Question

7. Do heavier cars really use more gasoline? Suppose a car is chosen at random. Let x be the weight of the car (in hundreds of pounds), and let y be the miles per gallon (mpg) At the 5% level of significance, test the claim is that heavier cars use more gasoline. Remember that a claim of correlation would be that X 29 46 29 47 23 40 34 52 Y 30 21 22 14 29 27 22 34 a. State Ho and Hi, and note which is equal to the original claim. H1: b. Use technology to calculate the value of the sample correlation coefficient r. (Round your value for r to be three decimal places). You may also want to note the value for P for (c) and the values of m and b for (e) [final answer] r= c. By using the P-value method, do we reject or fail to reject the null hypothesis? final answer]: P- [final answer] Comparison that supports your conclusion: [final answer] Conclusion about the null hypothesis: d. What is the conclusion about the original claim? [final answer] Conclusion about the original claim: e. Find the equation of the least-squares line y-mr+b (Round your value for m and b to three decimal places) final answer] Equation of line:

Answer 1

a) We are claiming that heavier cars use more gasoline, i.e. the correlation between weight and gasoline usage is positive, or in other words, the correlation between weight and miles per gallon is negative.

H₀: $\rho$ >= 0

H₁: $\rho$ < 0 (this is the original claim)

b) Correlation coefficient can be calculated using the CORREL() function in Excel and comes out to be:

r = -0.147

c) P-value for r = -0.147 and n=8 is

P = 0.728

Since, P > 0.05

We cannot reject the null hypothesis at the 0.05 significance level.

d) Hence, the original claim that heavier cars use more gasoline is not significant at the 5% significance level.

e) Running regression on X vs Y in Excel (go to Data tab -> Data Analysis -> Regression, choosing X and X-column and Y as Y-column), we get the following output:

	Coefficients	Standard Error	t Stat	P-value	Lower 95%	Upper 95%
Intercept	28.26809651	9.594245321	2.94635957	0.02573426	4.791823937	51.74436909
X	-0.090482574	0.247763646	-0.36519713	0.72749221	-0.696738376	0.515773229

Hence, the equation of best-fit regression line is:

Y = - 0.905 * X + 28.268