Question 153491
a)Since you don't know {{{sigma}}} and your sample size is <30, assume your samples are  normally distributed. 
Use the t-distribution as the best estimate. Degrees of freedom are n-1 or 9. 
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The calculated mean,{{{x[m]}}}, is 3.3048.
The calculated standard deviation, {{{s}}}, is 0.132
For {{{alpha=0.10}}} and {{{DOF=9}}}, {{{t=1.833}}}.
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The confidence interval is 
{{{x[m]-t(s/sqrt(n))< mu < x[m]+t(s/sqrt(n))}}} 
{{{3.3048-1.833(0.132/sqrt(10))< mu < 3.3048+1.833(0.132/sqrt(10))}}} 
{{{3.228<mu <3.381}}}
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b) Working backwards,
{{{t(s/sqrt(n))=0.03}}}
{{{t(0.132/sqrt(n))=0.03}}}
{{{t/sqrt(n)= 0.227}}}
t is a function of n so you have to iterate to find n.
I set up an iteration cell in EXCEL using TINV and varying n.
n=54 gives {{{t/sqrt(n)= 0.227743}}}
54 samples required to give {{{0 +- 0.03}}}
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=((z*(sigma))/0.03)^2}}}
z=1.65 for 90%
Use the calculated 0.132 as an estimate for {{{sigma}}}
{{{n=((1.65*(0.132))/0.03)^2}}}
{{{n=(7.26)^2}}}
{{{n=52.7}}} or {{{n=53}}}
Good, that's close.

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c) Factors include length errors from cutting machine, air in the mixture, humidity in the plant, density changes in the mixture, etc.