Question

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

A simple random sample with

n = 59

provided a sample mean of 26.5 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.

(b)

Develop a 95% confidence interval for the population mean.

(c)

Develop a 99% confidence interval for the population mean.

Answer 1

Solution :

Given that,

(a)

t /2,df = 1.672

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

= 1.672 * ( 4.4/ 59)

Margin of error = E = 1.0

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

- E < < + E

26.5 - 1.0 < < 26.5 + 1.0

26.5 < < 27.5

(26.5 , 27.5)

(b)

t /2,df = 2.002

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

= 2.002 * (4.4 / 59)

Margin of error = E = 1.1

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

- E < < + E

26.5 - 1.1 < < 26.5 + 1.1

25.4 < < 27.6

(25.4 , 27.6)

(c)

t /2,df = 2.663

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

= 2.663 * (4.4 / 59)

Margin of error = E = 1.5

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

- E < < + E

26.5 - 1.5 < < 26.5 + 1.5

25 < < 28

(25 , 28)