Question

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

Short answer

The 95% confidence interval estimate of the standard deviation of the population from which the sample was obtained is \(2.9\;{\rm{mi/h}}\; < \sigma < 6.9\;{\rm{mi/h}}\).

Step-by-step solution

Given information

The sample number of observations is\(n = 12\).

The level of confidence is 95%.

Compute the critical values

The degrees of freedom are computed asfollows:

\(\begin{array}{c}df = n - 1\\ = 12 - 1\\ = 11\end{array}\)

The level of confidence is 95%,whichimplies 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 \(\chi _L^2 = 3.8157\) and \(\chi _R^2 = 21.92\).

Compute the mean and standard deviation

The confidence interval for the standard deviation is given asfollows:

\(\sqrt {\frac{{\left( {n - 1} \right){s^2}}}{{\chi _R^2}}} < \sigma < \sqrt {\frac{{\left( {n - 1} \right){s^2}}}{{\chi _L^2}}} \)

Let x represents the sample observations.

The mean value is computed asfollows:

\(\begin{array}{c}\bar x = \frac{{\sum x }}{n}\\ = \frac{{62 + 61 + 61 + 57 + ... + 60 + 67}}{{12}}\\ = 60.667\end{array}\)

The standard deviation is computed asfollows:

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

Construct the confidence interval

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

\(\begin{array}{c}\sqrt {\frac{{\left( {n - 1} \right){s^2}}}{{\chi _R^2}}} < \sigma < \sqrt {\frac{{\left( {n - 1} \right){s^2}}}{{\chi _L^2}}} \\\sqrt {\frac{{\left( {12 - 1} \right){{4.075}^2}}}{{21.92}}} < \sigma < \sqrt {\frac{{\left( {12 - 1} \right){{4.075}^2}}}{{3.8157}}} \\2.887 < \sigma < 6.919\\2.9 < \sigma < 6.9\end{array}\)

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