Question

A simple random sample of n = 54 provided a sample mean of 22.5 and a sample standard deviation of 4.4.

Develop a 90% confidence interval for the population mean. (Round to two decimal places) __________and__________

Develop a 95% confidence interval for the population mean. (Round to two decimal places) __________and__________

Develop a 99% confidence interval for the population mean. (Round to two decimal places) __________and__________

Answer 1

Solution :

Given that,

Point estimate = sample mean = = 22.5

sample standard deviation = s = 4.4

sample size = n = 54

Degrees of freedom = df = n - 1 = 54 - 1 = 53

a) At 90% confidence level

= 1 - 90%

=1 - 0.90 =0.10

/2 = 0.05

t/2,df = t_0.05,53 = 1.674

Margin of error = E = t_/2,df * (s /n)

= 1.674 * (4.4 / 54)

Margin of error = E = 1.00

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

  ± E

= 22.5  ± 1.00

( 21.50, 23.50 )

b) At 95% confidence level

= 1 - 95%

=1 - 0.95 =0.05

/2 = 0.025

t/2,df = t_0.025,53 = 2.006

Margin of error = E = t_/2,df * (s /n)

= 2.006 * (4.4 / 54)

Margin of error = E = 1.20

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

  ± E

= 22.5  ± 1.20

( 21.30, 23.70 )

c) At 99% confidence level

= 1 - 99%

=1 - 0.99 =0.01

/2 = 0.005

t/2,df = t_0.005,53 = 2.672

Margin of error = E = t_/2,df * (s /n)

= 2.672 * (4.4 / 54)

Margin of error = E = 1.60

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

  ± E

= 22.5  ± 1.60

( 20.90, 24.10)