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Rudiy27
3 years ago
10

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

r district.
A) If a 95% confidence interval with a margin of error of no more than 0.04 is desired, give a close estimate of the minimum sample size necessary to achieve the desired margin of error.
B) If a 95% confidence interval with a margin of error of no more than 0.02 is desired, give a close estimate of the minimum sample size necessary to achieve the desired margin of error.
Mathematics
1 answer:
mojhsa [17]3 years ago
8 0

Answer:

(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.

Step-by-step explanation:

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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