document.write( "Question 257013: Hello again,\r
\n" ); document.write( "\n" ); document.write( "Here is another question that I have:\r
\n" ); document.write( "\n" ); document.write( "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.\r
\n" ); document.write( "\n" ); document.write( "a.Find the margin of error for a 99% confidence interval. Round your answer to the nearest hundredths\r
\n" ); document.write( "\n" ); document.write( "b.Find a 99% confidence interval for the population mean - the symbol is suppose to be a mean.\r
\n" ); document.write( "\n" ); document.write( "This is what I have come up with n=19, x=22.4, s=3.8, c=0.99
\n" ); document.write( "d.f.=15\r
\n" ); document.write( "\n" ); document.write( "2.947*38/square root19=2.5
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Algebra.Com's Answer #189026 by stanbon(75887)\"\" \"About 
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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\n" ); document.write( "a.Find the margin of error for a 99% confidence interval. Round your answer to the nearest hundredths
\n" ); document.write( "b.Find a 99% confidence interval for the population mean - the symbol is suppose to be a mean.
\n" ); document.write( "This is what I have come up with n=19, x=22.4, s=3.8, c=0.99
\n" ); document.write( "d.f.=15
\n" ); document.write( "2.947*3.8/square root19=2.5
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\n" ); document.write( "Comment:
\n" ); document.write( "I think the df is 18
\n" ); document.write( "So the margin of error is 2.87844..*3.8/sqrt(19) = 2.5
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\n" ); document.write( "sample mean = 22.4
\n" ); document.write( "99% C.I.: 22.4-2.5 < u < 22.4+2.5
\n" ); document.write( "99% CI: 19.9 < u < 24.9
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\n" ); document.write( "Cheers,
\n" ); document.write( "Stan H.\r
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