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 width of the 90% confidence interval?
A)$232.60
B)$364.30
C)$465.23
D)$728.60

Question

Quizplus · Accepted Answer

The formula for the confidence interval of a population mean is:
$CI=\bar{x}\pm z_{\frac{\alpha}{2}}\cdot\frac{\sigma}{\sqrt{n}}$
Where $\bar{x}$ is the sample mean, $\sigma$ is the population standard deviation, $n$ is the sample size, $z_{\frac{\alpha}{2}}$ is the z-score corresponding to the level of confidence, and $\alpha$ is the significance level.
Substituting the given values:
$CI=15,000\pm z_{0.05}\cdot\frac{1,000}{\sqrt{50}}$
The z-score corresponding to the 90% confidence level can be found from a standard normal distribution table or calculator. It is approximately 1.645.
$CI=15,000\pm1.645\cdot\frac{1,000}{\sqrt{50}}\approx15,000\pm232.6$
Therefore, the width of the 90% confidence interval is $2\cdot232.6=465.23$. The best choice is (C).

An Economist Is Interested in Studying the Incomes of Consumers