document.write( "Question 1047800: 5.7 Suppose that the population standard deviation (sigma) for a normally distributed standardized test of achievement is known to be 7.20. You can use these to estimate the standard error of the mean for a randomly drawn sample size of 16. \r
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\n" ); document.write( "\n" ); document.write( "But now suppose that we do not feel comfortable assuming that s = 7.20. Use the scores above to
\n" ); document.write( "a. estimate the standard error of the sample mean ()\r
\n" ); document.write( "\n" ); document.write( "b. find the 95% confidence interval for the mean.
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\n" ); document.write( "c. find the 99% confidence interval for the mean
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Algebra.Com's Answer #663361 by stanbon(75887)\"\" \"About 
You can put this solution on YOUR website!
Suppose that the population standard deviation (sigma) for a normally distributed standardized test of achievement is known to be 7.20. You can use these to estimate the standard error of the mean for a randomly drawn sample size of 16.
\n" ); document.write( "6 5 6 12 5 10 11 13
\n" ); document.write( "12 10 9 20 23 28 20 18
\n" ); document.write( "But now suppose that we do not feel comfortable assuming that s = 7.20. Use the scores above to
\n" ); document.write( "a. estimate the standard error of the sample mean ()
\n" ); document.write( "s = 6.7
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\n" ); document.write( "b. find the 95% confidence interval for the mean.
\n" ); document.write( "x-bar = 13
\n" ); document.write( "ME = 1.96*13/sqrt(16) = 7.37
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\n" ); document.write( "95% CI:: 13-7.37 < u < 13+7.37
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\n" ); document.write( "\n" ); document.write( "c. find the 99% confidence interval for the mean
\n" ); document.write( "x-bar = 13
\n" ); document.write( "ME = 2.5758*13/sqrt(16) = 8.37
\n" ); document.write( "---
\n" ); document.write( "99% CI: 13-8.37 < u < 13+8.37
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\n" ); document.write( "Cheers,
\n" ); document.write( "Stan H.
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