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# An engineer selects 10 components at random and measures their strengths. It is reported that the average strength of the components is between 72.3 and 74.5 with 99$\%$ confidence. (a) What is the sample standard deviation of the 10 component strengths? (b) If a 99$\%$ two-sided confidence interval is desired with a length no longer than 1.0, about how many additional components would you recommend be tested?

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$\textbf{Two-sided interval}$ for population mean $\mu$ with n observation and level of confidence $1-\alpha$ is defined by:

\begin{align*} \mu\in (\overline{x}-\frac{t_{\frac{\alpha}{2},n-1}\cdot s}{\sqrt{n}}, \overline{x}+ \frac{t_{\frac{\alpha}{2},n-1}\cdot s}{\sqrt{n}}), \end{align*}

where $\overline{x},s$ is estimates of mean and standard deviation, respectively.

We calculate $t_{\frac{\alpha}{2},n-1}=TINV(\frac{\alpha}{2},n-1)$ or using table of Student distribution.

Length of this two-sided interval is:

\begin{align*} \text{length}&=\overline{x}+ \frac{t_{\frac{\alpha}{2},n-1}\cdot s}{\sqrt{n}}-\overline{x}+\frac{t_{\frac{\alpha}{2},n-1}\cdot s}{\sqrt{n}}\\ &=2\cdot \frac{t_{\frac{\alpha}{2},n-1}\cdot s}{\sqrt{n}} \end{align*}

In our case length of confidence interval is $\text{length}=74.5-72.3=2.2$, n=10, $\alpha=0.01$.

#### (a)

\begin{align*} \text{length}&=2.2,\\ 2\cdot \frac{t_{\frac{0.01}{2},n-1}\cdot s}{\sqrt{n}}&=2.2,\\ 2\cdot \frac{t_{\frac{0.01}{2},9}\cdot s}{\sqrt{10}}&=2.2,\\ 3.25\cdot s&=\frac{2.2\cdot \sqrt{10}}{2}\\ s&=\frac{2.2\cdot \sqrt{10}}{3.25\cdot 2}\\ s&=\textbf{1.07}. \end{align*}

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