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g given\begin{array}{l}D=15 \\S=0.008 \quad \Rightarrow s^{2}=0.000064 \\\alpha=0.01 \quad(99 \%)\end{array}99 \% confidence interval of 62 .\begin{array}{l}\frac{(n-1) s^{2}}{x^{2} \alpha / 2}<6^{2}<\frac{(n-1) s^{2}}{x^{2}-\alpha / 2} \\n-1=d 9 \operatorname{ses} \sigma \text { of freedom }=15-1=14 \\\alpha=0.01\end{array}whese\begin{array}{l}x_{\alpha / 2}^{2} \Rightarrow x_{0.01 / 2}^{2} \Rightarrow x_{0.005}^{2} \\x_{\alpha / 2}^{2}=x_{0.005}^{2}=31.32\end{array}and x_{1}^{2}-\alpha / 2 \Rightarrow x_{1}^{2}-0.01 / 2 \Rightarrow x^{2}-0.005\begin{array}{c}x_{1}^{2}+\alpha_{12}=x^{2}{ }_{0}^{2} 995=4.08 \\\frac{14 \times 0.000069}{31.32}<6^{2}<\frac{14 \times 0.000064}{4.08} \\=0.000029<6^{2}<0.00020\end{array}99 \% confintencre interval of 6^{2} 1 \mathrm{j}\begin{array}{l}\text { lower bound }=0.000029 \\\text { upper brund }=0.00020 .\end{array} ...