SOLUTION: A hospital emergency room has collected a sample of n = 40 to estimate the mean number of visits per day. It has found the standard deviation is 32. Using a 90 percent confidence l

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Question 865077: A hospital emergency room has collected a sample of n = 40 to estimate the mean number of visits per day. It has found the standard deviation is 32. Using a 90 percent confidence level, what is its margin of error?
Found 2 solutions by jim_thompson5910, stanbon:
Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!
Using this table, the critical value for the 90% confidence level is 1.645 (note: look in the row that starts with infinity and look for the value above 90%)


So z = 1.645


Given Information:


n = 40
s = 32


Margin Of Error (ME)


ME+=+z%2A%28s%2Fsqrt%28n%29%29


ME+=+1.645%2A%2832%2Fsqrt%2840%29%29


ME+=+8.32311480156 Using a calculator here.


So the margin of error is approximately 8.32311480156


Don't forget to round that if needed. According to the instructions posted that I see, there are no rounding instructions, but they may want you to round.

Answer by stanbon(75887) About Me  (Show Source):
You can put this solution on YOUR website!
A hospital emergency room has collected a sample of n = 40 to estimate the mean number of visits per day. It has found the standard deviation is 32. Using a 90 percent confidence level, what is its margin of error?
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sample mean:: x-bar = 40
ME = 1.645(32/sqrt(40)) = 8.32
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90% CI:: 40-8.32 < u < 40 + 8.32
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Cheers,
Stan H.
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