document.write( "Question 629242: A snack food manufacturer selected a sample of 15 bags of pretzels off the assembly line and weighed their contents. If the sample mean is 10.0 and the sample standard deviation is 0.15, find the 95% confidence interval estimate for the true mean weight of bags of this type of pretzel.
\n" ); document.write( " A. (9.92, 12.34)
\n" ); document.write( "
\n" ); document.write( " B. (8.54, 11.46)
\n" ); document.write( "
\n" ); document.write( " C. (10.42, 10.58)
\n" ); document.write( "
\n" ); document.write( " D. (9.92, 10.08)
\n" ); document.write( "
\n" ); document.write( "

Algebra.Com's Answer #396118 by ewatrrr(24785)\"\" \"About 
You can put this solution on YOUR website!
 
\n" ); document.write( "Hi
\n" ); document.write( "15 bags of pretzels off the assembly line and weighed their contents.
\n" ); document.write( "If the sample mean is 10.0 and the sample standard deviation is 0.15
\n" ); document.write( "find the 95% confidence interval estimate for the true mean weight of bags
\n" ); document.write( "ME = \"+2.14%28.15%2Fsqrt%2815%29%29\" = .08
\n" ); document.write( "10 - .08 < \"mu\" < 10 + .08
\n" ); document.write( "9.92 < \"mu\" < 10.08\r
\n" ); document.write( "\n" ); document.write( "t - test table
\n" ); document.write( "DF Probability, p
\n" ); document.write( " 0.1 0.05 0.01 0.001
\n" ); document.write( "14 1.76 \"highlight%282.14%29\" 2.98 4.14
\n" ); document.write( " \n" ); document.write( "
\n" );