# To monitor complex chemical processes, chemical engineers will consider key process indicators, which may be just yield but most often depend on several quantities. Before trying to improve a process, 9 measurements were made on a key performance indicator. Find the sample variance s 2 of the sample and find a 95% confidence interval for \sigma 2 and find a 98\% confidence interval for \sigma.

To monitor complex chemical processes, chemical engineers will consider key process indicators, which may be just yield but most often depend on several quantities. Before trying to improve a process, 9 measurements were made on a key performance indicator. 123, 106, 114, 128, 113, 109, 120, 102, 111. Assume that the key performance indicator has a normal distribution. (a) Find the sample variance s 2 of the sample. (b) Find a $95\mathrm{%}$ confidence interval for $\sigma 2$. (c) Find a $98\mathrm{%}$ confidence interval for $\sigma$.
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Step 1
(a)
to find variance value
$123+106+114+128+113+109+120+102+111=1026$
$1026\cdot 1026=1052676$
...and divide by the number of items. we have 9 items , $\frac{1052676}{9}=116964$
set this number aside for a moment.
3) take your set of original numbers from step 1, and square them individually this time
${123}^{2}+{106}^{2}+{114}^{2}+{128}^{2}+{113}^{2}+{109}^{2}+{120}^{2}+{102}^{2}+{111}^{2}=117520$
4) subtract the amount in step 2 from the amount in step 3
$117520-116964=556$
5) subtract 1 from the number of items in your data set, $9-1=8$
6) divide the number in step 4 by the number in step 5. this gives you the variance $\frac{556}{8}=69.5$
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to find standard deviation value
take the square root of your answer from step 6. this gives you the standard deviation 8.3367
Step 2
(b)
CONFIDENCE INTERVAL FOR VARIANCE
$ci=\left(n-1\right)\frac{{s}^{2}}{{\psi }^{2}}right<{\sigma }^{2}<\left(n-1\right)\frac{{s}^{2}}{{\psi }^{2}}$ left where,
${s}^{2}=$ variance
${\psi }^{2}$ right $=\frac{1-\text{confidence level}}{2}$
${\psi }^{2}$ left $=1-{\psi }^{2}$ right
$n=$ sample size
since $\alpha =0.05$
${\psi }^{2}$ right $=\frac{1-\text{confidence level}}{2}=\frac{1-0.95}{2}=\frac{0.05}{2}=0.025$
${\psi }^{2}$ left $=1-{\psi }^{2}$ right $=1-0.025=0.975$
the two critical values ${\psi }^{2}$ left, ${\psi }^{2}$ right at 8 df are 17.5345 , 2.18 variance $\left({s}^{2}\right)=69.5$
sample size $\left(n\right)=9$
confidence interval $=\left[8\cdot \frac{69.5}{17.5345}<{\sigma }^{2}<8\cdot \frac{69.5}{2.18}\right]$
$=\left[\frac{556}{17.5345}<{\sigma }^{2}<\frac{556}{2.1797}\right]$
$\left[31.7089,255.081\right]$
Step 3
(c)
to calculate confidence interval for standard deviation
$ci=\left(n-1\right)\frac{{s}^{2}}{{\psi }^{2}}$ right $<{\sigma }^{2}<\left(n-1\right)\frac{{s}^{2}}{{\psi }^{2}}$ left
where,
$s=$ standard deviation
${\psi }^{2}$ right $=\frac{1-confidence\le vel}{2}$