document.write( "Question 1202095: If n=17,¯x(x-bar)=34, and s=4, construct a confidence interval at a 99% confidence level.
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document.write( " Assume the data came from a normally distributed population.
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document.write( "Give your answers to one decimal place.
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document.write( "_____<μ<_____ \n" );
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Algebra.Com's Answer #836753 by math_tutor2020(3817) ![]() You can put this solution on YOUR website! \n" ); document.write( "Answer: 31.2 < mu < 36.8\r \n" ); document.write( " \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "Work Shown:\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "mu = μ = Greek letter representing population mean \n" ); document.write( "n = 17 = sample size \n" ); document.write( "xbar = 34 = sample mean \n" ); document.write( "s = 4 = sample standard deviation \n" ); document.write( "df = degrees of freedom = n-1 = 17-1 = 16\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "Because the population standard deviation (sigma) is not known, and because n > 30 isn't the case, we must use the T distribution.\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "Refer to this T table \n" ); document.write( "https://www.sjsu.edu/faculty/gerstman/StatPrimer/t-table.pdf \n" ); document.write( "Locate the row labeled df = 16 \n" ); document.write( "Locate the columnn labeled \"99% confidence\". The confidence labels are at the bottom. \n" ); document.write( "The approximate value t = 2.921 is at this row and column intersection. It is the t critical value. \n" ); document.write( "Specialized stats calculators such as this one \n" ); document.write( "https://www.omnicalculator.com/statistics/critical-value \n" ); document.write( "can find the t critical value. \n" ); document.write( "Make sure to do a two-tailed test. \n" ); document.write( "Also set the significance level to 0.01 (recall that alpha = 1-C where C is the confidence level)\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "Compute the margin of error \n" ); document.write( "E = t*s/sqrt(n) \n" ); document.write( "E = 2.921*4/sqrt(17) \n" ); document.write( "E = 2.833786 \n" ); document.write( "That result is approximate.\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "Then, \n" ); document.write( "L = lower boundary \n" ); document.write( "L = xbar - E \n" ); document.write( "L = 34 - 2.833786 \n" ); document.write( "L = 31.166214 \n" ); document.write( "L = 31.2 \n" ); document.write( "and \n" ); document.write( "U = upper boundary \n" ); document.write( "U = xbar + E \n" ); document.write( "U = 34 + 2.833786 \n" ); document.write( "U = 36.833786 \n" ); document.write( "U = 36.8\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "The 99% confidence interval in the format L < mu < U is roughly 31.2 < mu < 36.8\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "That can be condensed to the format (L, U) so we get (31.2, 36.8)\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "We are 99% confident that the population mean (mu) is somewhere between 31.2 and 36.8 \n" ); document.write( " \n" ); document.write( " |