Question

In a study of annual apartment rental costs in a city, a random sample of 50 apartments has a mean annual rental cost of $11,535 and a standard deviation of $975. Construct a 99% confidence interval for the mean annual apartment rental cost of all apartments in the city.                                                           [3 marks]

Answer 1

Solution :

Given that,

Point estimate = sample mean = = $11535

Population standard deviation =    = $975
Sample size = n =50

At 99% confidence level the z is ,

  = 1 - 99% = 1 - 0.99 = 0.01

/ 2 = 0.01 / 2 = 0.005

Z/2 = Z0.005 = 2.576   ( Using z table )

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

= 2.576* (975 / 50)

= 355.1939

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

- E < < + E

11535-355.1939 < < 11535+355.1939

11179.8061< < 11890.1939

(11179.8061, 11890.1939 )