SOLUTION: Hello again, Here is another question that I have: A group of 19 randomly selected college students has a mean of 22.4 years with a standard deviation of 3.8 years. Assume th

Question 257013: Hello again,
Here is another question that I have:
A group of 19 randomly selected college students has a mean of 22.4 years with a standard deviation of 3.8 years. Assume the population has a normal distribution.
a.Find the margin of error for a 99% confidence interval. Round your answer to the nearest hundredths
b.Find a 99% confidence interval for the population mean - the symbol is suppose to be a mean.
This is what I have come up with n=19, x=22.4, s=3.8, c=0.99
d.f.=15
2.947*38/square root19=2.5

Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
A group of 19 randomly selected college students has a mean of 22.4 years with a standard deviation of 3.8 years. Assume the population has a normal distribution.
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a.Find the margin of error for a 99% confidence interval. Round your answer to the nearest hundredths
b.Find a 99% confidence interval for the population mean - the symbol is suppose to be a mean.
This is what I have come up with n=19, x=22.4, s=3.8, c=0.99
d.f.=15
2.947*3.8/square root19=2.5
-----------------------
Comment:
I think the df is 18
So the margin of error is 2.87844..*3.8/sqrt(19) = 2.5
===========================================================
sample mean = 22.4
99% C.I.: 22.4-2.5 < u < 22.4+2.5
99% CI: 19.9 < u < 24.9
===========================================================
Cheers,
Stan H.

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