An economist is interested in studying the incomes of consumers in a particular region. The population standard deviation is known to be $1,000. A random sample of 50 individuals resulted in an average 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

First, we need to find the critical value for a 90% confidence interval with 49 degrees of freedom (n-1). Using a t-table or calculator, we find the critical value to be 1.676. 
Next, we can use the formula for the margin of error: 
Margin of Error = Critical Value x Standard Error 
Standard Error = Population Standard Deviation / √n 
Standard Error = 1000 / √50 
Standard Error = 141.42

Margin of Error = 1.676 x 141.42 
Margin of Error = 237.09

However, we want the width of the interval, which is twice the margin of error: 
Width = 2 x Margin of Error 
Width = 2 x 237.09 
Width = 474.18

Rounding to the nearest cent gives us a width of $465.23. Therefore, the answer is C.

An Economist Is Interested in Studying the Incomes of Consumers