Question

ou 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. Which interval is wider? If convenient, use technology to construct the confidence intervals.

A random sample of

3939

gas grills has a mean price of

$632.10632.10.

Assume the population standard deviation is

$55.6055.60.

The 90% confidence interval is (____,___)

Answer 1

Solution :

Given that,

= 632.10

= 55.60

n = 39

(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 * (55.60 / 39)

= 6.25

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

- E < < + E

632.10 - 11.65 < < 632.10 + 11.65

620.45 < < 643.75

(620.45 , 643.75)

(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 * (55.60 / 39)

= 17.45

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

- E < < + E

632.10 - 17.45 < < 632.10 + 17.45

614.65 < < 649.55

The 95% confidence interval is : ($614.65 , $649.55)

95% confidence interval is wider .