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How large a sample n would we need to estimate p with a margin of error of .07 with 95% confidence

User Shawnay
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Final answer:

A sample size of approximately 7840 is needed to estimate the population proportion with a margin of error of 0.07 with 95% confidence, using the error bound formula for proportions with p value assumed at 0.5 for maximum sample size requirement.

Step-by-step explanation:

To determine how large a sample size n is needed to estimate a population proportion p with a margin of error of 0.07 with 95% confidence, we can use the error bound formula for the proportion. In this context, the Z-score corresponding to a 95% confidence level is approximately 1.96.

The error bound (EBM) formula for a proportion is given by:

EBM = Z √{·[(p)(1-p)/n]}

Where:

Solving the EBM formula for n to achieve a margin of error of 0.07:

0.07 = 1.96 √{·[p(1-p)/n]}

n = [1.96² √{·(p(1-p))}] / 0.07²

Since we don't have a specific p value, we use 0.5 for p to maximize the product p(1-p), which gives the largest sample size.

n = [1.96² √{·(0.5)(0.5)}] / 0.07²

n = 196² / 0.07²

n = 38416 / 0.0049

n ≈ 7840

A sample size of approximately 7840 would be needed to estimate p with a margin of error of 0.07 with 95% confidence.

User Sabreen
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