91影视

The sample number of observations is $n=12$.

The level of confidence is 95%.

The degrees of freedom are computed as follows:

$$\begin{gathered} df=n-1 \\ =12-1 \\ =11 \end{gathered}$$

The level of confidence is 95%,which implies the level of significance is 0.05.

Using the Chi-square table, the critical values at 0.05 level of significance and 11 degrees of freedom are role="math" localid="1648112099073" ${\mathit{蠂}}_{\mathbf{L}}^{\mathbf{2}}\mathbf{=}\mathbf{3}\mathbf{.}\mathbf{8157}$and ${\mathit{蠂}}_{\mathbf{R}}^{\mathbf{2}}\mathbf{=}\mathbf{21}\mathbf{.}\mathbf{92}$.

The confidence interval for the standard deviation is given as follows:

$\sqrt{\frac{\left(n-1\right){\mathbf{s}}^{\mathbf{2}}}{{\mathbf{蠂}}_{\mathbf{R}}^{\mathbf{2}}}}\mathbf{<}\mathit{蟽}\mathbf{<}\sqrt{\frac{\left(n-1\right){\mathbf{s}}^{\mathbf{2}}}{{\mathbf{蠂}}_{\mathbf{L}}^{\mathbf{2}}}}$

Let x represents the sample observations.

The mean value is computed as follows:

$$\begin{gathered} \stackrel{炉}{x}=\frac{鈭�x}{n} \\ =\frac{62+61+61+57+...+60+67}{12} \\ =60.667 \end{gathered}$$

The standard deviation is computed as follows:

$$\begin{gathered} s=\sqrt{\frac{鈭�{\left(x-\stackrel{炉}{x}\right)}^2}{n-1}} \\ =\sqrt{\frac{{\left(62-60.667\right)}^2+{\left(61-60.667\right)}^2+{\left(61-60.667\right)}^2+...+{\left(67-60.667\right)}^2}{12-1}} \\ =4.075 \end{gathered}$$

The 95% confidence interval estimate of the standard deviation of the population from which the sample was obtained is computed as follows:

$$\begin{gathered} \sqrt{\frac{\mathbf{(}\mathbf{n}\mathbf{-}\mathbf{1}\mathbf{)}{\mathbf{s}}^{\mathbf{2}}}{{\mathbf{蠂}}_{\mathbf{R}}^{\mathbf{2}}}}\mathbf{<}\mathit{蟽}\mathbf{<}\sqrt{\frac{\mathbf{(}\mathbf{n}\mathbf{-}\mathbf{1}\mathbf{)}{\mathbf{s}}^{\mathbf{2}}}{{\mathbf{蠂}}_{\mathbf{L}}^{\mathbf{2}}}} \\ \sqrt{\frac{\left(12-1\right){4.075}^2}{21.92}}<蟽<\sqrt{\frac{\left(12-1\right){4.075}^2}{3.8157}} \\ 2.887<蟽<6.919 \\ 2.9<蟽<6.9 \end{gathered}$$

Therefore, the 95% confidence interval estimate of the standard deviation of the population from which the sample was obtained is $\mathbf{2}\mathbf{.}\mathbf{9}\mathbf{mi}\mathbf{/}\mathbf{h}\mathbf{<}\mathbf{蟽}\mathbf{<}\mathbf{6}\mathbf{.}\mathbf{9}\mathbf{mi}\mathbf{/}\mathbf{h}$.