Assume that appartment rents are Normally distributed. You randomly choose 16 advertisements for appartments from your local newspaper and calculate that their mean rent is $508 with a standard devation of $78. Construct a 90% confidence interval for the mean monthly rent of all appartments.
Given:
n = 16, = 508, standard deviation (S) = 78, Confidence level (C) = 0.90, = 1 - 0.90 = 0.10
90% Confidence interval:
Where,
Critical value:
Therefore,
We are 90% confidence that the population mean is lies in that interval.
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