The average selling price of a smartphone purchased by a ran

The average selling price of a smartphone purchased by a random sample of 4 customers was $305. Assume the population standard deviation was $34.

A) Construct a 95% confidence interval to estimate the average selling price in the population with this sample and what is the lower and upper limit.

B) What is the margin of error for this interval?

Solution

(A) Given a=1-0.95=0.05, Z(0.025) = 1.96 (from standard normal table)

So the lower limit is

xbar - Z*s/vn =305-1.96*34/sqrt(4) =271.68

So the upper limit is

xbar +Z*s/vn =305+1.96*34/sqrt(4) =338.32

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(B) the margin of error = Z*s/vn

=1.96*34/sqrt(4)

=33.32