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.

Using the listed lemon/crash data, find the best predicted crash fatality rate for a year in which there are 500 metric tons of lemon imports. Is the prediction worthwhile?

The given data provides the information of the ‘lemon imports’ and ‘crash fatality rate’, as follows.

the formula for the estimated regression line is

\(y = {b_0} + {b_1}x\).

Here,

\({b_0}\)is the Y-intercept,

\({b_1}\)is the slope,

\(x\)is the explanatory variable, and

\(\hat y\) is the response variable (predicted value).

Let X denote the lemon imports and Y denote the crash fatality rate.

The calculations required to compute the slope and intercept are as follows.

The sample size is \(\left( n \right) = 5\).

The slope is computed as

\(\begin{array}{c}{b_1} = \frac{{n\left( {\sum {xy} } \right) - \left( {\sum x } \right)\left( {\sum y } \right)}}{{n\left( {\sum {{x^2}} } \right) - {{\left( {\sum x } \right)}^2}}}\\ = \frac{{5 \times 28571.7 - 1863 \times 77.2}}{{5 \times 762589 - {{1863}^2}}}\\ = - 0.00282\end{array}\).

The intercept is computed as

\(\begin{array}{c}{b_0} = \frac{{\left( {\sum y } \right)\left( {\sum {{x^2}} } \right) - \left( {\sum x } \right)\left( {\sum {xy} } \right)}}{{n\left( {\sum {{x^2}} } \right) - {{\left( {\sum x } \right)}^2}}}\\ = \frac{{77.2 \times 762589 - 1863 \times 28571.7}}{{5 \times 762589 - {{1863}^2}}}\\ = 16.4909\end{array}\).

Thus, the estimated regression equation is

\(\begin{array}{c}\hat y = {b_0} + {b_1}x\\ = 16.5 - 0.0028x\end{array}\).

Refer to exercise 19 of section 10-1 for the following result.

1) The scatter plot shows an approximate linear relationship between the variables.

2) The P-value is 0.0010.

As theP-value is less than the level of significance (0.05), the null hypothesis is rejected.

Therefore, the correlation is statistically significant.

Referring to figure 10-5, the criteria for a good regression model are satisfied.

Thus, the regression equation can be used for prediction.

Substitute 500 metric tons of lemons in the estimated linear regression model for the crash fatality rate.

\(\begin{array}{c}\hat y = 16.5 - 0.0028x\\ = 16.5 - 0.0028\left( {500} \right)\\ = 15.1\end{array}\).

Therefore, the best-predicted crash fatality rate for a year in which there are 500 metric tons of lemon imports will be 15.1.

The predictions do not seem worthwhile as the relationship between the two variables does not make sense.