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 a = 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 230 266 358 482 534 Crash Fatality Rate 15.8 15.6 15.5 15.2 14.9 What are the null and alternative hypotheses? O A. Ho: p=0 Hip#0 0C. Hp-p= 0 H:p<0 OB. Ho: p= 0 H:p> 0 OD. Ho: p=0 Hip=0 Construct a scatterplot. Choose the correct graph below. OA. OB AY 17- OC. OD y Q y y 17-1 Q 17- 17- 16- Q o 16- co a 16- o o O 2 o o O o o o 15- LY 15- 15- LY o 15- C LY х 14 14+ 14 0 200 400 600 200 400 14+ 0 600 200 400 600 0 200 400 600 The linear correlation coefficient ris (Round to three decimal places as needed.) The test statistic tis (Round to three decimal places as needed.) The P-value is (Round to three decimal places as needed.)

Answer 1

Solution:

Given data:

x	y
230	15.8
266	15.6
358	15.5
482	15.2
534	14.9

Null and alternative hypotheses:

$H0:\rho=0$

$Ha:\rho\neq 0$

Scatter plot is,

15.9 15.8 15.7 15.6 15.5 15.4 15.3 15.2 15.1 15 14.9 14.8 200 300 400 500 600 х

Correlation coefficient can be find using excel function, =CORREL(A2:A6,B2:B6)

Hence, linear correlation coefficient -0.976

The sample size is n = 5 so then the number of degrees of freedom is df = n-2 = 5 - 2 = 3.

Test statistic is:

n-2 t=r1 1- p2 -0.976 5 - 2 1--0.9762 = -7.763

P-value approach:

p = P(|t| > -7.763) = 0.004 ...Using excel function, =TDIST(7.763,3,2)

Because the p-value is less than the significance level 0.05, there is sufficient evidence to support the claim that there is a linear correlation between lemon imports and crash fatality rates for a significance level of 0.05

The results suggest that imported lemons cause car fatalities.

Done