Question

A simple random sample with n=52 provided a sample mean of 24.0 and a sample standard deviation of 4.5.

a. Develop a 90% confidence interval for the population mean (to 1 decimal).

  ,  

b. Develop a 95% confidence interval for the population mean (to 1 decimal).

  ,  

c. Develop a 99% confidence interval for the population mean (to 1 decimal).

  ,  

d. What happens to the margin of error and the confidence interval as the confidence level is increased?

Answer 1

Solution :

a.

t _/2,df = 1.675

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

= 1.675 * (4.5 / 52)

Margin of error = E = 1.0

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

- E < < + E

24.0 - 1.0 < < 24.0 + 1.0

23.0 < < 25.0

A 90% confidence interval for the population mean :(23.0 , 25.0)

b.

t _/2,df = 2.008

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

= 2.008 * (4.5 / 52)

Margin of error = E = 1.3

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

- E < < + E

24.0 - 1.3 < < 24.0 + 1.3

22.7 < < 25.3

A 95% confidence interval for the population mean :(22.7 , 25.3)

c.

t _/2,df = 2.676

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

= 2.676 * (4.5 / 52)

Margin of error = E = 1.7

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

- E < < + E

24.0 - 1.7 < < 24.0 + 1.7

22.3 < < 25.7

A 99% confidence interval for the population mean : (22.3, 25.7)

d.

Margin of error and confidance interval also increased