Question

1. Suppose we collect a random sample of n = 9 and find an average income of $49,000 with a sample standard deviation s = $12,000. Provide each of the following using this information.

1. A 95% confidence interval estimate of the population mean µ. What is the value for the margin of error? Interpret your results.
2. A 90% confidence interval estimate of the population mean µ.
3. A 99% confidence interval estimate of the population mean µ.

Answer 1

Solution :

Given that,

Point estimate = sample mean = = $ 49000

sample standard deviation = s = $ 12000

sample size = n = 9

Degrees of freedom = df = n - 1 = 9 - 1 = 8

a) At 95% confidence level

= 1 - 95%

=1 - 0.95 =0.05

/2 = 0.025

t/2,df = t_0.025,8 = 2.306

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

= 2.306 * ( 12000/ 9)

Margin of error = E = $ 9224

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

  ± E  

= $ 49000 ± $ 9224

= ( $ 39776, $ 58224 )

We are 95% confidence that the true mean collect income between $ 39776 and $ 58224.

b) At 90% confidence level

= 1 - 90%

=1 - 0.90 =0.10

/2 = 0.05

t/2,df = t_0.05,8 = 1.860

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

= 1.860 * ( 12000/ 9)

Margin of error = E = $ 7440

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

  ± E  

= $ 49000 ± $ 7440

= ( $ 41560, $ 56440 )

We are 90% confidence that the true mean collect income between $ 41560 and $ 56440.

c) At 99% confidence level

= 1 - 99%

=1 - 0.99 =0.01

/2 = 0.005

t/2,df = t_0.005,8 = 3.355

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

= 3.355 * ( 12000/ 9)

Margin of error = E = $ 13420

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

  ± E  

= $ 49000 ± $ 13420

= ( $ 35580, $ 62420 )

We are 99% confidence that the true mean collect income between $ 35580 and $ 62420.