Question

1. A loan officer compares the interest rates for 48-month fixed-rate auto loans and 48-month variable-rate auto loans. Two independent, random samples of auto loan rates are selected. A sample of eight 48-month fixed−rate auto loans had the following loan rates (all written as percentages): 8.75 7.63 7.26 9.43 7.86 7.20 8.09 8.60 while a sample of five 48−month variable−rate auto loans had loan rates as follows: 7.60 7.00 6.79 7.36 6.99

(a) Set up the null and alternative hypotheses needed to determine whether the mean rates for 48-month fixed-rate and variable-rate auto loans differ. H0: µf − µv = 48 versus Ha: µf − µv ≠ 48

(b) Use the data analysis tool in Excel to test the hypotheses you set up in part a. Assuming that the normality and equal variances assumptions hold, use the Excel output and critical values to test these hypotheses by setting α equal to .10, .05, .01, and .001. How much evidence is there that the mean rates for 48−month fixed and variable−rate auto loans differ? (Round your answer to 3 decimal places.) t = with 11 df Reject H0 at α = , but not at α =

(c) Use the p−value to test these hypotheses by setting α equal to .10, .05, .01, and .001. How much evidence is there that the mean rates for 48−month fixed− and variable−rate auto loans differ? (Round your answer to 4 decimal places.) p−value = Reject H0 at α = but not at α =

(d) Use a hypothesis test to establish that the difference between the mean rates for fixed− and variable−rate 48−month auto loans exceeds .4. Use α equal to .05. (Round your t answer to 3 decimal places and other answers to 1 decimal place.) H0: µf − µv versus Ha: µf − µv t = H0 with a = .05.

Answer 1

(a) The hypothesis being tested is:

H0: µ1 = µ2

H1: µ1 ≠ µ2

(b) t = 2.572 with 11 df Reject H0 at α = 0.1, 0.05, but not at α = 0.01, 0.001

(c) p−value = 0.0259, Reject H0 at α = 0.1, 0.05, but not at α = 0.01, 0.001

(d) t = 1.494

Fixed	Variable
8.1025	7.1480	mean
0.7779	0.3257	std. dev.
8	5	n
	11	df
	0.95450	difference (Fixed - Variable)
	0.42362	pooled variance
	0.65086	pooled std. dev.
	0.37105	standard error of difference
	0	hypothesized difference
	2.572	t
	.0259	p-value (two-tailed)

Fixed	Variable
8.1025	7.1480	mean
0.7779	0.3257	std. dev.
8	5	n
	11	df
	0.95450	difference (Fixed - Variable)
	0.42362	pooled variance
	0.65086	pooled std. dev.
	0.37105	standard error of difference
	0.4	hypothesized difference
	1.494	t
	.0816	p-value (one-tailed, upper)