Question

A simple random sample with

n = 56

provided a sample mean of 29.5 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.

Answer 1

Solution :

Given that,

Point estimate = sample mean = = 29.5

sample standard deviation = s = 4.4

sample size = n = 56

Degrees of freedom = df = n - 1 = 56 - 1 = 55

a) At 90% confidence level

= 1 - 90%  

= 1 - 0.90 =0.10

/2 = 0.05

t/2,df = 1.673

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

= 1.673 * ( 4.4/ 56)

Margin of error = E = 1.67

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

  ± E

29.5  ± 1.67

(27.83 , 31.17)

b) At 95% confidence level

= 1 - 95%  

= 1 - 0.95 =0.05

/2 = 0.025

t/2,df = 2.004

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

= 2.004 * ( 4.4/ 56)

Margin of error = E = 1.18

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

  ± E

28.32 ± 1.18

(27.14 , 29.5)

c) At 99% confidence level

= 1 - 99%  

= 1 - 0.99 =0.01

/2 = 0.005

t/2,df = 2.668

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

= 2.668 * ( 4.4/ 56)

Margin of error = E = 1.57

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

  ± E

28.32 ± 1.57

(26.75 , 29.89)