90% = mean ± 1.645 SEm
Use same process indicated in your first post.
A sample of 106 golfers showed that their average score on a particular golf course was 87.98 with a standard deviation of 5.39.
Answer each of the following (show all work and state the final answer to at least two decimal places.):
(A) Find the 90% confidence interval of the mean score for all 106 golfers.
(B) Find the 90% confidence interval of the mean score for all golfers if this is a sample of 130 golfers instead of a sample of 106.
(C) Which confidence interval is larger and why?
1 answer