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 = 98 (for 980,000 registered boats) with a 95% confidence level.

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

Some important values inferred from the question are as follows.

\(\begin{array}{c}{\rm{Confidence}}\;{\rm{level}} = 95\% \\{x_0} = 98\\n = 24\end{array}\).

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

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

The regression equation of y on x is

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

Substituting the value of\({x_0} = 98\), the following value of\(\hat y\)is obtained:

\(\begin{array}{c}\hat y = - 49.048987 + 1.4062442\left( {98} \right)\\ = 88.7629446\end{array}\).

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

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

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.

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

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

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, \({s_e} = 9.6605284\)

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

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

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

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

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

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

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

\(\begin{array}{c}E = {t_{\frac{\alpha }{2}}}{s_e}\sqrt {1 + \frac{1}{n} + \frac{{n{{\left( {{x_0} - \bar x} \right)}^2}}}{{n\left( {\sum {{x^2}} } \right) - {{\left( {\sum x } \right)}^2}}}} \\ = \left( {2.0739} \right)\left( {9.6605284} \right)\sqrt {1 + \frac{1}{{24}} + \frac{{24{{\left( {98 - 85.25} \right)}^2}}}{{24\left( {177128} \right) - \left( {4186116} \right)}}} \\ = 21.02937478\end{array}\)