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 percent with a standard deviation of 9.3 percent.What is the 95 percent confidence interval for this single-sample t test?
A) [10.47, 15.19]
B) [11.64, 16.36]
C) [12.99, 15.01]
D) [6.94, 11.66]

Question

Quizplus · Accepted Answer

To calculate the 95% confidence interval for a single-sample t test, we can use the formula:
Confidence interval = sample mean ± (t-value x standard error)

First, we need to calculate the t-value:
t = (sample mean - population mean) / (standard deviation / sqrt(sample size))
t = (14 - 12.83) / (9.3 / sqrt(62))
t = 1.30

Using a t-table with 61 degrees of freedom and a 95% confidence level, we find the t-value to be 1.998.

Next, we need to calculate the standard error:
SE = standard deviation / sqrt(sample size)
SE = 9.3 / sqrt(62)
SE = 1.18

Now we can plug in the values and calculate the confidence interval:
CI = 14 ± (1.998 x 1.18)
CI = [11.64, 16.36]

Therefore, the best choice is B because it includes the calculated 95% confidence interval.

Assume That the National Credit Card Interest Rate Is 12