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1. Interpret a 99% confidence interval of (5.355, 5.445) for a population mean.
2.A nursing school wants to estimate the true mean annual income of its alumni. It randomly samples 200 of its alumni. The mean annual income was $52,500 with a standard deviation of $1,800. Find a 95% confidence interval for the true mean annual income of the nursing school alumni. Write a statement about the confidence level and the interval you find.
3.An auditor wants to estimate what proportion of a bank’s commercial loan files are incomplete. The auditor randomly samples 60 files and finds 12 are incomplete. Using a 95% confidence interval, estimate the true proportion of incomplete files for ALL the bank’s commercial loans. Write a statement about the confidence level and the interval you find.
Submitted: 91 days and 21 hours ago.
Category: Math
Value: $30
Status: CLOSED
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Level/Year: college
Subject: elementry statisics
Already Tried:
I need all 3 question answered with work, so I can figure out other problems please help me, thanks
Accepted Answer
Hi,
1. We are 99% confident that the population mean is between 5.355 and 5.445.
2.
Lower limit: 52,500 - 1.96*1,800/sqrt(200) = 52250.5
Upper limit: 52,500 + 1.96*1,800/sqrt(200) = 52749.5
We are 95% confident that the true men annual income of the nursing school alumni is between $52,250.5 and $52,749.5.
3.
xhat = x/n = 12/60 = 0.2
Lower limit: 0.2 - 1.96*sqrt(0.2(1-0.2)/60) = 0.0988
Upper limit: 0.2 + 1.96*sqrt(0.2(1-0.2)/60) = 0.3012
We are 95% confident that the true proportion of incomplete files for all the bank's commercial loans is between 0.0988 and 0.3012.
Expert:
abozer
Pos. Feedback:
99.9 %
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Answered:
8/21/2009
Math Answer Team
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