Question

You 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 49 gas grills has a mean price of $631.60. Assume the population standard deviation is $55.70.

Answer 1

Solution :

Given that,

=631.60

= 55.70

n = 49

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 * (55.70 / 49  )

= 13.10

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

- E < < + E

631.60 - 13.10 < < 631.60 + 13.10

59.8 < < 644.70

(618.50, 644.70 )

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 * (55.70 / 49  )

= 15.60

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

- E < < + E

631.60 - 15.60 < < 631.60 + 15.60

616.00 < < 647.20

(616.00, 647.20 )

95% confidence interval is​ wider