posted by Edward .
How do I go about tackling this problem?
The American Medical Association wishes to determine the percentage of obstetricians who are considering leaving the profession becaose of the rapidly increasing number of lawsuits against obstetricians. How large a sample should be taken to find the answer to within + or = 3% at the 95% confidence level?
Try this formula:
n = [(z-value)^2 * p * q]/E^2
Note: I'm using * to mean multiply.
With your values:
= [(1.96)^2 * .5 * .5]/.03^2
Note: Round to the next highest whole number.
Information: n = sample size needed; .5 for p and .5 for q are used if no value is stated in the problem. E = maximum error, which is .03 (3%) in the problem. Z-value is found using a z-table (for 95%, the value is 1.96).
I hope this will help.