Question

Regression and Predictions. Exercises 13–28 use the same data sets as Exercises 13–28 in Section 10-1. In each case, find the regression equation, letting the first variable be the predictor (x) variable. Find the indicated predicted value by following the prediction procedure summarized in Figure 10-5 on page 493.

Use the listed boat/manatee data. In a year not included in the data below, there were 970,000 registered pleasure boats in Florida. Find the best predicted number of manatee fatalities resulting from encounters with boats. Is the result reasonably close to 79, which was the actual number of manatee fatalities?

Pleasure Boats	99	99	97	95	90	90	87	90	90
Manatee Fatalities	92	73	90	97	83	88	81	73	68

Short answer

The regression equation is\(\hat y = 13.6 + 0.744x\).

Thebest predicted number of manatee fatalities resulting from encounters with boatsis 83.

The prediction is not reasonably close to 79.

Step-by-step solution

Given information

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

Calculate the mean values

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.

Calculate the standard deviation of x and y

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.

Calculate the correlation coefficient

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.

Calculate the slope of the regression line

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.

Calculate the intercept of the regression line

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.

Form a regression equation

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 \(\hat y = 13.6 + 0.744x\).

Analyze the model

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.

Predict the value

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.