Question

You would like to rent an unfurnished one–bedroom apartment in Braamfontein, Johannesburg next
year. The mean monthly rent for a random sample of 50 apartments advertised on Property24 is
R2500: Assume that the population standard deviation is R600 and that the monthly rentals are
normally distributed. Create a 90% confidence interval for the mean monthly rent.

Answer 1

n = 50     

x-bar = 2500     

s = 600     

% = 90     

Standard Error, SE = σ/√n =    600 /√50 = 84.85281374

z- score = 1.644853627     

Width of the confidence interval = z * SE =     1.64485362695147 * 84.8528137423857 = 139.5704584

Lower Limit of the confidence interval = x-bar - width =      2500 - 139.570458441201 = 2360.429542

Upper Limit of the confidence interval = x-bar + width =      2500 + 139.570458441201 = 2639.570458

The confidence interval is [2360.43, 2639.57]