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 100 individuals resulted in a mean income of $15,000. What is the width of the 90% confidence interval?
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At 90% confidence interval the critical value is z0.05 = 1.645
The 90% confidence interval is
+/-
z0.05 * 
= 15000 +/- 1.645 * 1000/
= 15000 +/- 164.5
= 14835.5, 15164.5
The width of the interval = 15164.5 - 14835.5 = 329
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