To calculate the required sample size for estimating a proportion with a desired level of confidence and margin of error, the following formula can be used:
n = (Z^2 * p * q) / E^2
Where:
- n = required sample size
- Z = Z-score corresponding to the desired confidence level (for 95% confidence level, Z = 1.96)
- p = estimated population proportion (0.20 in this case)
- q = 1 - p
- E = desired margin of error (0.06 in this case)
Plugging in the values:
n = (1.96^2 * 0.20 * 0.80) / 0.06^2
n = (3.8416 * 0.16) / 0.0036
n = 0.614656 / 0.0036
n ≈ 170.18
Therefore, a sample size of approximately 171 patients would be needed to estimate the proportion of trauma patients who die in the hospital with a 95% confidence level and a margin of error of 0.06, given that the estimated population proportion is 0.20.
A health planning agency wishes to know, for a certain geographic region, what proportion of patients admitted to hospitals for the treatment of trauma die in the hospital. A 95% confidence interval is desired, the width of the interval must be 0.06, and the population proportion, from other evidence, is estimated to be .20. How large a sample is needed?
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