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. From a random sample of 37 business​ days, the mean closing price of a certain stock was ​$118.77. Assume the population standard deviation is ​$11.26.

The​ 90% confidence interval is

Answer 1

Solution:
Given that, x̄=$118.77,σ=$11.26,n=37
(1–α)%=90%
α=0.10
α/2=0.05
Zα/2=1.645 ...from standard normal table.
Margin of error=E=Zα/2 ×(σ/√n)
=1.645×(11.26/√37)
=3.0451
Margin of error=E= 3.0451
90% confidence interval for population mean is given as
x̄± Margin of error=(118.77-3.0451,118.77+3.0451)
=(115.7249,121.8151)
Lower limit =115.7249
Upper limit=121.8151
We are 90% confident that the population mean is lies between 115.7249 and 121.8151

Now,
(1–α)%=95%
α=0.05
α/2=0.025
Zα/2=1.96 ...from standard normal table.
Margin of error=E=Zα/2 ×(σ/√n)
=1.96×(11.26/√37)
=3.6282
Margin of error=E= 3.6282
95% confidence interval for population mean is given as
x̄± Margin of error=(118.77-3.6282,118.77+3.6282)
=(115.1418, 122.3982)
Lower limit =115.1418
Upper limit=122.3982
We are 95% confident that the population mean is lies between 115.1418 and 122.3982
As we compare the both confidence intervals we find that 95% confidence interval is more wider than the 90% confidence interval.