Question

A simple random sample of 70 items from a population with σ, = 7 resulted in a sample mean of 33.

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

Solution :

Given that,

= 33

= 7

n = 70

a) At 90% confidence level the z is ,

  = 1 - 90% = 10 - 0.90 = 0.1

/ 2 = 0.1 / 2 = 0.05

Z_/2 = Z_0.05 = 1.645

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

= 1.645 * (7 / 70)

= 1.38

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

- E < < + E

33 - 1.38 < < 33 + 1.38

31.62 < < 34.38

(31.62, 34.38)

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 * (7 / 70)

= 1.64

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

- E < < + E

33 - 1.64 < < 33 + 1.64

31.36 < < 34.64

(31.36, 34.64)

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/ 70)

= 2.16

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

- E < < + E

33 - 2.16 < < 33 + 2.16

30.84 < < 35.16

(30.84, 35.16)