Question

Construct the indicated confidence intervals for (a) the population variance \sigma^{2} and (b) the population standard deviation \sigma Assume the sample is from a normally distributed population. c=0.95, s^{2}=11.56, n=30

Confidence intervals
ANSWERED
asked 2021-05-29
Construct the indicated confidence intervals for (a) the population variance
\(\displaystyle\sigma^{{{2}}}\)
and (b) the population standard deviation
\(\displaystyle\sigma\)
Assume the sample is from a normally distributed population.
\(\displaystyle{c}={0.95},{s}^{{{2}}}={11.56},{n}={30}\)

Expert Answers (1)

2021-05-30
Determine the critical values using table 6 with df=n-1=30-1=29
\(\displaystyle{x}^{{{2}}}_{\left\lbrace{1}-{0.025}\right\rbrace}={x}^{{{2}}}_{\left\lbrace{0.975}\right\rbrace}={16.047}\)
The boundaries of the confidence interval for the standart deviation are then:
\(\displaystyle\sqrt{{{\frac{{{n}-{1}}}{{{x}^{{{2}}}_{\left\lbrace{\frac{{\alpha}}{{{2}}}}\right\rbrace}}}}}}\dot{{\lbrace}}{s}=\sqrt{{{\frac{{{30}-{1}}}{{{45.722}}}}}}\dot{{\lbrace}}\sqrt{{{11.56}}}\approx{2.708}\)
\(\displaystyle\sqrt{{{\frac{{{n}-{1}}}{{{x}^{{{2}}}_{1}-{\left\lbrace{\frac{{\alpha}}{{{2}}}}\right\rbrace}}}}}}\dot{{\lbrace}}{s}=\sqrt{{{\frac{{{30}-{1}}}{{{16.047}}}}}}\dot{{\lbrace}}\sqrt{{{11.56}}}\approx{4.571}\)
2. The boundaries of the confidence interval for the variance is then the square of the boundaries of the confidence intervals for the standard deviation:
\(\displaystyle{\left({2.708}^{{{2}}},{4.571}^{{{2}}}\right)}={\left({7333},{50.894}\right)}\)
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