Question

1. Apartment rental rates: You want to rent an unfurnished one-bedroom apartment for next semester. The mean monthly rent for a random sample of 10 apartments advertised in the local newspaper is $960. Assume that the population standard deviation is $80.

a. Find a 95% confidence interval for the mean monthly rent for unfurnished one-bedroom apartments available for rent in this community. Interpret your result.

b. Find and compare the margin of error for intervals with 90, 95, and 99% confidence. Which one has largest margin of error?

Answer 1

Solution :

Given that,

Point estimate = sample mean = = 960

Population standard deviation =    = 80

Sample size = n = 10

(A)

At 95% confidence level

= 1 - 95%  

= 1 - 0.95 = 0.05

/2 = 0.025

Z/2 = Z0.025 = 1.96

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

= 1.96 * ( 80 /  10 )

= 49.58

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

- E < < + E

960 - 49.58 <   < 960 + 49.58

910.42 <   < 1009.58

(910.42 , 1009.58 )

(b)

At 90% confidence level

= 1 - 90%  

= 1 - 0.90 =0.10

/2 = 0.05

Z/2 = Z0.05 = 1.645

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

= 1.645 * ( 80 /  10 )

= 41.62

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

- E < < + E

960 - 41.62 <   < 960 + 41. 62

918.38 <   < 1001.62

(918.38 , 1001.62 )

At 95% confidence level

= 1 - 95%  

= 1 - 0.95 = 0.05

/2 = 0.025

Z/2 = Z0.025 = 1.96

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

= 1.96 * ( 80 /  10 )

= 49.58

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

- E < < + E

960 - 49.58 <   < 960 + 49.58

910.42 <   < 1009.58

(910.42 , 1009.58 )

At 99% confidence level

= 1 - 99%  

= 1 - 0.99 = 0.01

/2 = 0.005

Z/2 = Z0.005 = 2.576

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

= 2.576 * ( 80 /  10 )

= 65.17

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

- E < < + E

960 - 65.17  <   < 960 + 65.17

894.83 <   < 1025.17

( 894.83, 1025.17 )

Largest  margin of error = 99 % = 65.17