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3 years ago

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will give brainliest!!!!!! will give brainliest!!!!!! will give brainliest!!!!!! will give brainliest!!!!!! will give brainliest

!!!!!! will give brainliest!!!!!! will give brainliest!!!!!! will give brainliest!!!!!! will give brainliest!!!!!! will give brainliest!!!!!!

2 answers:

aliya0001 [1]3 years ago

4 0

The answer to this question is D

Rom4ik [11]3 years ago

3 0

I think the answer is d I did the math but I’m not shure

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A candidate for a US Representative seat from Indiana hires a polling firm to gauge her percentage of support among voters in he

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(A) The minimum sample size required achieve the margin of error of 0.04 is 601.

(B) The minimum sample size required achieve a margin of error of 0.02 is 2401.

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Let us assume that the percentage of support for the candidate, among voters in her district, is 50%.

(A)

The margin of error, <em>MOE</em> = 0.04.

The formula for margin of error is:

$MOE=z_{\alpha /2}\sqrt{\frac{p(1-p)}{n}}$

The critical value of <em>z</em> for 95% confidence interval is: $z_{\alpha/2}=1.96$

Compute the minimum sample size required as follows:

$MOE=z_{\alpha /2}\sqrt{\frac{p(1-p)}{n}}\\0.04=1.96\times \sqrt{\frac{0.50(1-0.50)}{n}}\\(\frac{0.04}{1.96})^{2} =\frac{0.50(1-0.50)}{n}\\n=600.25\approx 601$

Thus, the minimum sample size required achieve the margin of error of 0.04 is 601.

(B)

The margin of error, <em>MOE</em> = 0.02.

The formula for margin of error is:

$MOE=z_{\alpha /2}\sqrt{\frac{p(1-p)}{n}}$

The critical value of <em>z</em> for 95% confidence interval is: $z_{\alpha/2}=1.96$

Compute the minimum sample size required as follows:

$MOE=z_{\alpha /2}\sqrt{\frac{p(1-p)}{n}}\\0.02=1.96\times \sqrt{\frac{0.50(1-0.50)}{n}}\\(\frac{0.02}{1.96})^{2} =\frac{0.50(1-0.50)}{n}\\n=2401.00\approx 2401$

Thus, the minimum sample size required achieve a margin of error of 0.02 is 2401.

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3 years ago