Problem
a. Why does "margin of error" have no relevance for confidence interval estimates of $\sigma$ or $\sigma^{2}$ ? b. What is wrong with expressing the confidence interval of $6.6 \mathrm{bpm}<\sigma<14.4 \mathrm{bpm}$ as $10.5 \mathrm{bpm} \pm 3.9 \mathrm{bpm}$ ? a. Choose the correct answer below. A. Unlike confidence interval estimates of $p$ or $\mu$, confidence interval estimates of $\sigma$ or $\sigma^{2}$ are created using a distribution that is symmetric, so there is no " $\pm \mathrm{E}$ " as in confidence interval estimates of $p$ or $\mu$. B. Unlike confidence interval estimates of $p$ or $\mu$, confidence interval estimates of $\sigma$ or $\sigma^{2}$ are not created using a distribution that is symmetric, so there is no " $\pm \mathrm{E}$ " as in confidence interval estimates of $p$ or $\mu$. C. Like confidence interval estimates of $p$ or $\mu$, confidence interval estimates of $\sigma$ or $\sigma^{2}$ are created using a distribution that is symmetric, so there is no " $\pm \mathrm{E}$ " as in confidence interval estimates of $p$ or $\mu$. D. Like confidence interval estimates of $p$ or $\mu$, confidence interval]estimates of $\sigma$ or $\sigma^{2}$ are not created using a distribution that is symmetric, so there is no " $\pm \mathrm{E}$ " as in confidence interval estimates of $p$ or $\mu$.
Solution
Step 1 ：Choose the correct answer below.
Step 2 ：Unlike confidence interval estimates of $p$ or $\mu$, confidence interval estimates of $\sigma$ or $\sigma^{2}$ are not created using a distribution that is symmetric, so there is no " $\pm \mathrm{E}$ " as in confidence interval estimates of $p$ or $\mu$.
Step 3 ：The correct answer is \(\boxed{\text{B}}\). Unlike confidence interval estimates of $p$ or $\mu$, confidence interval estimates of $\sigma$ or $\sigma^{2}$ are not created using a distribution that is symmetric, so there is no " $\pm \mathrm{E}$ " as in confidence interval estimates of $p$ or $\mu$.
