document.write( "Question 1039782: A food snack manufacturer samples 9 bags of pretzels off the assembly line and weighed their contents. If the sample mean is 10.4 and the sample standard deviation is 0.15, find the 95% confidence interval of the true mean.
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\n" ); document.write( "(10.30, 10.50)
\n" ); document.write( " B.
\n" ); document.write( "(10.05, 10.75)
\n" ); document.write( " C.
\n" ); document.write( "(10.35, 10.45)
\n" ); document.write( " D.
\n" ); document.write( "(10.28, 10.52)
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Algebra.Com's Answer #654950 by jim_thompson5910(35256)\"\" \"About 
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Use a table like this one to find the t critical value is t = 2.306\r
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\n" ); document.write( "\n" ); document.write( "How to find this value? Look at the row that has df = 8 and look at the column that corresponds to 95% confidence. The value 2.306 is found at this row and column intersection\r
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\n" ); document.write( "\n" ); document.write( "Note: degrees of freedom = df = n-1 = 9-1 = 8\r
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\n" ); document.write( "\n" ); document.write( "We have the following information\r
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\n" ); document.write( "\n" ); document.write( "xbar = 10.4 is the sample mean
\n" ); document.write( "s = 0.15 is the standard deviation
\n" ); document.write( "n = 9 is the sample size
\n" ); document.write( "t = 2.306 is the t critical value (found above)\r
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\n" ); document.write( "\n" ); document.write( "Let's compute the lower and upper boundaries
\n" ); document.write( "L = lower boundary
\n" ); document.write( "U = upper boundary\r
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\n" ); document.write( "\n" ); document.write( "L = xbar - t*s/sqrt(n)
\n" ); document.write( "L = 10.4 - 2.306*0.15/sqrt(9)
\n" ); document.write( "L = 10.2847
\n" ); document.write( "L = 10.28\r
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\n" ); document.write( "\n" ); document.write( "U = xbar + t*s/sqrt(n)
\n" ); document.write( "U = 10.4 + 2.306*0.15/sqrt(9)
\n" ); document.write( "U = 10.5153
\n" ); document.write( "U = 10.52\r
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\n" ); document.write( "\n" ); document.write( "The 95% confidence interval, for the population mean mu, is therefore (L,U) = (10.28, 10.52) which is choice D
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