Questions: Find the regression equation, letting the first variable be the predictor (x) variable. Using the listed lemon/crash data, where lemon imports are in metric tons and the fatality rates are per 100,000 people, find the best predicted crash fatality rate for a year in which there are 475 metric tons of lemon imports. Is the prediction worthwhile? Use a significance level of 0.05. Lemon Imports: 230, 267, 354, 476, 515 Crash Fatality Rate: 15.9, 15.7, 15.4, 15.3, 15 Find the equation of the regression line. ŷ = + (x) (Round the y-intercept to three decimal places as needed. Round the slope to four decimal places as needed.)

Solution Steps

The means of the lemon imports ($\bar{x}$) and crash fatality rates ($\bar{y}$) are calculated as follows:

$$\bar{x} = \frac{1}{n} \sum_{i=1}^{n} x_i = 368.4$$

$$\bar{y} = \frac{1}{n} \sum_{i=1}^{n} y_i = 15.46$$

The correlation coefficient ($r$) is found to be:

$$r = -0.9622$$

The numerator for the slope ($\beta$) is calculated as:

$$\text{Numerator for } \beta = \sum_{i=1}^{n} x_i y_i - n \bar{x} \bar{y} = 28308.3 - 5 \cdot 368.4 \cdot 15.46 = -169.02$$

The denominator for the slope is:

$$\text{Denominator for } \beta = \sum_{i=1}^{n} x_i^2 - n \bar{x}^2 = 741306 - 5 \cdot 368.4^2 = 62713.2$$

Thus, the slope ($\beta$) is:

$$\beta = \frac{-169.02}{62713.2} = -0.0027$$

The intercept ($\alpha$) is calculated using the formula:

$$\alpha = \bar{y} - \beta \bar{x} = 15.46 - (-0.0027) \cdot 368.4 = 16.4529$$

The equation of the regression line is:

$$\hat{y} = 16.4529 + (-0.0027)x$$

To predict the crash fatality rate for 475 metric tons of lemon imports, we substitute $x = 475$ into the regression equation:

$$\hat{y} = 16.4529 + (-0.0027) \cdot 475 = 15.1704$$

The correlation coefficient ($r = -0.9622$) indicates a strong negative linear relationship between lemon imports and crash fatality rates. Since the absolute value of the correlation coefficient is greater than 0.05, the prediction is considered worthwhile.

Final Answer