Answer:
The 95% confidence intervals of given margin error
(0.42 , 0.52)
Explanation:
Step 1:-
The 95% confidence intervals of given margin error
x⁻ ± M.E
The 95% confidence intervals of given margin error
( [tex]x^{-} - \frac{2s}{\sqrt{n} }[/tex] , [tex](x^{-} +\frac{2s}{\sqrt{n} },[/tex])
Now given random sample of potential voters in an upcoming election 47%
That is The sample of the mean ( x⁻ ) = 0.47
the margin error for the estimate was 5%
That is margin error of the sample M.E = [tex]\frac{2s}{\sqrt{n} }[/tex] = 5% =0.05
Step 2:-
The 95% confidence intervals of given margin error
( [tex]x^{-} - \frac{2s}{\sqrt{n} }[/tex] , [tex](x^{-} +\frac{2s}{\sqrt{n} },[/tex])
[tex](0.47 -0.05,0.47+0.05})[/tex]
(0.42 , 0.52)
The interval of the 95% confidence interval is (42%,53%)
The given parameters are:
The 95% confidence intervals of margin of error is calculated using:
[tex]CI = (\bar x \pm E)[/tex]
So, we have:
[tex]CI = (47\% \pm 5\%)[/tex]
Expand the above equation
[tex]CI = (47\% - 5\%,47\% + 5\%)[/tex]
[tex]CI = (42\%,53\%)[/tex]
Hence, the interval of the 95% confidence interval is (42%,53%)
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