QUESTION IMAGE
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 21 mortgage institutions, the mean interest rate was 3.54% and the standard deviation was 0.44%. assume the interest rates are normally distributed.
which distribution should be used to construct the confidence interval?
a. use a t - distribution because it is a random sample, σ is unknown, and the interest rates are normally distributed.
b. use a normal distribution because n < 30 and the interest rates are normally distributed.
c. use a normal distribution because the interest rates are normally distributed and σ is known.
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.
Step1: Determine the degrees of freedom
The degrees of freedom \(df=n - 1\). Given \(n = 21\), so \(df=21-1 = 20\).
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, for \(df = 20\) and \(\alpha/2=0.025\), \(t_{\alpha/2}=2.086\).
Step3: Calculate the margin of error \(E\)
The formula for the margin of error for a t - distribution is \(E=t_{\alpha/2}\frac{s}{\sqrt{n}}\). Given \(\bar{x}=3.54\), \(s = 0.44\), \(n = 21\), and \(t_{\alpha/2}=2.086\).
Step4: Calculate the confidence interval
The confidence interval formula is \(\bar{x}-E<\mu<\bar{x}+E\).
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A. The 95% confidence interval is \((3.34,3.74)\)