As a major manufacturer of golf clubs, a major corporation wants to estimate the proportion of golfers who are right-handed. assume that the manufacturer has an estimate found from a previous study in scotland, which suggested that 75% of golfers are right-handed. how many golfers must be surveyed if they want the resulting confidence interval to be within 0.02 at 95% confidence? a. 2401 b. 2400 c. 1800 d. 9310 e. 1801

E. 1801

Step-by-step explanation:

1) Notation and definitions

random sample needed (variable of interest)

estimated proportion of interest.

true population proportion of interest.

ME=0.02 represent the margin of error desired

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".  

The margin of error is the range of values below and above the sample statistic in a confidence interval.  

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".  

The population proportion have the following distribution

2) Solution to the problem

In order to find the critical value we need to take in count that we are finding the interval for a proportion, so on this case we need to use the z distribution. Since our interval is at 95% of confidence, our significance level would be given by and . And the critical value would be given by:

The margin of error for the proportion interval is given by this formula:  

   (a)  

And on this case we have that and we are interested in order to find the value of n, if we solve n from equation (a) we got:  

  (b)  

And replacing into equation (b) the values from part a we got:

And rounded up we have that n=1801

E. 1801

it should be a.

hope it