A simple random sample is drawn from a population that is known to be normally d

Falak Kinney 2021-09-19 Answered
A simple random sample is drawn from a population that is known to be normally distributed. The sample variance, s2, is determined to be 12.6
A. Construct a \(\displaystyle{90}\%\) confidence interval for \(\displaystyle\sigma\) if the sample size, n, is 20. (Hint: use the result obtained from part (a)) 7.94 (LB) & 23.66 (UB))
B. Construct a \(\displaystyle{90}\%\) confidence interval for \(\displaystyle\sigma\) if the sample size, n, is 30. (Hint: use the result obtained from part (b)) 8.59 (LB) & 20.63 (UB)

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Expert Answer

Laith Petty
Answered 2021-09-20 Author has 16052 answers
Step 1
Here \(\displaystyle\sigma{2}={12.6}\) and \(\displaystyle{c}={90}\%\) which is 0.9.
Therefore \(\displaystyle{\frac{{{1}-{c}}}{{{2}}}}={\frac{{{1}-{0.9}}}{{{2}}}}={0.05}\) and \(\displaystyle{1}-{\left({\frac{{{1}-{c}}}{{{2}}}}\right)}={0.95}\)
Part A
For \(\displaystyle{n}={20}\) the degrees of freedom are:
\(\displaystyle{d}{f}={20}-{1}={19}\)
So the \(\displaystyle\chi\) squared table values are:
\(\displaystyle{X}_{{{0.95}}}={10.117}\) and \(\displaystyle{X}_{{{0.05}}}={30.144}\)
So the upper and lower bounds are given by:
\(\displaystyle{U}{B}={\left({\frac{{{19}}}{{{10.117}}}}\right)}{\left({12.6}\right)}\)
\(\displaystyle{U}{B}={23.66}\)
\(\displaystyle{L}{B}={\left({\frac{{{19}}}{{{30.144}}}}\right)}{\left({12.6}\right)}\)
\(\displaystyle{L}{B}={7.94}\)
Part B
For \(\displaystyle{n}={30}\) the degrees of freedom are:
\(\displaystyle{d}{f}={30}-{1}={29}\)
So the \(\displaystyle\chi\) squared table values are:
\(\displaystyle{X}_{{{0.95}}}={17.708}\) and \(\displaystyle{X}_{{{0.05}}}={42.557}\)
So the upper and lower bounds are given by:
\(\displaystyle{U}{B}={\left({\frac{{{29}}}{{{17.708}}}}\right)}{\left({12.6}\right)}\)
\(\displaystyle{U}{B}={20.63}\)
\(\displaystyle{L}{B}={\left({\frac{{{29}}}{{{42.557}}}}\right)}{\left({12.6}\right)}\)
\(\displaystyle{L}{B}={8.59}\)
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