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.

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.
select the correct choice below and, if necessary, fill in any answer boxes to complete your choice.
a. the 95% confidence interval is (, ).
(round to two decimal places as needed.)
b. neither distribution can be used to construct the confidence interval.

Explanation:

Step1: Determine the degrees of freedom

The degrees of freedom \(df=n - 1\), where \(n = 13\). So \(df=13-1 = 12\).

Step2: Find the critical value \(t_{\alpha/2}\)

For a 95% confidence interval, \(\alpha=1 - 0.95=0.05\), and \(\alpha/2=0.025\). Using the t - distribution table or a calculator, \(t_{0.025,12}=2.179\).

Step3: Calculate the margin of error \(E\)

The formula for the margin of error for a t - distribution is \(E=t_{\alpha/2}\frac{s}{\sqrt{n}}\), where \(s = 0.36\), \(n = 13\).

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Step4: Calculate the confidence interval

The confidence interval for the population mean \(\mu\) is \(\bar{x}-E<\mu<\bar{x}+E\), where \(\bar{x}=3.48\).

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Answer:

A. The 95% confidence interval is \((3.26,3.70)\)