document.write( "Question 153491: A random sample of 10 miniature Tootsie Rolls was taken from a bag. Each piece was weighed on a very accurate scale. The results in grams were
\n" ); document.write( "3.087 3.131 3.241 3.241 3.270 3.353 3.400 3.411 3.437 3.477
\n" ); document.write( "a). Construct a 90 percent confidence interval for the true mean weight. (b) What sample size would be necessary to estimate the true weight with an error of ± 0.03 grams with 90 percent confidence? (c) Discuss the factors which might cause variation in the weight of Tootsie Rolls during manufacture.
\n" ); document.write( "

Algebra.Com's Answer #112997 by Fombitz(32388)\"\" \"About 
You can put this solution on YOUR website!
a)Since you don't know \"sigma\" and your sample size is <30, assume your samples are normally distributed.
\n" ); document.write( "Use the t-distribution as the best estimate. Degrees of freedom are n-1 or 9.
\n" ); document.write( ".
\n" ); document.write( ".
\n" ); document.write( ".
\n" ); document.write( "The calculated mean,\"x%5Bm%5D\", is 3.3048.
\n" ); document.write( "The calculated standard deviation, \"s\", is 0.132
\n" ); document.write( "For \"alpha=0.10\" and \"DOF=9\", \"t=1.833\".
\n" ); document.write( ".
\n" ); document.write( ".
\n" ); document.write( ".
\n" ); document.write( "The confidence interval is
\n" ); document.write( "\"x%5Bm%5D-t%28s%2Fsqrt%28n%29%29%3C+mu+%3C+x%5Bm%5D%2Bt%28s%2Fsqrt%28n%29%29\"
\n" ); document.write( "\"3.3048-1.833%280.132%2Fsqrt%2810%29%29%3C+mu+%3C+3.3048%2B1.833%280.132%2Fsqrt%2810%29%29\"
\n" ); document.write( "\"3.228%3Cmu+%3C3.381\"
\n" ); document.write( ".
\n" ); document.write( ".
\n" ); document.write( ".
\n" ); document.write( "b) Working backwards,
\n" ); document.write( "\"t%28s%2Fsqrt%28n%29%29=0.03\"
\n" ); document.write( "\"t%280.132%2Fsqrt%28n%29%29=0.03\"
\n" ); document.write( "\"t%2Fsqrt%28n%29=+0.227\"
\n" ); document.write( "t is a function of n so you have to iterate to find n.
\n" ); document.write( "I set up an iteration cell in EXCEL using TINV and varying n.
\n" ); document.write( "n=54 gives \"t%2Fsqrt%28n%29=+0.227743\"
\n" ); document.write( "54 samples required to give \"0+%2B-+0.03\"
\n" ); document.write( "For this large a value for n, we can use the normal distribution as a check of the value since as n gets large the t distribution approaches the normal distribution.
\n" ); document.write( "\"n=%28%28z%2A%28sigma%29%29%2F0.03%29%5E2\"
\n" ); document.write( "z=1.65 for 90%
\n" ); document.write( "Use the calculated 0.132 as an estimate for \"sigma\"
\n" ); document.write( "\"n=%28%281.65%2A%280.132%29%29%2F0.03%29%5E2\"
\n" ); document.write( "\"n=%287.26%29%5E2\"
\n" ); document.write( "\"n=52.7\" or \"n=53\"
\n" ); document.write( "Good, that's close.\r
\n" ); document.write( "\n" ); document.write( ".
\n" ); document.write( ".
\n" ); document.write( ".
\n" ); document.write( "c) Factors include length errors from cutting machine, air in the mixture, humidity in the plant, density changes in the mixture, etc.
\n" ); document.write( "
\n" );