Question

5. You want to rent an unfurnished two-bedroom apartment in Edmonton. The mean monthly rent for a random sample of 10 apartments advertised on Kijiji is $1150 with a standard deviation of $50. Assume monthly 2-bedroom apartment rents in Edmonton follow a normal distribution.

a) For each of the given confidence levels, find the corresponding confidence interval for the mean monthly rent for an unfurnished two-bedroom apartment in Edmonton. Suppose, also, that a full survey of the population of two-bedroom apartments taken at the same time reveals that the true mean monthly rent for an unfurnished two-bedroom apartment in Edmonton is $1200. Do the calculated intervals capture µ? Why do you think each interval cover or not cover µ?

Confidence Level	Calculation x±tα2,n-1sn	Interval	Does interval cover $1200?	Why this result?
(5m) 95%
(5m) 99%

b) (4m) Find the probability that you would observe a value as extreme as (as low as or lower than) $1150 a month if the hypothesis that the mean rent for an unfurnished two-bedroom rent in Edmonton is $1200 a month is actually true. Do you think that there is strong evidence that the true mean is now lower than $1200? Why?

Probability:

Strong Evidence that the true mean now lower than $1200? _____ Why? _________________________________

Answer 1

Sample mean = 1150

Standard deviation = 50

number of samples = 10

Significance level = 1 - confidence interval

a) significance level = 1-0.95 = 0.05

The critical value from two sided z-table is 1.96

To calculate mean limits with 95% confidence interval,

1150 - 1.96*50/(10^)-0.5 < mean < 1150 + 1.96*50/(10^)-0.5

1119 < mean < 1181

interval of 62, and it doesn't cover 1200

b) with significance level = 1-0.99 = 0.01

The critical value from two sided z-table is 2.58

To calculate mean limits with 99% confidence interval,

1150 - 2.58*50/(10^)-0.5 < mean < 1150 + 2.58*50/(10^)-0.5

1109.2 < mean < 1190.7

interval of 81.8, and it doesn't cover 1200