Construct a $99 \%$ confidence interval for the population standard deviation $\sigma$ if a sample of size 11 has standard deviation $s = 15$.
A) $9.45

Question

Construct a $99 \%$ confidence interval for the population standard deviation $\sigma$ if a sample of size 11 has standard deviation $s = 15$.
A) $9.45 < \sigma < 32.31$
B) $9.17 < \sigma < 29.40$
C) $9.85 < \sigma < 29.66$
D) $9.62 < \sigma < 30.83$

Quizplus · Accepted Answer

To construct a $99\%$ confidence interval for the population standard deviation $\sigma$, we use the formula involving the chi-square distribution: $\sqrt{\frac{(n-1)s^2}{\chi^2_{\alpha/2}}}< \sigma < \sqrt{\frac{(n-1)s^2}{\chi^2_{1-\alpha/2}}}$, where $n$ is the sample size, $s$ is the sample standard deviation, and $\chi^2_{\alpha/2}$ and $\chi^2_{1-\alpha/2}$ are the critical values from the chi-square distribution for $n-1$ degrees of freedom. For a $99\%$ confidence interval and $n=11$, the degrees of freedom is $10$. The critical values for $\chi^2$ at $0.005$ and $0.995$ for $10$ degrees of freedom are approximately $2.558$ and $23.209$, respectively. Plugging these values into the formula gives the interval $9.45 < \sigma < 32.31$.

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Construct A $99 \%$ Confidence Interval for the Population Standard Deviation $\sigma$