## Related questions with answers

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?

Solution

Verified$\textbf{Two-sided interval}$ 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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