Answer:
2401
Step-by-step explanation:
Given that we want to estimate average income in a population.
The standard deviation of income in the population is $1,000
We want confidence interval around our estimate to be +/- $40
i.e Margin of error = 40
We know that margin of error = Std error * Z critical for 95%
i.e. [tex]40 = 1.96 * std error[/tex]
Std error = [tex]\frac{std dev}{\sqrt{n} } \\=\frac{1000}{\sqrt{n} }[/tex]
Together we get
[tex]40=1.96(\frac{1000}{\sqrt{n} } )\\\sqrt{n} =\frac{1000*1.96}{40} \\=49\\n=2401\\[/tex]
