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

Algebra.Com
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.
RELATED QUESTIONS

Please help..I have taking stats online and trying to teach myself...I am very confused... (answered by stanbon)
Could you please give me an idea on how to work this problem, I am so sketchy on where to (answered by stanbon)
Hello, I am having a very difficult time figuring this statistics problem. I appreciate... (answered by rapaljer,Fombitz)
a sample of 85 golfers showed that their average score on a particular golfers course was (answered by stanbon)
A sample of 106 golfers showed that their average score on a particular golf course was... (answered by stanbon)
. A sample of 40 golfers showed that their average score on a particular golf course was... (answered by trelle)
Could someone please assist in helping me with this?... A sample of 142 golfers showed (answered by ewatrrr)
A sample of 89 golfers showed that their average score on a particular golf course was... (answered by stanbon)