A sample of 143 golfers showed that their average score on a particular golf course was 80.35 with a standard deviation of 3.33.
Answer each of the following (show all work
and state the final answer to at least two decimal places.):
(A) Find the 95% confidence interval of the mean score for all 143 golfers.
(B) Find the 95% confidence interval of the mean score for all golfers if this is a sample of 90 golfers instead of a sample of 143.
(C) Which confidence interval is smaller and why? (Points : 6)

Question

A sample of 143 golfers showed that their average score on a particular golf course was 80.35 with a standard deviation of 3.33.  
Answer each of the following (show all work
and state the final answer to at least two decimal places.):
(A) Find the 95% confidence interval of the mean score for all 143 golfers.
(B) Find the 95% confidence interval of the mean score for all golfers if this is a sample of 90 golfers instead of a sample of 143.
(C) Which confidence interval is smaller and why? (Points : 6)

kropot72 · Answer

The formula to find ther answer to (A) is:
$$xbar-1.96\frac{\sigma}{\sqrt{n}}< \mu < xbar+\frac{\sigma}{\sqrt{n}}$$
You just need to substitute. Can you do that?