Question

1. In 1991 the average interest rate charged by U.S. credit card issuers was 18.8 percent. A financial officer wishes to study whether the increased competition in the credit card business has reduced interest rates. To do this, the officer will test a hypothesis about the current mean interest rate, , charged by all U.S. credit card issuers. To perform the hypothesis, the officer randomly selects n = 15 credit cards and obtains the sample mean interest rate 16.8 percent and the standard deviation of 1.54. Is this an one-sample or two-sample test? Is this test for population mean or proportion? Is this a large sample or small sample? State the null and alternative hypotheses: is this a left-, right-, or two-tailed test? Find the critical value(s) at 0.05 level of significance. Compute the test statistic. Decision: do you reject or fail to reject the null hypothesis? What is the P-value? (A range is sufficient.) State your conclusion to the original research question: Compute the corresponding 99% confidence interval:

Answer 1

1) This is one sample test t test

2) Population mean

3) Small sample as sample size is les than 30

4) Ho: $\mu$ = 18.8

Ha: $\mu$ < 18.8 (Claim)

5) Left tailed test

6) critical values :

t critical value for left tailed test = - 1.76131014

7) Assuming that the data is normally distributed. Sample size is < 30 &  also as the population sd is not given we will use t stat

n = 15

sample mean = 16.8

sample sd = 1.54

t T- s/n 16.8 – 18.8 1.54/15

t = - 5.030

8) As t stat falls in the rejection area, we reject the Null hypothesis.

9) p value = 0.0001

10) As Ho is rejected, we have sufficient evidence to believe that the average interest rate for charged by credit card issues in US has reduced.

11) 99% CI

t critical value = TINV(0.01, 14) = 2.977

$E = t * \frac{s}{\sqrt{n}} = 2.977 * \frac{1.54}{\sqrt{15}}= 1.184$

CI = 16.8 +/- 1.184

CI = 15.616 , 17.984