Assume that the national credit card interest rate is 12.83 percent.A study of 81 college students finds that their average interest rate is 15 percent with a standard deviation of 8.7 percent.What is the 95 percent confidence interval for this single-sample t test?
A) [10.54, 19.54]
B) [11.64, 16.36]
C) [12.99, 15.01]
D) [13.08, 16.92]

Question

Quizplus · Accepted Answer

The 95% confidence interval for a single-sample t-test can be calculated using the formula: $ \bar{x} \pm t_{\alpha/2} \cdot \frac{s}{\sqrt{n}} $, where $ \bar{x} $ is the sample mean, $ t_{\alpha/2} $ is the t-score for the desired confidence level (which depends on the degrees of freedom, $ n-1 $), $ s $ is the sample standard deviation, and $ n $ is the sample size. Given $ \bar{x} = 15\% $, $ s = 8.7\% $, and $ n = 81 $, and knowing that the degrees of freedom $ df = n-1 = 80 $ which for large samples approximates the z-score for 95% confidence (about 1.96), the calculation yields the interval [13.08, 16.92].

Assume That the National Credit Card Interest Rate Is 12