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# The sugar content of the syrup in canned peaches is normally distributed. A random sample of n = 10 cans yields a sample standard deviation of s = 4.8 milligrams. Calculate a 95% two-sided confidence interval for σ.

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If $s^2$ is the sample variance from a random sample of $n$ observations from a normal distribution with unknown variance $\sigma^2$, then a $\textbf{$100(1 - \alpha)\%$ confidence interval for $\sigma$}$ is

\begin{align} \sqrt{\frac{(n-1)s^2}{\chi^2_{\frac{\alpha}{2},n-1}}}\leq\sigma\leq \sqrt{\frac{(n-1)s^2}{\chi^2_{1-\frac{\alpha}{2},n-1}}} \end{align}

where $\chi^2_{\frac{\alpha}{2},n-1}$ and $\chi^2_{1-\frac{\alpha}{2},n-1}$ are the upper and lower $100\frac{\alpha}{2}$ percentage points of the chi-square distribution with $n - 1$ degrees of freedom, respectively.

We know that confidence level is $95\%$, simple size $n= 10$ and simple standard deviation $s=4.8$, than

\begin{align*} \frac{\alpha}{2}=\frac{1-0.95}{2}=0.025,\quad \text{and}\quad 1-\frac{\alpha}{2}=1-\frac{1-0.95}{2}=0.975. \end{align*}

Therefore,

\begin{align} \chi^2_{\frac{\alpha}{2},n-1}&=\chi^2_{0.025,9}=19.02\\ \chi^2_{1-\frac{\alpha}{2},n-1}&=\chi^2_{0.975,9}=2.70. \end{align}

Now, from Equations (1),(2) and (3) we get

$\sqrt{\frac{9\times 4.8^2}{19.02}}\leq\sigma\leq \sqrt{\frac{9\times 4.8^2}{2.70}}$

Then, a $95\%$ two-sided confidence interval for $\sigma$ is

$\boxed{3.3018\leq\sigma\leq 8.7636}$

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