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The paired data forthe 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.

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

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\({x_0} = 98\), 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.

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