The estimate of the population proportion is to be within plus or minus .10, with a 99 percent level of confidence. The best estimate of the population proportion is .45. How large a sample is required?
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To determine the required sample size, we can use the formula for calculating sample size for estimating population proportions:
n = (Z^2 * p * (1-p)) / E^2
Where:
- n is the required sample size
- Z is the Z-score corresponding to the desired level of confidence (99% confidence level corresponds to a Z-score of 2.575)
- p is the estimated population proportion (0.45 in this case)
- E is the margin of error (0.10 in this case)
Plugging in the values, we get:
n = (2.575^2 * 0.45 * (1-0.45)) / 0.10^2
Simplifying this equation, we have:
n = (6.6306 * 0.45 * 0.55) / 0.01
n ≈ 1620.2025
Since we can't have a fraction of a sample, we round up the sample size to the next whole number. Therefore, the required sample size is approximately 1621.
Here is one formula you might use for this problem:
n = [(z-value)^2 * p * q]/E^2
With your data:
n = [(2.575)^2 * .45 * .55]/.10^2
I'll let you finish the calculation (round to the next highest whole number).
Note: n = sample size needed; .45 for best estimate of the population proportion; .55 for q, which is (1 - p). E = maximum error, which is .10 from the problem. Z-value is found using a z-table (for 99%, the value is 2.575).
I hope this helps.