In Exercises 9–16, assume that each sample is a simple random sample obtained from a population with a normal distribution.

Highway Speeds Listed below are speeds (mi/h) measured from southbound traffic onI-280 near Cupertino, California (based on data from Sig Alert). This simple random sample was obtained at 3:30 PM on a weekday. Use the sample data to construct a 95% confidence interval estimate of the population standard deviation. Does the confidence interval describe the standard deviation for all times during the week?

62 61 61 57 61 54 59 58 59 69 60 67

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" ${\mathit{\chi }}_{\mathbf{L}}^{\mathbf{2}}\mathbf{=}\mathbf{3}\mathbf{.}\mathbf{8157}$and ${\mathit{\chi }}_{\mathbf{R}}^{\mathbf{2}}\mathbf{=}\mathbf{21}\mathbf{.}\mathbf{92}$.

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

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

Let x represents the sample observations.

The mean value is computed as follows:

$$\begin{gathered} \overline{x}=\frac{\sum x}{n} \\ =\frac{62+61+61+57+...+60+67}{12} \\ =60.667 \end{gathered}$$

The standard deviation is computed as follows:

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

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

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

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