Question

A simple random sample of 90 items from a population with ? = 7 resulted in a sample mean of 31. If required, round your answers to two decimal places.

a. Provide a 90% confidence interval for the population mean.
______ to ______

b. Provide a 95% confidence interval for the population mean.
_____ to _____

c. Provide a 99% confidence interval for the population mean.

_____ to _____

Answer 1

Given, sig=7 and xbar =31

And n = sample size =90

a) At 90% confidence Za=1.645

So therefore 90% confidence interval is

xbar - Za*(sig/sqrt(n)) < Mu < bar + Za*(sig/sqrt(n))

Where Mu is Population Mean

31 - 1.645*(7/sqrt(90)) < Mu < 31+1.645*(7/sqrt(90))

31 - 1.22 < Mu < 31+1.22

29.78 < Mu < 32.22

b) At 95% confidence Za=1.96

So therefore 95% confidence interval is

xbar - Za*(sig/sqrt(n)) < Mu < bar + Za*(sig/sqrt(n))

Where Mu is Population Mean

31 - 1.96*(7/sqrt(90)) < Mu < 31+1.96*(7/sqrt(90))

31 - 1.45 < Mu < 31+1.45

29.55 < Mu < 32.45

C) At 99% confidence Za=2.33

So therefore 99% confidence interval is

xbar - Za*(sig/sqrt(n)) < Mu < bar + Za*(sig/sqrt(n))

Where Mu is Population Mean

31 - 2.33*(7/sqrt(90)) < Mu < 31+2.33*(7/sqrt(90))

31 - 1.73 < Mu < 31+1.73

29.27 < Mu < 32.73