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.

323e

To tackle this problem, follow these steps:

Step 1: Identify the formula to use.
The formula to determine the sample size needed is:
n = [(z-value)^2 * p * q] / E^2

Step 2: Determine the values needed for the formula.
In this case, we have the following values:
- z-value: For a 95% confidence level, use 1.96.
- p: The estimated proportion of obstetricians considering leaving the profession is not provided, so we can assume it is 0.5.
- q: Use the complement of p, which is also 0.5.
- E: The maximum error (margin of error) is given as 3%, which can be written as 0.03 as a decimal.

Step 3: Substitute the values into the formula and calculate.
Using the formula:
n = [(1.96)^2 * 0.5 * 0.5] / 0.03^2
n = [3.8416 * 0.25] / 0.0009
n = 0.9604 / 0.0009
n ≈ 1067.11

Step 4: Round up to the next whole number.
Since you can't have a fraction of a participant, round up to the next whole number.
n ≈ 1068

Step 5: Interpret the result.
Therefore, to find the answer within a + or - 3% error at a 95% confidence level, a sample size of approximately 1068 obstetricians should be taken.

I hope this clear explanation helps you tackle the problem!

To tackle this problem, you can use the formula:

n = [(z-value)^2 * p * q]/E^2

where:
- n represents the sample size needed
- z-value is the standard score found using a z-table, which corresponds to the desired confidence level. In this case, the 95% confidence level corresponds to a z-value of 1.96.
- p represents the estimated proportion of obstetricians considering leaving the profession. The problem does not provide a specific value, so we can assume it to be 0.5 (or 50%).
- q represents the complement of p, which is 1 - p. Therefore, q would also be assumed to be 0.5.
- E represents the maximum error we are willing to accept. The problem specifies a maximum error of 3%, so E would be 0.03.

Plugging in the given values, we have:

n = [(1.96)^2 * 0.5 * 0.5]/0.03^2

Calculating this expression gives us the required sample size. However, it's important to round it up to the next highest whole number, as sample sizes must be whole numbers.

I hope this explanation helps you tackle the problem!