SOLUTION: Could you please give me an idea on how to work this problem, I am so sketchy on where to start and what to do. Thank you.
A sample of 35 golfers showed that their average sc
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-> SOLUTION: Could you please give me an idea on how to work this problem, I am so sketchy on where to start and what to do. Thank you.
A sample of 35 golfers showed that their average sc
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Question 264825: Could you please give me an idea on how to work this problem, I am so sketchy on where to start and what to do. Thank you.
A sample of 35 golfers showed that their average score on a particular golf course was 80 with a standard deviation of 6.
Answer each of the following (show all work):
(A) Find the 95% confidence interval of the mean score for all 35 golfers.
(B) Find the 95% confidence interval of the mean score for all golfers if this is a sample of 50 golfers instead of a sample of 35.
(C) Which confidence interval is smaller and why?
You can put this solution on YOUR website! A sample of 35 golfers showed that their average score on a particular golf course was 80 with a standard deviation of 6.
Answer each of the following (show all work):
(A) Find the 95% confidence interval of the mean score for all 35 golfers.
The sample mean = xbar = 80
The standard error se = invNorm(0.025)*s/sqrt(n) = 1.96*6/sqrt(35) = 1.9878
95% CI: xbar-se < u < xbar+se
95% CI: 80-1.9878 < u < 80+1.9878
95% CI: 78.0122 < u < 81.9878
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(B) Find the 95% confidence interval of the mean score for all golfers if this is a sample of 50 golfers instead of a sample of 35.
xbar = 80
standard error = 1.96*6/sqrt(50) = 1.6631
95% CI: 78.337 < u < 81.6631
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(C) Which confidence interval is smaller and why?
The last one which has a larger sample size.
Why? As sample size increases the standard error decreases
because sqrt(n) is in the denominator. As n get larger
the s/sqrt(n) gets smaller so the CI interval gets smaller;
the CI interval is always twice the standard error in width.
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If any of the above is not clear, please say so.
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Cheers,
Stan H.
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