Question

You want to obtain a sample to estimate the proportion of a population that possess a particular genetic marker. based on previous evidence, you believe approximately p ∗ = 78 % of the population have the genetic marker. you would like to be 95% confident that your estimate is within 1% of the true population proportion. how large of a sample size is required?

Answer 1

To estimate the proportion of a population with a certain genetic marker to within 1% of the true population proportion with 95% confidence, approximately 1462 individuals need to be sampled.

Step-by-step explanation:

To determine the sample size required to estimate the population proportion within a specific margin of error with a given level of confidence, you can use the formula for the sample size of a proportion:

n = (Z^2 * p * (1 - p)) / E^2

where:

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In this case, with a confidence level of 95%, the Z-score (Z) is approximately 1.96 (from the Z-table for the standard normal distribution). The proportion (p) is believed to be 78% or 0.78, and the margin of error (E) is 1% or 0.01.

Plugging the values into the formula, we get:

n = (1.96^2 * 0.78 * (1 - 0.78)) / 0.01^2

After calculating, the required sample size (n) is:

n = (3.8416 * 0.78 * 0.22) / 0.0001
n = (0.658416 * 0.22) / 0.0001
n ≈ 1461.33

Therefore, you would need to sample approximately 1462 individuals (since we round up to the next whole number when dealing with individuals in a sample) to be 95% confident that the estimated population proportion is within 1% of the true population proportion.