document.write( "Question 1202553: A beer producer wants to find out whether its existing bottling equipment is working to standards. 151 bottles were selected for testing: the average beer fill level is x̄ = 50.1mm and the sample variance s^2 = 4mm^2.\r
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document.write( "Determine the 99% confidence interval for the mean value.
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document.write( "Determine the 90% confidence interval for the variance. \n" );
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Algebra.Com's Answer #837463 by Theo(13342) You can put this solution on YOUR website! sample size is 151. \n" ); document.write( "sample mean is 50.1 millimeters. \n" ); document.write( "sample variance is 4 millimeters. \n" ); document.write( "sample standard deviation is square root of 4 = 2 millimeters. \n" ); document.write( "standard error is 2/sqrt(151) = 1.628 rounded to 4 decimal places.\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "critical t-score at 99% two-tailed confidence inteval with 150 degrees of freedom is plus or minus 2.609 rounded to 3 decimal places.\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "critical t-score at 90% two-tailed confidence interval with 150 degrees of freedom is plus or minus 1.656 rounded to 3 decimal places.\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "mean of the sample is assumed to be 50.1. \n" ); document.write( "mean of the variance is assumed to be 4.\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "use the t-score formula to find the raw score of the confidence interval. \n" ); document.write( "formula is t = (x-m)/s \n" ); document.write( "t is the t-score. \n" ); document.write( "x is the critical raw score. \n" ); document.write( "m is the mean of the sample. \n" ); document.write( "s is the standard error.\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "for the mean, the formula becomes: \n" ); document.write( "2.609 = (x - 50.1) / 1.628 on the high side. \n" ); document.write( "solve for x to get x = 2.609 * 1.628 + 50.1 = 54.35 rounded to 2 decimal places. \n" ); document.write( "-2.609 = (x - 50.1) / 1.628 on the low side. \n" ); document.write( "solve for x to get x = -2.609 * 1.628 + 50.1 = 45.85 rounded to 2 decimal places.\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "for the variance, the formula becomes: \n" ); document.write( "1.656 = (x - 4) / 1.628 on the high side. \n" ); document.write( "solve for x to get x = 1.656 * 1.628 + 4 = 6.7 rounded to 2 decimal places. \n" ); document.write( "-1.656 = (x - 4) / 1.628 on the low side. \n" ); document.write( "solve for x to get x = -1.656 * 1.628 + 4 = 1.30 rounded to 2 decimal places.\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "your 99% confidence interval for the mean is 45.85 to 54.35. \n" ); document.write( "your 99% confidence interval for the variance is 1.30 to 6.70.\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "i'm pretty sure about the mean of the sample. \n" ); document.write( "i think that's the way to analyze the variance of the sample, but i'm not surre.\r \n" ); document.write( " \n" ); document.write( " \n" ); document.write( "\n" ); document.write( " \n" ); document.write( " |