 # 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. UkusakazaL 2021-08-10 Answered
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 $=\frac{1-confidence\le vel}{2}$