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 (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 = versus Ha: µf − µv ≠

(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

Soln

Let us call For Fixed-rate auto Loans Group as f and Variable-rate auto Loans Group as v

a)

Null and Alternate Hypothesis

H₀: µ_f − µ_v = 0

H_a: µ_f − µ_v <> 0

b)

Steps in Excel

• Import data in Excel
• Click on Data Tab and then on Data Analysis Tool
• Select data and put hypothesized value as 0
• Click OK
• Repeat above steps but change alpha values to get different value for different alpha

Excel Output

Result

Since for alpha = 0.01, the test statistic (2.572) is less than critical value (3.106), we fail to reject the null hypothesis

Hence we reject null hypothesis when alpha = 0.05 or 0.1 but not when alpha = 0.01

c)

Result

Since for alpha = 0.01, the p-value (0.026) is greater than 0.01, we fail to reject the null hypothesis

Hence we reject null hypothesis when alpha = 0.05 or 0.1 but not when alpha = 0.01

d)

Null and Alternate Hypothesis

H₀: µ_f − µ_v = 0.4

H_a: µ_f − µ_v > 0.4

Test Statistic

t-Test: Two-Sample Assuming Equal Variances Fixed 8.103 0.605 Variable 7.148 0.106 Mean Variance Observations Pooled Variance Hypothesized Mean Difference df t Stat P(T<=t) one-tail t Critical one-tail P(T<=t) two-tail t Critical two-tail 0.424 0.4 11 1.494 0.082 1.796 0.163 2.201

t = 1.494

p-value = 0.082

Result and Conclusion

Since the p-value (0.082) is greater than 0.05, we fail to reject the null hypothesis