Question

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: 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% Figure 10.7 Excel Output of Testing the Equality of Mean Loan Rates for Fixed and Variable 48-Month Auto Loans t-Test: Two-Sample Assuming Equal Variances Fixed-Rate (%) Variable-Rate (%) Mean 8.10250 7.14800 Variance .605079 .10607 Observations 8 5 Pooled Variance .423621 Hypothesized Mean Difference 0 df 11 t Stat 2.572442 P(T<=t) one-tail .012969 t Critical one-tail 1.79588 P(T<=t) two-tail .025937 t Critical two-tail 2.200985 (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 = versus Ha: µf − µv ≠ (b) Figure 10.7 gives the Excel output of using the equal variances procedure 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 = (Click to select) , but not at α = (Click to select) (Click to select) evidence that rates differ. (c) Figure 10.7 gives the p−value for testing the hypotheses you set up in part a. 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 = (Click to select) but not at α = (Click to select) (Click to select) evidence. (d) Calculate a 95 percent confidence interval for the difference between the mean rates for fixed− and variable−rate 48−month auto loans. Can we be 95 percent confident that the difference between these means exceeds .4 percent? (Round your answers to 3 decimal places. Negative value should be indicated by a minus sign.) Confidence interval = [ , ]. (Click to select) , the entire interval is (Click to select) .40. (e) Use a hypothesis test to establish that the difference between the mean rates for fixed− and variable−rate 48−month auto loans exceeds .4 percent. Use α equal to .05. (Round your t answer to 3 decimal places and other answers to 1 decimal place.) H0: µf − µv (Click to select) versus Ha: µf − µv (Click to select) t = (Click to select) H0 with a = .05.

Answer 1

Result:

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: 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% Figure 10.7 Excel Output of Testing the Equality of Mean Loan Rates for Fixed and Variable 48-Month Auto Loans t-Test: Two-Sample Assuming Equal Variances

Fixed-Rate (%) Variable-Rate (%)

Mean 8.10250 7.14800

Variance .605079 .10607

Observations 8 5

Pooled Variance .423621

Hypothesized Mean Difference 0

df 11

t Stat 2.572442

P(T<=t) one-tail .012969

t Critical one-tail 1.79588

P(T<=t) two-tail .025937

t Critical two-tail 2.200985

(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 =0 versus Ha: µf − µv ≠0

(b) Figure 10.7 gives the Excel output of using the equal variances procedure 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 =2.572 with 11 df Reject H0 at α = 0.05 , but not at α = 0.01 (sufficient) evidence that rates differ.

(c) Figure 10.7 gives the p−value for testing the hypotheses you set up in part a. 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 =0.0259 Reject H0 at α = 0.05 () but not at α = 0.01 ) (sufficient) evidence.

(d) Calculate a 95 percent confidence interval for the difference between the mean rates for fixed− and variable−rate 48−month auto loans. Can we be 95 percent confident that the difference between these means exceeds .4 percent? (Round your answers to 3 decimal places. Negative value should be indicated by a minus sign.)

Confidence interval = [ 0.138, 1.771 ]. (Click to select) , the entire interval is (contains) .40.

for the Difference Between Two Means
Data
Confidence Level	95%
Intermediate Calculations
Degrees of Freedom	11
t Value	2.2010
Interval Half Width	0.8167
Confidence Interval
Interval Lower Limit	0.1378
Interval Upper Limit	1.7712

(e) Use a hypothesis test to establish that the difference between the mean rates for fixed− and variable−rate 48−month auto loans exceeds .4 percent. Use α equal to .05. (Round your t answer to 3 decimal places and other answers to 1 decimal place.)

H0: µf − µv =0 (Click to select) versus Ha: µf − µv >0.4

(Click to select) t = 1.494 (Fail to Reject) H0 with a = .05.

Pooled-Variance t Test for the Difference Between Two Means
(assumes equal population variances)
Data
Hypothesized Difference	0.4
Level of Significance	0.05
Population 1 Sample
Sample Size	8
Sample Mean	8.1025
Sample Standard Deviation	0.777867965
Population 2 Sample
Sample Size	5
Sample Mean	7.148
Sample Standard Deviation	0.325683896
Intermediate Calculations
Population 1 Sample Degrees of Freedom	7
Population 2 Sample Degrees of Freedom	4
Total Degrees of Freedom	11
Pooled Variance	0.4236
Standard Error	0.3710
Difference in Sample Means	0.9545
t Test Statistic	1.4944
Upper-Tail Test
Upper Critical Value	1.7959
p-Value	0.0816
Do not reject the null hypothesis