Question

use the standard normal distribution or the t - distribution to construct a 95% confidence interval for the population mean. justify your decision. if neither distribution can be used, explain why. interpret the results.
in a random sample of 13 mortgage institutions, the mean interest rate was 3.48% and the standard deviation was 0.36%. assume the interest rates are normally distributed.
which distribution should be used to construct the confidence interval?
a. use a normal distribution because the interest rates are normally distributed and σ is known.
b. use a normal distribution because n < 30 and the interest rates are normally distributed.
c. use a t - distribution because it is a random sample, σ is unknown, and the interest rates are normally distributed.
d. use a t - distribution because the interest rates are normally distributed and σ is known.
e. cannot use the standard normal distribution or the t - distribution because σ is unknown, n < 30, and the interest rates are not normally distributed.

Explanation:

Step1: Check conditions for distributions

• For normal distribution: We need to know the population standard deviation \(\sigma\) (not given here, we have sample standard deviation \(s = 0.36\)).
• For t - distribution: The requirements are a random sample, the population is normally distributed (given that interest rates are normally distributed), and the population standard deviation \(\sigma\) is unknown (we only know \(s\)). Here \(n=13\) (random sample), population is normal, \(\sigma\) is unknown.

Answer:

C. Use a t - distribution because it is a random sample, \(\sigma\) is unknown, and the interest rates are normally distributed.