Question

Use the sample information x⎯⎯x¯ = 40, σ = 3, n = 18 to calculate the following confidence intervals for μ assuming the sample is from a normal population.

(a) 90 percent confidence. (Round your answers to 4 decimal places.)
  
The 90% confidence interval is from  to  

(b) 95 percent confidence. (Round your answers to 4 decimal places.)
  
The 95% confidence interval is from  to  

(c) 99 percent confidence. (Round your answers to 4 decimal places.)
  
The 99% confidence interval is from  to  

(d) Describe how the intervals change as you increase the confidence level.
  

Answer 1

Solution :

Given that,

Point estimate = sample mean = = 40

Population standard deviation = = 3

Sample size = n = 18

a)

At 90% confidence level the z is ,

= 1 - 90% = 1 - 0.90 = 0.10

/ 2 = 0.10 / 2 = 0.05

Z_/2 = Z_0.05 = 1.645

Margin of error = E = Z_/2* ( /n)

= 1.645 * (3 / 18 )

= 1.1632

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

- E < < + E

40 - 1.1632 < < 40 + 1.1632

38.8368 < < 41.1632

The 90% confidence interval is (38.8368 , 41.1632)

b)

At 95% confidence level the z is ,

= 1 - 95% = 1 - 0.95 = 0.05

/ 2 = 0.05 / 2 = 0.025

Z_/2 = Z_0.025 = 1.96

Margin of error = E = Z_/2* ( /n)

= 1.96 * (3 / 18 )

= 1.3859

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

- E < < + E

40 - 1.3859 < < 40 + 1.3859

38.6141< < 41.3859

The 95% confidence interval is (38.6141 , 41.3859)

c)

At 99% confidence level the z is ,

= 1 - 99% = 1 - 0.99 = 0.01

/ 2 = 0.01 / 2 = 0.005

Z_/2 = Z_0.005 = 2.576

Margin of error = E = Z_/2* ( /n)

= 2.576 * (3 / 18 )

= 1.8215

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

- E < < + E

40 - 1.8215 < < 40 + 1.8215

38.1785 < < 41.8215

The 99% confidence interval is (38.1785 , 41.8215)

d)

Increasing the confidence will increase the margin of error resulting in a wider interval.