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September 20, 2014

September 20, 2014

Posted by **Edward** on Sunday, April 29, 2007 at 4:24pm.

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.

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