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. Interpret the results and compare the widths of the confidence intervals.

From a random sample of 74 ​dates, the mean record high daily temperature in a certain city has a mean of 84.20°F. Assume the population standard deviation is

15.06°F.

The​ 90% confidence interval is

The​ 95% confidence interval is

Answer 1

Solution :

Given that,

= 84.20

= 15.06

n = 74

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* (15.06 / 74 )

= 2.88

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

- E < < + E

84.20 - 2.88 < < 84.20 + 2.88

81.32 < < 87.08

(81.32 , 87.08)

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 * (15.06 / 74 )

= 3.43

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

- E < < + E

84.20 - 3.43 < < 84.20 + 3.43

80.77 < < 87.63

( 80.77 , 87.63)