A manufacturer of golf equipment wishes to estimate the number of left-handed golfers. A previous study indicates that the proportion of left-handed golfers is 8%. How large a sample is needed in order to be 98% confident that the sample proportion will not differ from the true proportion by more than 2%?

A sample of 997 is needed.

Step-by-step explanation:

In a sample with a number n of people surveyed with a probability of a success of , and a confidence level of , we have the following confidence interval of proportions.

In which

z is the z-score that has a p-value of .

The margin of error is of:

A previous study indicates that the proportion of left-handed golfers is 8%.

This means that

98% confidence level

So , z is the value of Z that has a p-value of , so .  

How large a sample is needed in order to be 98% confident that the sample proportion will not differ from the true proportion by more than 2%?

This is n for which M = 0.02. So

Rounding up:

A sample of 997 is needed.

you should at least use a calculater

step-by-step explanation:

p

step-by-step explanation: