Question

You are given the sample mean and the population standard deviation. Use this information to construct the​ 90% and​ 95% confidence intervals for the population mean. Interpret the results and compare the widths of the confidence intervals. If​ convenient, use technology to construct the confidence intervals.

A random sample of 60 home theater systems has a mean price of ​$147.00. Assume the population standard deviation is ​$19.40.

Construct a​ 90% confidence interval for the population mean. The​ 90% confidence interval is _____. ​(Round to two decimal places as​ needed.)

pt 2:

You are given the sample mean and the population standard deviation. Use this information to construct the​ 90% and​ 95% confidence intervals for the population mean. Interpret the results and compare the widths of the confidence intervals.

From a random sample of 42 business​ days, the mean closing price of a certain stock was ​$119.95. Assume the population standard deviation is ​$9.93

The​ 90% confidence interval is ____. ​(Round to two decimal places as​ needed.)

pt 3:

Find the critical value (tc) for the confidence level c=0.9898 and sample size n=77.

tc=_____. ​(Round to the nearest thousandth as​ needed.)

Answer 1

Solution:

pt : 1 :- Given that n = 60, x = 147, σ = 19.40

90% Confidence level for Z = 1.645

95% Confidence level for Z = 1.96

=> 90% Confidence interval for the population = X +/- Z*σ/sqrt(n)
= 147 +/- 1.645*19.40/sqrt(60)
= (142.88 , 151.12)

=> 95% Confidence interval for the population = X +/- Z*σ/sqrt(n)
= 147 +/- 1.96*19.40/sqrt(60)
= (142.09 , 151.91)

pt 2 :- Given that n = 42, x = 119.95, σ = 9.93

=> 90% Confidence interval for the population = X +/- Z*σ/sqrt(n)
= 119.95 +/- 1.645*9.93/sqrt(42)
= (117.43 , 122.47)

=> 95% Confidence interval for the population = X +/- Z*σ/sqrt(n)
= 119.95 +/- 1.96*9.93/sqrt(42)
= (116.95 , 122.95)

pt 3 :- the confidence level c=0.98 and sample size n=7.

df = n-1 = 7-1 = 6

tc = 3.143