Question

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?

Answer

A. 232.60
B. 364.30
C. 465.23
D. 728.60

Answer 1

A. 232.6

Step-by-step explanation:

Standard deviation of population = $1000

Z value for 90% confidence interval is 1.645

width = z *
`(standard deviation)/(√(sample size) )`

width = 1.645 *
`(1000)/(√(50) )` = 232.6