Question

Assume that the national credit card interest rate is 12.83 percent. A study of 62 college students finds that their average interest rate is 14% with a standard deviation of 9.3%. What is the 95% confidence interval for this single-sample t test?

Group of answer choices

[10.47, 15.19]

[11.64, 16.36]

[12.99, 15.01]

[6.94, 11.66]

Answer 1

The 95% confidence interval for the average credit card interest rate among college students is [11.64%, 16.36%], calculated using a sample mean of 14%, a standard deviation of 9.3%, a sample size of 62, and an approximate critical t-value of 2 for a two-tailed test with 61 degrees of freedom.

Step-by-step explanation:

To calculate the 95% confidence interval for the average credit card interest rate among college students, you use a formula that incorporates the sample mean, the sample standard deviation, and the appropriate t-value for a 95% confidence level. Because we have a relatively small sample (n=62), which is less than 30, we will use the t-distribution rather than the z-distribution.

First, calculate the standard error of the mean (SEM) using the formula:

• SEM = s / √n

Where s is the sample standard deviation and n is the sample size. Inserting the values we get:

• SEM = 9.3% / √62 = 1.18%

Next, determine the critical t-value for a two-tailed test with 61 degrees of freedom (n-1 = 62-1) at the 95% confidence level. Based on the t-distribution table, we'll assume an approximate t-value of 2 (Considering you don't have access to statistical tables, this value may have a small error. Always consult the t-distribution table for exact values).

Then the confidence interval is calculated as:

• Confidence interval = sample mean ± (t-value * SEM)
• Confidence interval = 14% ± (2 * 1.18%)
• Confidence interval = 14% ± 2.36%
• Confidence interval = [11.64%, 16.36%]

Thus, the 95% confidence interval for the average interest rate of the college students' credit cards is [11.64%, 16.36%].