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%.

To determine the sample size needed to estimate the number of left-handed golfers with a given level of confidence and margin of error, we can use the formula for sample size calculation for proportions.

The formula is:
n = (Z^2 * p * q) / 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 from previous study
- q is 1 minus the estimated proportion (since we are estimating the proportion of non-left-handed golfers)
- E is the maximum acceptable margin of error

In this case, the confidence level is 90% (which corresponds to a Z-score of approximately 1.645), the estimated proportion from previous study is 10% (or 0.10), and the maximum acceptable margin of error is 2% (or 0.02). Therefore, we can calculate the required sample size as follows:

n = (1.645^2 * 0.10 * (1 - 0.10)) / 0.02^2

n = (2.706025 * 0.09) / 0.0004

n ≈ 60.150625 / 0.0004

n ≈ 150,376.5625

Since you can't have a fractional sample size, you would need a sample size of at least 150,377 to be 90% confident that the sample proportion will not differ from the true proportion by more than 2%.