Question

A bank manager wants to know the mean amount of mortgage paid per month by homeowners in an area. A random sample of 118 homeowners selected from this area showed that they pay an average of $1581 per month for their mortgages. The population standard deviation of such mortgages is $209.

Find a 95% confidence interval for the mean amount of mortgage paid per month by all homeowners in this area.

Round your answers to two decimal places.

______ to ________ dollars

Answer 1

Solution :

Given that,

Point estimate = sample mean =     =1581

Population standard deviation =    = 209

Sample size n =118

At 95% confidence level the z is ,

= 1 - 95% = 1 - 0.95 = 0.05

/ 2 = 0.05 / 2 = 0.025

Z/2 = Z0.025 = 1.96   ( Using z table )

Margin of error = E =   Z/2    * ( /n)
=1.96 * ( 209 / 118 )

= 37.71
At 95% confidence interval
is,

- E < < + E

1581 - 37.71 <   <1581 + 37.71

154.29 <   < 1618.71

___154.29___ to __1618.71______ dollars