Question

 

A simple random sample of 50 items from a population with

σ = 8

resulted in a sample mean of 32. (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

Solution :

Given that,

Point estimate = sample mean = = 32

Population standard deviation = = 8

Sample size = n = 50

(a)

At 90% confidence level the z is ,

   = 1 - 90% = 1 - 0.90 = 0.10

/ 2 = 0.10 / 2 = 0.05

Z_/2 = Z_0.05 = 1.645

Margin of error = E = Z_/2* ( /n)

= 1.645 * (8 / 50)

= 1.86

At 90% confidence interval estimate of the population mean is,

- E < < + E

32 - 1.86 < < 32 + 1.86

30.14 < < 33.86

A 90% confidence interval for the population mean 30.14 to 33.86

(b)

At 95% confidence level the z is ,

= 1 - 95% = 1 - 0.95 = 0.05

/ 2 = 0.05 / 2 = 0.025

Z_/2 = Z_0.025 = 1.96

Margin of error = E = Z_/2* ( /n)

= 1.96 * (8 / 50)

= 2.22

At 95% confidence interval estimate of the population mean is,

- E < < + E

32 - 2.22 < < 32 + 2.22

29.78 < < 34.22

A 95% confidence interval for the population mean 29.78 to 34.22

(c)

At 99% confidence level the z is ,

= 1 - 99% = 1 - 0.99 = 0.01

/ 2 = 0.01 / 2 = 0.005

Z_/2 = Z_0.005 = 2.576

Margin of error = E = Z_/2* ( /n)

= 2.576 * (8 / 50)

= 2.91

At 99% confidence interval estimate of the population mean is,

- E < < + E

32 - 2.91 < < 32 + 2.91

29.09 < < 34.91

A 99% confidence interval for the population mean 29.09 to 34.91