Question

From a random sample of potential voters in an upcoming election 47% indicated they intended to vote for candidate R A 95 percent confidence interval was constructed from the sample and the margin error for the estimate was 5%

Answer 1

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

  ( $x^{-} - \frac{2s}{\sqrt{n} }$        , $(x^{-} +\frac{2s}{\sqrt{n} },$)

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 = $\frac{2s}{\sqrt{n} }$ = 5% =0.05

Step 2:-

The 95% confidence intervals of given margin error

  ( $x^{-} - \frac{2s}{\sqrt{n} }$        , $(x^{-} +\frac{2s}{\sqrt{n} },$)

$(0.47 -0.05,0.47+0.05})$

(0.42 , 0.52)