SOLUTION: Please help? I'm so lost on this problem. Can anyone help? A sample of 140 golfers showed that their average score on a particular golf course was 93.38 with a standard deviatio

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Question 620938: Please help? I'm so lost on this problem. Can anyone help?
A sample of 140 golfers showed that their average score on a particular golf course was 93.38 with a standard deviation of 4.36.
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 140 golfers.
(B) Find the 90% confidence interval of the mean score for all golfers if this is a sample of 95 golfers instead of a sample of 140.
(C) Which confidence interval is smaller and why?

Answer by jim_thompson5910(35256) (Show Source):
You can put this solution on YOUR website!
I'll do part A) to get you going.

A)

Lower Bound:

LB = xbar - z*sigma/sqrt(n)

LB = 93.38 - 1.64485*4.36/sqrt(140)

LB = 93.38 - 1.64485*4.36/11.8321596

LB = 93.38 - 7.171546/11.8321596

LB = 93.38 - 0.606106

LB = 92.773894

-----------------------

Upper Bound:

LB = xbar + z*sigma/sqrt(n)

LB = 93.38 + 1.64485*4.36/sqrt(140)

LB = 93.38 + 1.64485*4.36/11.8321596

LB = 93.38 + 7.171546/11.8321596

LB = 93.38 + 0.606106

LB = 93.986106

So the 90% confidence interval of the mean score for all 140 golfers is (92.774, 93.986)

Note: I rounded to three places because it said to round to "at least two decimal places", so I just picked one number higher

Another note: The value z = 1.64485 is the value of z such that 90% of the distribution under the standard normal curve is between z = -1.64485 and z = 1.64485. Use a calculator or a table to find this value.