Question

Let a random sample of 100 homes sold yields a sample mean sale price of $100,000 and a sample standard deviation of $5,000. Find a 99% confidence interval for the average sale price given the information provided above.

Calculate the following:

1) Margin of error = Answer

2) x̄ ± margin error = Answer < μ < Answer

Table1 -

Common Z-values for confidence intervals

Confidence Level Zα/2

90% 1.645

95% 1.96

99% 2.58

Answer 1

Solution :

Given that,

1 ) = $100,000

s = $5,000

n = 100

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

At 99% confidence level the z is ,

= 1 - 99% = 1 - 0.99 = 0.01

/ 2 = 0.01 / 2 = 0.005

t _/2,df = t_0.005,99 =1.660

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

= 1.660 * (5,000/ 100)

Margin of erro = 830

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

- E < < + E

100000 - 830< < 100000 + 830

99170 < <100830

2 ) Using standard normal table,

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

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 = 1.960

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