You randomly select and weigh 27 samples of an allergy medication. The sample standard deviation is 1.22 milligrams. Assuming the weights are normally distributed, construct 90% confidence intervals for the population variance and standard deviation. Please interpret your result.

You randomly select and weigh 27 samples of an allergy medication. The sample standard deviation is 1.22 milligrams. Assuming the weights are normally distributed, construct 90% confidence intervals for the population variance and standard deviation. Please interpret your result.

Question
Confidence intervals
asked 2020-12-15
You randomly select and weigh 27 samples of an allergy medication. The sample standard deviation is 1.22 milligrams. Assuming the weights are normally distributed, construct \(90\%\) confidence intervals for the population variance and standard deviation. Please interpret your result.

Answers (1)

2020-12-16
Step 1
From the given information, \(\displaystyle{S}={1.22}\) milligrams and \(\displaystyle{n}={27}\).
The level of confidence is 0.90.
\(\displaystyle{1}-\alpha={0.90}\)
\(\displaystyle\alpha={1}-{0.90}\)
\(\displaystyle\alpha={0.10}\)
Step 2
Then, a \(90\%\) interval for the population variance is
\(\displaystyle{C}{I}={\left(\frac{{{\left({n}-{1}\right)}{S}^{2}}}{{{X}_{{{1}-\frac{a}{{n}},{n}-{1}}}^{{2}}}}<\sigma^{2}<\frac{{{\left({n}-{1}\right)}{S}^{2}}}{{{X}_{{\frac{a}{{n}},{n}-{1}}}^{{2}}}}\right)}\)</span>
\(\displaystyle={\left(\frac{{{\left({27}-{1}\right)}{1.22}^{2}}}{{{X}_{{{1}-\frac{0.10}{{2}},{27}-{1}}}^{{2}}}}<\sigma^{2}<\frac{{{\left({27}-{1}\right)}{1.22}^{2}}}{{{X}_{{\frac{0.10}{{2}},{27}-{1}}}^{{2}}}}\right)}\)</span>
\(\displaystyle={\left(\frac{{{26}\times{1.22}^{2}}}{{{X}_{{{0.95},{26}}}^{{2}}}}<\sigma^{2}<\frac{{{26}\times{1.22}^{2}}}{{{X}_{{{0.5},{26}}}^{{2}}}}\right)}\)</span>
\(\displaystyle={\left(\frac{{{26}\times{1.22}^{2}}}{{38.855}}<\sigma^{2}<\frac{{{26}\times{1.22}^{2}}}{{15.379}}\right)}\)</span>
\(\displaystyle={\left({0.9952}<\sigma^{2}<{2.5163}\right)}\)</span>
Interpretation:
There is \(90\%\) confidence that the population variance lies between 0.9952 and 2.5163.
Step 3
The \(90\%\) confidence interval for the population standard deviation is
\(\displaystyle{C}{I}={\left(\frac{{{\left({n}-{1}\right)}{S}^{2}}}{{{X}_{{{1}-\frac{a}{{n}},{n}-{1}}}^{{2}}}}<\sigma<\frac{{{\left({n}-{1}\right)}{S}^{2}}}{{{X}_{{\frac{a}{{n}},{n}-{1}}}^{{2}}}}\right)}\)</span>
\(\displaystyle={\left(\sqrt{{\frac{{{\left({27}-{1}\right)}{1.22}^{2}}}{{X}_{{{1}-\frac{0.10}{{2}},{27}-{1}}}}}}<\sigma<\sqrt{{\frac{{{\left({27}-{1}\right)}{1.22}^{2}}}{{{X}_{{\frac{0.10}{{2}},{27}-{1}}}^{{2}}}}}}\right)}\)</span>
\(\displaystyle={\left(\sqrt{{\frac{{{26}\times{1.22}^{2}}}{{{X}_{{{0.95},{26}}}^{{2}}}}}}<\sigma<\sqrt{{\frac{{{26}\times{1.22}^{2}}}{{{X}_{{{0.05},{26}}}^{{2}}}}}}\right)}\)</span>
\(\displaystyle{\left(\sqrt{{\frac{{{26}\times{1.22}^{2}}}{{38.885}}}}<\sigma<\sqrt{{\frac{{{26}\times{1.22}^{2}}}{{15.379}}}}\right)}\)</span> [(Using excel function), (=CHISQ.INV(0.95,26)), (Using excel function), (=CHISQ.INV(0.05,26))]
\(\displaystyle={\left({0.9976}<\sigma<{1.5863}\right)}\)</span>
Interpretation:
There is \(90\%\) confidence that the population standard deviation lies between 0.9976 and 1.5863.
0

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