Answer: 0.0179
Step-by-step explanation:
We know that , the standard error of the sampling distribution of sample proportion (p) is given by :-
[tex]S.E.=\sqrt{\dfrac{p(1-p)}{n}}[/tex]
where , n= sample size
Let p be the proportion of households who have incomes in excess of $60,000 .
As per given , we have
p= 20% = 0.20
n= 500
Then,
[tex]S.E.=\sqrt{\dfrac{0.20(1-0.20)}{500}}=\sqrt{\dfrac{0.20\times0.80}{500}}\\\\=\sqrt{0.00032}\\\\=0.01788854382\approx0.0179[/tex]
Hence, the standard error of the sampling distribution of sample proportion of households who have incomes in excess of $60,000 is 0.0179.
The standard error of the sampling distribution of sample proportion of households who have incomes in excess of $60,000 will be 0.0179.
Since an actual census showed that 20% of the households in Michigan have incomes in excess of $60,000.
So here the standard error should be
[tex]\sqrt (.2\times (1-.2)\div 500) \\\\= 0.0179[/tex]
Hence, the standard error should be 0.0179.
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