Question

listed below are annual data for various years. the data are weights (metric tons) of imported lemons and car crash fatality rates per 100,000 population. construct a scatterplot, find the value of the linear correlation coefficient r, and find the p - value using α = 0.05. is there sufficient evidence to conclude that there is a linear correlation between lemon imports and crash fatality rates? do the results suggest that imported lemons cause car fatalities?

lemon imports	228	265	359	482	534

the linear correlation coefficient is r = \\(\square\\).
(round to three decimal places as needed.)

the test statistic is t = \\(\square\\).
(round to three decimal places as needed.)

the p - value is \\(\square\\).
(round to three decimal places as needed.)

because the p - value is \\(\square\\) than the significance level 0.05, there \\(\square\\) sufficient evidence to support the claim that there is a linear correlation between lemon imports and crash fatality rates for a significance level of α = 0.05.

do the results suggest that imported lemons cause car fatalities?
a. the results suggest that imported lemons cause car fatalities.

Explanation:

Step1: Define Variables

Let \( x \) be Lemon Imports (\( 228, 265, 359, 482, 534 \)) and \( y \) be Crash Fatality Rate (\( 15.8, 15.7, 15.4, 15.3, 14.9 \)). \( n = 5 \).

Step2: Calculate Sums

• \( \sum x = 228 + 265 + 359 + 482 + 534 = 1868 \)
• \( \sum y = 15.8 + 15.7 + 15.4 + 15.3 + 14.9 = 77.1 \)
• \( \sum xy = (228×15.8)+(265×15.7)+(359×15.4)+(482×15.3)+(534×14.9) \)

\( = 3602.4 + 4160.5 + 5528.6 + 7374.6 + 7956.6 = 28622.7 \)

• \( \sum x^2 = 228^2 + 265^2 + 359^2 + 482^2 + 534^2 \)

\( = 51984 + 70225 + 128881 + 232324 + 285156 = 768570 \)

• \( \sum y^2 = 15.8^2 + 15.7^2 + 15.4^2 + 15.3^2 + 14.9^2 \)

\( = 249.64 + 246.49 + 237.16 + 234.09 + 222.01 = 1189.39 \)

Step3: Compute Correlation Coefficient \( r \)

Use formula:
\( r = \frac{n\sum xy - \sum x \sum y}{\sqrt{[n\sum x^2 - (\sum x)^2][n\sum y^2 - (\sum y)^2]}} \)

Substitute values:
Numerator: \( 5×28622.7 - 1868×77.1 = 143113.5 - 143922.8 = -809.3 \)
Denominator part 1: \( 5×768570 - 1868^2 = 3842850 - 3490524 = 352326 \)
Denominator part 2: \( 5×1189.39 - 77.1^2 = 5946.95 - 5944.41 = 2.54 \)
Denominator: \( \sqrt{352326×2.54} \approx \sqrt{894908.04} \approx 946.0 \) (approx)
\( r = \frac{-809.3}{946.0} \approx -0.855 \)

Step4: Test Statistic \( t \)

Formula: \( t = r\sqrt{\frac{n - 2}{1 - r^2}} \)
Substitute \( r = -0.855 \), \( n = 5 \):
\( t = -0.855\sqrt{\frac{3}{1 - 0.731}} = -0.855\sqrt{\frac{3}{0.269}} \approx -0.855×3.37 \approx -2.881 \) (absolute value for P - value)

Step5: Find P - value

Degrees of freedom \( df = n - 2 = 3 \). Two - tailed test.
Using t - distribution table or calculator, for \( |t| = 2.881 \), \( df = 3 \), P - value ≈ 0.057 (approx, more accurately with calculator: ~0.057)

Step6: Compare P - value with \( \alpha = 0.05 \)

P - value (≈0.057) > 0.05, so not sufficient evidence.

Step7: Causation

Correlation ≠ causation. Even if correlated, lemons don’t cause crashes (likely coincidence or lurking variable).

Answer:

• Linear correlation coefficient \( r \approx \boldsymbol{-0.855} \)
• Test statistic \( t \approx \boldsymbol{-2.881} \) (or 2.881, absolute value)
• P - value ≈ \(\boldsymbol{0.057}\) (varies slightly by method)
• P - value is greater than 0.05, so is not sufficient evidence.
• Do results suggest causation? No (correlation ≠ causation).