Question

I am having so much trouble with this class, and have to pass. Can someone help?
A sample of 143 glofers showed that their average score on a particular golf course was 80.35 with a standard deviation of 3.33.Answer by showing work the following to alt least two decimal points
A. find the 95% confidence interval of the mean B. Find the 95% confidence interval of the mean score for all golfers if this is a sample of 90 golfers instead of 143.
C. Which confidence interval is smaller and why?

Answer 1

the confidence interval sets up a Z score as a critical point. 1-95% = 5% split between 2 tails

.05/2 = .025 per tail, so we need to determine the Z score for an area of .025

Answer 2

should be: 1.96