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The paired data for the variables 鈥榥umber of registered boats鈥� and 鈥榥umber of manatee deaths鈥� are provided.

Some important values inferred from the question are as follows.

\(\begin{array}{c}Confidence\;Level = 99\% \\{x_0} = 85\\n = 24\end{array}\)

Let x denote the variable 鈥榬egistered boats鈥�.

Let y denote the variable 鈥榥umber of manatee deaths鈥�.

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\({x_0} = 85\), the following value of\(\hat y\)is obtained:

\(\begin{array}{c}\hat y = - 49.048987 + 1.4062442\left( {85} \right)\\ = 70.481772\end{array}\).

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

\(\begin{array}{c}C{\rm{onfidence}}\;{\rm{level}} = 99\% \\100\left( {1 - \alpha } \right) = 99\\1 - \alpha = 0.99\\ = 0.01\end{array}\).

Therefore,

\(\begin{array}{c}\frac{\alpha }{2} = \frac{{0.01}}{2}\\ = 0.005\end{array}\).

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 a level of significance equal to 0.005 and a degree of freedom equal to 22 is 2.8188.

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.

\(\begin{array}{c}E = {t_{\frac{\alpha }{2}}}{s_e}\sqrt {1 + \frac{1}{n} + \frac{{n{{\left( {{x_0} - \bar x} \right)}^2}}}{{n\left( {\sum {{x^2}} } \right) - {{\left( {\sum x } \right)}^2}}}} \\ = \left( {2.8188} \right)\left( {9.6605284} \right)\sqrt {1 + \frac{1}{{24}} + \frac{{24{{\left( {85 - 85.25} \right)}^2}}}{{24\left( {177128} \right) - \left( {4186116} \right)}}} \\ = 27.79293058\end{array}\)

Thus, the prediction interval becomes

\(\begin{array}{c}P.I. = \left( {\hat y - E,\hat y + E} \right)\\ = \left( {70.481772 - 27.79293058,70.481772 + 27.79293058} \right)\\ = \left( {42.7,98.3} \right)\end{array}\).

Therefore, the 99% prediction interval for the number of manatee deaths when the number of registered boats is equal to 850,000 is (42.7 manatees,98.3 manatees).