Question

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 525 metric tons of lemon imports. is the prediction worthwhile? use a significance level of 0.05.
lemon imports 228 267 358 491 542
crash fatality rate 16.1 16 15.6 15.6 15.1
find the equation of the regression line.
$hat{y} = \square + (\square)x$
(round the y-intercept to three decimal places as needed. round the slope to four decimal places as needed.)

Explanation:

Step1: Calculate necessary sums

First, we need to find the sums of \( x \) (lemon imports), \( y \) (crash fatality rate), \( xy \), and \( x^2 \).

Given \( x \): \( 228, 267, 358, 491, 542 \)

\( y \): \( 16.1, 16, 15.6, 15.6, 15.1 \)

\( n = 5 \) (number of data points)

Sum of \( x \): \( \sum x = 228 + 267 + 358 + 491 + 542 = 1886 \)

Sum of \( y \): \( \sum y = 16.1 + 16 + 15.6 + 15.6 + 15.1 = 78.4 \)

Sum of \( xy \): \( (228\times16.1)+(267\times16)+(358\times15.6)+(491\times15.6)+(542\times15.1) \)

\( 228\times16.1 = 3670.8 \)

\( 267\times16 = 4272 \)

\( 358\times15.6 = 5584.8 \)

\( 491\times15.6 = 7659.6 \)

\( 542\times15.1 = 8184.2 \)

\( \sum xy = 3670.8 + 4272 + 5584.8 + 7659.6 + 8184.2 = 29371.4 \)

Sum of \( x^2 \): \( 228^2 + 267^2 + 358^2 + 491^2 + 542^2 \)

\( 228^2 = 51984 \)

\( 267^2 = 71289 \)

\( 358^2 = 128164 \)

\( 491^2 = 241081 \)

\( 542^2 = 293764 \)

\( \sum x^2 = 51984 + 71289 + 128164 + 241081 + 293764 = 786282 \)

Step2: Calculate slope (\( b \)) and y-intercept (\( a \))

The formula for the slope \( b \) of the regression line is:

\( b = \frac{n\sum xy - \sum x \sum y}{n\sum x^2 - (\sum x)^2} \)

The formula for the y-intercept \( a \) is:

\( a = \bar{y} - b\bar{x} \), where \( \bar{x} = \frac{\sum x}{n} \) and \( \bar{y} = \frac{\sum y}{n} \)

First, calculate \( \bar{x} \) and \( \bar{y} \):

\( \bar{x} = \frac{1886}{5} = 377.2 \)

\( \bar{y} = \frac{78.4}{5} = 15.68 \)

Now calculate \( b \):

\( n\sum xy - \sum x \sum y = 5\times29371.4 - 1886\times78.4 \)

\( 5\times29371.4 = 146857 \)

\( 1886\times78.4 = 1886\times78 + 1886\times0.4 = 147108 + 754.4 = 147862.4 \)

\( 146857 - 147862.4 = -1005.4 \)

\( n\sum x^2 - (\sum x)^2 = 5\times786282 - (1886)^2 \)

\( 5\times786282 = 3931410 \)

\( 1886^2 = (1800 + 86)^2 = 1800^2 + 2\times1800\times86 + 86^2 = 3240000 + 309600 + 7396 = 3556996 \)

\( 3931410 - 3556996 = 374414 \)

So, \( b = \frac{-1005.4}{374414} \approx -0.002685 \) (rounded to four decimal places: -0.0027)

Now calculate \( a \):

\( a = \bar{y} - b\bar{x} = 15.68 - (-0.002685)\times377.2 \)

\( (-0.002685)\times377.2 \approx -1.013 \) (wait, no: \( b \) is negative, so \( -b\bar{x} \) is positive)

Wait, \( b \approx -0.002685 \), so \( -b\bar{x} = 0.002685\times377.2 \approx 1.013 \)

\( a = 15.68 + 1.013 = 16.693 \) (rounded to three decimal places)

Wait, let's recalculate \( b \) more accurately:

\( -1005.4 \div 374414 \approx -0.002685 \), which is approximately -0.0027 when rounded to four decimal places.

And \( a = 15.68 - (-0.002685)\times377.2 = 15.68 + (0.002685\times377.2) \)

\( 0.002685\times377.2 \approx 0.002685\times300 = 0.8055, 0.002685\times77.2 = 0.2073 \), total \( 0.8055 + 0.2073 = 1.0128 \)

So \( a = 15.68 + 1.0128 = 16.6928 \approx 16.693 \)

So the regression equation is \( \hat{y} = 16.693 - 0.0027x \)

Answer:

The equation of the regression line is \( \hat{y} = \boxed{16.693} + (\boxed{-0.0027})x \)