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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document.write( " 6 5 6 12 5 10 11 13
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document.write( "12 10 9 20 23 28 20 18\r
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document.write( "But now suppose that we do not feel comfortable assuming that s = 7.20. Use the scores above to
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document.write( "a. estimate the standard error of the sample mean ()\r
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document.write( "b. find the 95% confidence interval for the mean.
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document.write( "c. find the 99% confidence interval for the mean
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Algebra.Com's Answer #663365 by ewatrrr(24785)![]() ![]() You can put this solution on YOUR website! sample mean = 13 \n" ); document.write( "sample SD = 6.9 (Generally rounded to one more decimal point than Data values have) \n" ); document.write( "Recommend following to check work on sample means and 'corrected' standard deviations. \n" ); document.write( "www.numberempire.com/statisticscalculator.php \n" ); document.write( "95% confidence interval \n" ); document.write( "Small Sample - We are using the T distribution because \n" ); document.write( "a) population standard deviation is not being used \n" ); document.write( "b) n = 16 < 30 \n" ); document.write( "use t-value = 2.13 \n" ); document.write( "SE = (2.13)(6.9/sqrt(16)) = (2.13)1.725 \n" ); document.write( "Round the confidence interval limits to the same number of decimal places as the sample standard deviation \n" ); document.write( "CI = 13 ± SE\r \n" ); document.write( "\n" ); document.write( "| \n" ); document.write( "99% confidence interval \n" ); document.write( "Small Sample \n" ); document.write( "use t-value = 2.95 \n" ); document.write( "SE = (2.95)(6.9/sqrt(16)) = (2.95)1.725 = 5.1 \n" ); document.write( "Round the confidence interval limits to the same number of decimal places as the sample standard deviation \n" ); document.write( "CI = 13 ± 5.1 \n" ); document.write( " |