Question

A random sample of n=100 observations produced a mean of x⎯⎯=30 with a standard deviation of s=5.

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

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

(c) Find a 90% confidence interval for μ
Lower-bound:  Upper-bound:

Answer 1

Solution :

Given that,

= 30

s = 5

n = 100

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

a ) At 99% confidence level the t is ,

= 1 - 99% = 1 - 0.99 = 0.01

/ 2 = 0.01 / 2 = 0.005

_{t /2 df} = t_0.005,99 = 2.626

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

= 2.626 * (5 / 100)

                              = 1.313

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

- E < < + E

30 - 1.313 < < 30 + 1.313

28.687 < < 31.313

Lower-bound: =28.687

  Upper-bound: = 31.313

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_0.025,99 = 1.984

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

= 1.984 * (5 / 100)

                             = 0.992

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

- E < < + E

30 - 0.992 < < 30 + 0.992

29.008 < < 30.992

Lower-bound: = 29.008

  Upper-bound: = 30.992

c ) At 90% confidence level the t is ,

= 1 - 90% = 1 - 0.90 = 0.1

/ 2 = 0.1 / 2 = 0.05

t _/2,df = t_0.05,99 = 1.660

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

= 1.660 * (5 / 100)

                              = 0.830

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

- E < < + E

30 - 0.830 < < 30 + 0.830

29.170 < < 30.830

Lower-bound: =29.170

  Upper-bound: = 30.830