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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

Respuesta :

Answer:

A. 232.6

Explanation:

Standard deviation of population = $1000

Z value for 90% confidence interval is 1.645

width = z * [tex]\frac{standard deviation}{\sqrt{sample size} }[/tex]

width = 1.645 * [tex]\frac{1000}{\sqrt{50} }[/tex] = 232.6

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