A manufacturer of golf equipment wishes to estimate the number of left-handed golfers. How large a sample is needed in order to be 90% confident that the sample proportion will not differ from the true proportion by more than 2%? A previous study indicates that the proportion of left-handed golfers is 10%.
i ưant
To determine the sample size needed, we can use the formula for sample size calculation for estimating a proportion:
n = (Z^2 * p * (1-p)) / E^2
where:
- n is the sample size needed
- Z is the Z-score corresponding to the desired confidence level (90% confidence level corresponds to a Z-score of 1.645)
- p is the estimated proportion of left-handed golfers from the previous study (10% or 0.10 as a decimal)
- E is the maximum allowable margin of error (2% or 0.02 as a decimal)
Plugging in the values, we can calculate the sample size:
n = (1.645^2 * 0.10 * (1-0.10)) / 0.02^2
n = (2.706025 * 0.10 * 0.9) / 0.0004
n = 0.24354 / 0.0004
n ≈ 608.85
Therefore, a sample size of approximately 609 would be needed in order to be 90% confident that the sample proportion will not differ from the true proportion by more than 2%.