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 232 265 357 483 530 Crash Fatality Rate 15.9 15.7 15.4 15.2 14.9 What are the null and alternative hypotheses? O A. Ho:p*0 Hip=0 0 C. 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 ов. OC. OD. A AY 177 AY 17- A 17- 17 O 18- 16- O 18- 16- 15- 15- o 15 15- 14- 14 14 14 200 400 300 200 400 800 200 400 600 200 400 600 The linear correlation coefficient r is (Round to three decimal places as needed.) The test statistict is (Round to three decimal places as needed.) The P-value is a (Round to three decimal places as needed.) Because the P-value is V than the significance level 0.05, there sufficient evidence to support the claim that there is a linear correlation between lemon imports and crash fatality rates for a significance level of a = 0.05. Do the results suggest that imported lemons cause car fatalities? O A. The results suggest that an increase in imported lemons causes in an increase in car fatality rates. OB. The results do not suggest any cause-effect relationship between the two variables. OC. The results suggest that an increase in imported lemons causes car fatality rates to remain the same OD. The results suggest that imported lemons cause car fatalities. Click to select your answer(s).

Answer 1

Scattered plot is 16 15.8 15.6 15.4 Series1 15.2 in 14.8 0 100 200 300 400 500 600 Option A is correct The data points are falling from the top hence there is negative correlation between the varaibles x and y

X 12 y 15.9 y2 252.81 Σ*y 3688.8 232 53824 265 15.7 70225 246.49 4160.5 357 15.4 127449 237.16 5497.8 483 15.2 233289 231.04 7341.6 530 14.9 280900 222.01 7897 1867 77.1 765687 1189.51 28585.7 Σx=1867, Σx2 =765687, Σy=77.1, Σy] =1189.51, Σxy=28585.7 S=Σx2 - Σα)2 =68549.2 m Sy=Σy2 . (Σv2 =0.6280 m Σα)(Σν). = Sxy=xy- -203.44 m

Correlation coefficient r Sxy = Sex Syy -203.44 V68549.2*0.628 = -0.980517 = -0.981 Test hypothesis are Ho: P = 0 H: P70 ryn-2 Test statistic t= = (-0.981) 5-2 V1-(-0.981) = -8.758 1-72 p-value=P(t|>-8.758)=0.0031 Since p-value<0.05 hence reject the null hypothesis Because the p-value is less than the significance level 0.05 there is sufficient evidence to support that there is linear correlation between ...... The result does not suggest any cause –effect relationship between the two variables Option B is correct