Question

A simple random sample of 50 items from a population with a standard deviation of 7, resulted in a sample mean of 38.

If required, round your answers to two decimal places.

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

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

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

Answer 1

Solution :

Given that,

= 50

= 7

n = 38

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 * (7 / 38 )

= 1.87

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

- E < < + E

50 - 1.87 < < 50 + 1.87

48.13 < < 51.87

(48.13, 51.87)

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 = 1.960

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

= 1.960 * (7 / 38 )

= 2.23

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

- E < < + E

50 - 2.23 < < 50 + 2.23

47.77 < < 52.23

(47.77, 52.23)

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 * (7 / 38 )

= 2.93

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

- E < < + E

50 - 2.93 < < 50 + 2.93

47.07 < < 52.93

(47.07, 52.93):