David wants to rent an unfurnished apartment for next semester. He took a random

Question

David wants to rent an unfurnished apartment for next semester. He took a random sample of 9 apartments adversised in the local Halifax paper, and recorded the rental rates. The rents (in $ per month) were: 500 650 600 505 450 550 515 495 640

c) A student at the University of Calgary carried out a similar study, but using a sample of 16 apartments. The value of the test statistic for that study was 2.71. What would be the p-value to test against the alternative that the mean rent is not equal to $500.

Explanation / Answer

Mean = 545
Standard deviation =69.9553

a)
The t critical value for 90% confidence interval is (with 8 degrees of freedom) : 1.860

Sample mean = 545
Standard deviation = 68.9553
Standard error of mean = / n
Standard error of mean = 68.9553 / 9
SE = 68.9553/3
Standard error of mean 22.9851
90% Confidence interval 545-(22.9851)(1.86) and 545+(22.9851)(1.86)

(502.25, 587.75)

b)
H0: = 500
HA: 500
Sample mean = 545
Standard deviation = 68.9553
Standard error of mean = s / n
Standard error of mean = 68.9553 / 9
SE = 68.9553/3
Standard error of mean 22.9851
t = (xbar- ) / SE
t = (545-500) / 22.9851
t = 1.9578

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