000 is (65.1 manatees, 106.8 manatees).