Question

Use the standard normal distribution or the​ t-distribution to construct a 99​% 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 3.59​% and the standard deviation was 0.42​%. Assume the interest rates are normally distributed.

Select the correct choice below​ and, if​ necessary, fill in any answer boxes to complete your choice.

A. The 99​% confidence interval is (_____​, _____ ​).

​(Round to two decimal places as​ needed.)

B.Neither distribution can be used to construct the confidence interval.

In a random sample of 25 ​people, the mean commute time to work was 30.8 minutes and the standard deviation was7.2 minutes. Assume the population is normally distributed and use a​ t-distribution to construct a 80​% confidence interval for the population mean μ.

What is the margin of error of μ​? ______

​(Round to one decimal place as​ needed.)

Answer 1

1)

Z - value for 99% interval = 2.58

Confidence interval: Mean - Z*SD/Sqrt(n) , Mean + Z*SD/Sqrt(n)

Confidence interval: 3.59% - 2.58*0.42% / Sqrt(13) , 3.59% + 2.58*0.42% / Sqrt(13)

Confidence interval: 3.29%, 3.89%

2)

n = 25, df = n - 1 = 24

Z(80%,24) = 1.318

Confidence interval: Mean - Z*SD/Sqrt(n) , Mean + Z*SD/Sqrt(n)

Confidence interval: 30.8 - 1.318*7.2/Sqrt(25) , 30.8 + 1.318*7.2/Sqrt(25)

Confidence interval: 28.9, 32.7

Margin of error =  Z*SD/Sqrt(n)

=  1.318*7.2/Sqrt(25)

= 1.9