Question

You want to rent an unfurnished one-bedroom apartment for next semester. You take a random sample of 10 apartments advertised in the local newspaper and record the rental rates. Here are the rents (in dollars per month). 455, 615, 305, 335, 510, 650, 590, 660, 675, 425

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

Answer 1

Solution:

Given that

The sample mean is

Mean   = (x / n)

= (455+615+305+335+510+650+590+660+675+425 / 10)

= 5220 /10

= 522

Mean   = 522

Sample standard deviation is s

s =   1/(n-1)(x -  )²

⁼  1/9 (455- 522 )²+ (615- 522 )²+ (305 - 522 )²+ (335 -522 )² +(510 - 522 )²+ (650 - 522 )² + (590- 522 )²  + (660 - 522 )²+ (675 - 522 )²+ (425 - 522 )²

= 1/9 (4489+8649+47089+34969+144+16384+4624 +19044 +23409 +9409 )

= 168210/9

= 18690

= 136.71

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

At 95% confidence level the t is ,

  = 1 - 95% = 1 - 0.95 = 0.05

/ 2 = 0.05 / 2 = 0.025

t _/2,df = t_0.025,9 =2.262

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

= 2.262 * (136.71 / 10)

= 97.79

Margin of error = 97.79

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

- E < < + E

522 - 97.79 < < 522 + 97.79

424.21 < < 619.79

(424.21, 619.79)