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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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Two-sided interval\textbf{Two-sided interval} for population mean μ\mu with n observation and level of confidence 1α1-\alpha is defined by:

μ(xtα2,n1sn,x+tα2,n1sn),\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 x,s\overline{x},s is estimates of mean and standard deviation, respectively.

We calculate tα2,n1=TINV(α2,n1)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:

length=x+tα2,n1snx+tα2,n1sn=2tα2,n1sn\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 length=74.572.3=2.2\text{length}=74.5-72.3=2.2, n=10, α=0.01\alpha=0.01.

(a)

length=2.2,2t0.012,n1sn=2.2,2t0.012,9s10=2.2,3.25s=2.2102s=2.2103.252s=1.07.\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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