Question

Your local newspaper contains a large number of advertisements for unfurnished one-bedroom apartments. You choose 16 at random and calculate that their mean monthly rent is $508 and that the standard deviation of their rents is $78. Please show all work and write legibly Construct a 90% confidence interval for the mean monthly rent of all advertised one-bedroom apartments. (Fill in the blanks below and give your answers to 2 decimal places.) The 90% 90%  confidence interval is ($__________ , $ ____________).

Answer 1

Solution :

Given that,

= 508

s = 78

n = 16

Degrees of freedom = df = n - 1 = 16 - 1 = 15

At 90% confidence level the t is ,

= 1 - 90% = 1 - 0.90 = 0.10

/ 2 = 0.10 / 2 = 0.05

t _/2,df = t_0.05,15 = 2.947

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

= 2.947 * (78 / 16)

= 57.47

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

- E < < + E

508 - 57.47 < < 508 + 57.47

450.53 < < 565.47

The 90% confidence interval is ($450.53 , $565.47)