Question

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

Speed Dating In a study of speed dating conducted at Columbia University, male subjects were asked to rate the attractiveness of their female dates, and a sample of the results is listed below (1 = not attractive; 10 = extremely attractive). Construct a 95% confidence interval estimate of the standard deviation of the population from which the sample was obtained.

7 8 2 10 6 5 7 8 8 9 5 9

Short answer

The 95% confidence interval estimate of the standard deviation of the population from which the sample was obtained is \(1.6 < \sigma < 3.8\).

Step-by-step solution

Given information

The sample number of male subjects that were asked to rate the attractiveness of their female dates 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:

\(\begin{array}{c}\bar x = \frac{{\sum x }}{n}\\ = \frac{{7 + 8 + 2 + 10 + ... + 9}}{{12}}\\ = 7\end{array}\)

The standard deviation is computed asfollows:

\(\begin{array}{c}s = \sqrt {\frac{{\sum {{{\left( {x - \bar x} \right)}^2}} }}{{n - 1}}} \\ = \sqrt {\frac{{{{\left( {7 - 7} \right)}^2} + {{\left( {8 - 7} \right)}^2} + {{\left( {2 - 7} \right)}^2} + ... + {{\left( {9 - 7} \right)}^2}}}{{12 - 1}}} \\ = 2.216\end{array}\)

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:

\(\begin{array}{c}\sqrt {\frac{{\left( {n - 1} \right){s^2}}}{{\chi _R^2}}} < \sigma < \sqrt {\frac{{\left( {n - 1} \right){s^2}}}{{\chi _L^2}}} \\\sqrt {\frac{{\left( {12 - 1} \right){{2.216}^2}}}{{21.92}}} < \sigma < \sqrt {\frac{{\left( {12 - 1} \right){{2.216}^2}}}{{3.8157}}} \\1.5698 < \sigma < 3.7625\\1.6 < \sigma < 3.8\end{array}\)

Therefore, the 95% confidence interval estimate of the standard deviation of the population from which the sample was obtained is\(1.6\; < \sigma < 3.8\;\).