An economist is interested in studying the incomes of consumers in a 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 upper end in a 99% confidence interval for the average income?
A)$15,052
B)$15,141
C)$15,330
D)$15,364

Question

Quizplus · Accepted Answer

The formula for a confidence interval for the population mean is:
$\bar{X} \pm z_{\alpha/2}\frac{\sigma}{\sqrt{n}}$
where $\bar{X}$ is the sample mean, $\sigma$ is the population standard deviation, $n$ is the sample size, and $z_{\alpha/2}$ is the z-score corresponding to the desired confidence level.
Substituting the given values, we get:
$15,000 \pm z_{0.005}\frac{1,000}{\sqrt{50}}$
Using a standard normal distribution table or calculator, we find that $z_{0.005}=2.576$. Substituting this value, we get:
$15,000 \pm 2.576 	imes \frac{1,000}{\sqrt{50}}$
Simplifying, we get:
$15,000 \pm 181.89$
Thus, the 99% confidence interval for the average income is $(14,818.11, 15,181.89)$. The upper end is $15,181.89$, which is closest to answer choice D ($15,364$).

An Economist Is Interested in Studying the Incomes of Consumers