Question

You randomly choose some unfurnished one-bedroom apartments from a large number of advertisements in your local newspaper. You calculated their mean monthly rent is $570 and their standard deviation is $150. Construct the following confidence intervals for the mean monthly rent of all advertised one-bedroom apartments.

• Confidence interval with C=90%, n=10:

Answer 1

Given that,

= 570

s =150

n = 10

Degrees of freedom = df = n - 1 = 10- 1 =9

At 90% confidence level the t is ,

= 1 - 90% = 1 - 0.90 = 0.1

/ 2 = 0.1 / 2 = 0.05

t /2,df = t0.05,9 =1.833    ( using student t table)

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

=1.833 * ( 150/ 10) = 86.9468

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

- E < < + E

570- 86.9468 < < 570+ 86.9468

483.0532 < < 656.9468

( 483.0532 , 656.9468)