Question

2.

A simple random sample with n = 54 provided a sample mean of 22.5 and a

sample standard deviation of 4.4.

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.

d.

What happens to the margin of error and the confidence interval as the

confidence level is increased?

Answer 1

2)

Solution :

Given that,

= 22.5

s = 4.4

n = 54

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

(a)

At 90% confidence level the is ,

= 1 - 90% = 1 - 0.90 = 0.10

/ 2 = 0.10 / 2 = 0.05

t _/2,df = t_0.025,53 = 1.674

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

= 1.674 * (4.4 / 54)

= 0.6

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

- E < < + E

22.5 - 0.6 < < 22.5 + 0.6

21.9 < < 23.1

(21.9 , 23.1)

(b)

At 95% confidence level the t is ,

= 1 - 95% = 1 - 0.95 = 0.05

/ 2 = 0.05 / 2 = 0.025

t _/2,df = t_0.005,53 = 2.006

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

= 2.006 * (4.4 / 54)

= 1.2

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

- E < < + E

22.5 - 1.2 < < 22.5 + 1.2

21.3 < < 23.7

(21.3 , 23.7)

(c)

At 99% confidence level the t is ,

= 1 - 99% = 1 - 0.99 = 0.01

/ 2 = 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)

= 1.6

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

- E < < + E

22.5 - 1.6 < < 22.5 + 1.6

20.9 < < 24.1

(20.9 , 24.1)