SOLUTION: 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 with th
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Answer each of the following show all work with th
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Question 441291: 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 with 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 106.
C) Which confidence interval is larger and why? Answer by ewatrrr(24785) (Show Source):
Hi
sample of 106 golfers showed that their average score on a particular golf
course was 87.98 with a standard deviation of 5.39.
A) Find the 90% confidence interval of the mean score for all 106 golfers.
ME = 1.645[5.39/sqrt(106)] = .8612
CI: 87.98-.8612 < u < 87.98+.8612
CI: 86.9988 < u < 88.8412
B) Find the 90% confidence interval of the mean score for all golfers if this is a sample of 130 golfers instead of 106.
ME = 1.645[5.39/sqrt(130)] = .7776
CI: 87.2024 < u < 88.7576
C) Which confidence interval is larger and why?
confidence interval is larger for the smaller sample size of 106 due to
the fact the ME is larger
Summary of z values for various confidence intervals
a a/2 crtical regions
90% 0.1 5% z <-1.645 z >+1.645
95% 0.05 2.50% z <-1.96 z >+1.96
98% 0.02 1% z <-2.33 z >+2.33
99% 0.01 0.50% z<-2.575 z >+2.575
