An economist is interested in studying the incomes of consumers in a particular country. The population standard deviation is known to be $1,000. A random sample of 50 individuals resulted in a mean income of $15,000. What is the width of the 90% confidence interval?
A) $232.60
B) $364.30
C) $465.23
D) $728.60

Question

Quizplus · Accepted Answer

We use the formula for the margin of error of a confidence interval:

Margin of Error = z*(sigma/sqrt(n))

Where:
z = the z-score corresponding to the desired confidence level (90% in this case), which is 1.645
sigma = the population standard deviation = $1,000
n = sample size = 50
sqrt = square root

Margin of Error = 1.645*($1,000/sqrt(50)) = $232.58 (rounded to two decimal places)

The width of the confidence interval is double the margin of error:

Width = 2*Margin of Error = 2*$232.58 = $465.16 (rounded to two decimal places)

Therefore, the correct choice is C, $465.23, which is the closest option to the calculated width.

An Economist Is Interested in Studying the Incomes of Consumers