An economist is interested in studying the incomes of consumers in a particular region.The population standard deviation is known to be $2,000.A random sample of 50 individuals resulted in an average income of $30,000.
-What is the upper end point in a 99% confidence interval for the average income?
A) $30,282
B) $30,1042
C) $30,728
D) $30,660

Question

Quizplus · Accepted Answer

We can use the formula for the confidence interval of the mean:

CI = X̄ ± Z* (σ/√n)

Where:
X̄ = sample mean = $30,000
Z* = z-score for 99% confidence level = 2.576 (from the standard normal distribution table)
σ = population standard deviation = $2,000
n = sample size = 50

Plugging in the values, we get:

CI = 30,000 ± 2.576 * (2,000/√50)
CI = 30,000 ± 578.07
CI = ($29,421.93, $30,578.07)

Since we want the upper end point of the confidence interval, we take the second value:
Upper end point = $30,000 + $578.07 = $30,728

Therefore, the answer is C) $30,728.

An Economist Is Interested in Studying the Incomes of Consumers