91Ó°ÊÓ

Values are given on two variables namely, pleasure boats and manatee fatalities.

Let x represent thepleasure boats.

Let y represent themanatee fatalities.

Themean value of xis given as,

\(\begin{array}{c}\bar x = \frac{{\sum\limits_{i = 1}^n {{x_i}} }}{n}\\ = \frac{{99 + 99 + .... + 90}}{9}\\ = 93\end{array}\)

Therefore, the mean value of x is 93.

Themean value of yis given as,

\(\begin{array}{c}\bar y = \frac{{\sum\limits_{i = 1}^n {{y_i}} }}{n}\\ = \frac{{92 + 73 + .... + 68}}{9}\\ = 82.778\end{array}\)

Therefore, the mean value of y is 82.778.

The standard deviation of xis given as,

\(\begin{array}{c}{s_x} = \sqrt {\frac{{\sum\limits_{i = 1}^n {{{({x_i} - \bar x)}^2}} }}{{n - 1}}} \\ = \sqrt {\frac{{{{\left( {99 - 93} \right)}^2} + {{\left( {99 - 93} \right)}^2} + ..... + {{\left( {90 - 93} \right)}^2}}}{{9 - 1}}} \\ = 4.528\end{array}\)

Therefore, the standard deviation of x is 4.528.

The standard deviation of y is given as,

\(\begin{array}{c}{s_y} = \sqrt {\frac{{\sum\limits_{i = 1}^n {{{({y_i} - \bar y)}^2}} }}{{n - 1}}} \\ = \sqrt {\frac{{{{\left( {92 - 82.778} \right)}^2} + {{\left( {73 - 82.778} \right)}^2} + ..... + {{\left( {68 - 82.778} \right)}^2}}}{{9 - 1}}} \\ = 9.871\end{array}\)

Therefore, the standard deviation of y is 9.871.

The correlation coefficient is given as,

\(r = \frac{{n\left( {\sum {xy} } \right) - \left( {\sum x } \right)\left( {\sum y } \right)}}{{\sqrt {\left( {\left( {n\sum {{x^2}} } \right) - {{\left( {\sum x } \right)}^2}} \right)\left( {\left( {n\sum {{y^2}} } \right) - {{\left( {\sum y } \right)}^2}} \right)} }}\)

The calculations required to compute the correlation coefficient are as follows,

The correlation coefficient is given as,

\(\begin{array}{c}r = \frac{{n\left( {\sum {xy} } \right) - \left( {\sum x } \right)\left( {\sum y } \right)}}{{\sqrt {\left( {\left( {n\sum {{x^2}} } \right) - {{\left( {\sum x } \right)}^2}} \right)\left( {\left( {n\sum {{y^2}} } \right) - {{\left( {\sum y } \right)}^2}} \right)} }}\\ = \frac{{9\left( {69407} \right) - \left( {837} \right)\left( {745} \right)}}{{\sqrt {\left( {\left( {9 \times 78005} \right) - {{\left( {837} \right)}^2}} \right)\left( {\left( {9 \times 62449} \right) - {{\left( {745} \right)}^2}} \right)} }}\\ = 0.3412\end{array}\)

Therefore, the correlation coefficient is 0.3412.

The slopeof the regression line is given as,

\(\begin{array}{c}{b_1} = r\frac{{{s_Y}}}{{{s_X}}}\\ = 0.3412 \times \frac{{9.871}}{{4.528}}\\ = 0.744\end{array}\)

Therefore, the value of slope is 0.744.

The interceptis computed as,

\(\begin{array}{c}{b_0} = \bar y - {b_1}\bar x\\ = 82.778 - \left( {0.744 \times 93} \right)\\ = 13.586\\ \approx 13.6\end{array}\)

Therefore, the value of intercept is 13.6.

Theregression equation is given as,

\(\begin{array}{c}\hat y = {b_0} + {b_1}x\\ = 13.6 + 0.744x\end{array}\)

Thus, the regression equation is \(\hat y = 13.6 + 0.744x\).

Referring to exercise 24 of section 10-1,

1)The scatter plot does not show a linear relationship between the variables.

2)The P-value is 0.369.

As the P-value is greater than the level of significance (0.05), this implies the null hypothesis fails to be rejected.

Therefore, the correlation is not significant.

Referring to figure 10-5, thecriteria for a good regression model are not satisfied.

Therefore, the regression equation cannot be used to predict the value of y.

The best predicted number of manatee fatalities resulting from encounters with boatsis the mean number ofManatee Fatalities; that is 82.778 approximately equal to 83.

Therefore, thebest predicted number of manatee fatalities resulting from encounters with boatsis 83.

The prediction value is 83 which is not reasonably close to 79.