Question

Each bound should be rounded to three decimal places.

A random sample of ?=100 observations produced a mean of ?⎯⎯⎯=32 with a standard deviation of ?=4

(a) Find a 95% confidence interval for ?μ
Lower-bound:  

Upper-bound:

(b) Find a 90% confidence interval for ?μ
Lower-bound:  

Upper-bound:

(c) Find a 99% confidence interval for ?μ
Lower-bound:  

Upper-bound:

Answer 1

Solution :

Given that,

Point estimate = sample mean = = 32

sample standard deviation = s = 4

sample size = n = 100

Degrees of freedom = df = n - 1 = 100 - 1 = 99

a) At 95% confidence level

= 1 - 95%

=1 - 0.95 =0.05

/2 = 0.025

t/2,df = t0.025,99 = 1.984

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

= 1.984 * ( 4/ 100)

Margin of error = E = 0.79

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

  ± E  

= 32  ± 0.79

= ( 31.21, 32.79 )

lower limit = 31.21

upper limit = 32.79

b) At 90% confidence level

= 1 - 90%

=1 - 0.90 =0.10

/2 = 0.05

t/2,df = t0.05,99 = 1.660

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

= 1.660 * ( 4/ 100)

Margin of error = E = 0.66

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

  ± E  

= 32  ± 0.66

= ( 31.34, 32.66 )

lower limit = 31.34

upper limit = 32.66

c) At 99% confidence level

= 1 - 99%

=1 - 0.99 =0.01

/2 = 0.005

t/2,df = t0.005,99 = 2.626

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

= 2.626 * ( 4/ 100)

Margin of error = E = 1.05

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

  ± E  

= 32  ± 1.05

= ( 30.95, 33.05 )

lower limit = 30.95

upper limit = 33.05