Question

You may need to use the appropriate appendix table or technology to answer this question.

A simple random sample with

n = 54

provided a sample mean of 23.5 and a sample standard deviation of 4.4. (Round your answers to one decimal place.)

(a)

Develop a 90% confidence interval for the population mean.

to

(b)

Develop a 95% confidence interval for the population mean.

to

(c)

Develop a 99% confidence interval for the population mean.

Answer 1

solution :

Given that,

Point estimate = sample mean = = 23.5

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 = 1.674

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

= 1.674* (4.4 / 54)

Margin of error = E = 1.0

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

  ± E  

23.5 ± 1.0

( 22.5 , 24.5 )

b) At 95% confidence level

= 1 - 95%

=1 - 0.95 =0.05

/2 = 0.025

t/2,df = 2.006

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

= 2.006* (4.4 / 54)

Margin of error = E = 1.2

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

  ± E  

23.5 ± 1.2

( 22.2 , 24.7 )

c) At 99% confidence level

= 1 - 99%

=1 - 0.99 =0.01

/2 = 0.005

t/2,df = 2.672

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

= 2.672* (4.4 / 54)

Margin of error = E = 1.6

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

  ± E  

23.5 ± 1.6

( 21.9 , 25.1 )