document.write( "Question 664090: You are given the sample mean and the standard deviation. Use this information to construct the 90% and 95% confidence intervals for the population mean. A random sample of 32 eight ounce drinks has a mean of 86.7 calories and a standard deviation of 43.9 calories.\r
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document.write( "The 90% CI is
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document.write( "The 95% CI is \n" );
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Algebra.Com's Answer #413125 by Theo(13342)![]() ![]() You can put this solution on YOUR website! sample size is equal to 32 \n" ); document.write( "sample mean is equal to 86.7 calories \n" ); document.write( "sample standard deviation is equal to 43.9 calories.\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "if this is a distribution of means, you have to calculate the standard error. \n" ); document.write( "standard error = standard deviation / square root of sample size. \n" ); document.write( "se = sd / sqrt(n) \n" ); document.write( "se = 43.9 / sqrt(32) = 7.76050\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "standard error is equal to 7.76050 calories.\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "since you are using the sample standard deviation rather than the population standard deviation, you have to use a T score rather than a Z score.\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "the degrees of freedom is equal to the sample size minus 1 which = 31.\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "The critical T value at 90% confidence interval with 31 degrees of freedom is equal to +/- 1.69552\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "The critical T value at 95% confidence interval with 31 degrees of freedom is equal to +/- 2.03951\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "you need to translate your T score to X score.\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "X score = T score * standard error of sample + mean of sample\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "at 90% confidence interval this becomes:\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "X score lower limit = -1.69552 * 7.76050 + 86.7 = 73.54192 \n" ); document.write( "X score upper limit = 1.69552 * 7.76050 + 86.7 = 99.85808\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "at 95% confidence interval this becomes:\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "X score lower limit = -2.03951 * 7.76050 + 86.7 = 70.87246 \n" ); document.write( "X score upper limit = 2.03951 * 7.76050 + 86.7 = 102.52762\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "you are 90% confident that the population score will be between 73.54192 and 99.85808.\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "you are 95% confident that the population score will be between 70.87246 and 102.52762\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "i'm reasonably confident this is correct. \n" ); document.write( "if you were using the population standard deviation then you would have used a Z score instead of a T score.\r \n" ); document.write( " \n" ); document.write( " \n" ); document.write( " \n" ); document.write( " \n" ); document.write( "\n" ); document.write( " \n" ); document.write( " |