The Academy of Orthopedic Surgeons states that 80% of women wear shoes that are too small for their feet. A researcher wants to be 98% confident that this portion is within 3% of the true proportion. How large a sample is necessary?
To calculate the sample size necessary, we need to use a formula for sample size for estimating a population proportion. The formula is:
n = (Z^2 * p * (1-p)) / E^2
Where:
- n is the required sample size
- Z is the Z-score corresponding to the desired confidence level
- p is the estimated proportion
- E is the desired margin of error
In this case, we want to be 98% confident, so the Z-score will be the value that corresponds to a 98% confidence level, which is approximately 2.33 (look this up in a Z-score table or use a calculator).
The estimated proportion is given as 80%, which can be expressed as 0.8.
The desired margin of error is 3% or 0.03.
Now we can substitute these values into the formula:
n = (2.33^2 * 0.8 * (1-0.8)) / 0.03^2
Simplifying the equation:
n = (5.4289 * 0.16) / 0.0009
n = 0.868624 / 0.0009
n ≈ 964.027
Therefore, a sample size of approximately 964 is necessary to achieve a 98% confidence level with a margin of error of 3% when estimating the proportion of women wearing shoes that are too small for their feet.
Try this formula:
n = [(z-value)^2 * p * q]/E^2
= [(2.33)^2 * .8 * .2]/.03^2
I'll let you finish the calculation. Round to the next highest whole number.
Note: n = sample size needed; .8 for p and .2 for q (q = 1 - p). E = maximum error, which is .03 (3%) in the problem. Z-value is found using a z-table (for 98%, the value is approximately 2.33).
I hope this will help.