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, female subjects were asked to rate the attractiveness of their male 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.

5 8 3 8 6 10 3 7 9 8 5 5 6 8 8 7 3 5 5 6 8 7 8 8 8 7

Short answer

The 95% confidence interval estimate of the standard deviation of the population from which the sample was obtained is \(1.5 < \sigma < 2.6\).

Step-by-step solution

Given information

The sample number of female subjects that were asked to rate the attractiveness of their male dates is\(n = 26\).

The level of confidence is 95%.

Compute the critical values

The degrees of freedom are computed asfollows:

\(\begin{array}{c}df = n - 1\\ = 26 - 1\\ = 25\end{array}\)

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 25 degrees of freedom are \(\chi _L^2 = 13.12\) and \(\chi _R^2 = 40.646\).

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:

\(\begin{array}{c}\bar x = \frac{{\sum x }}{n}\\ = \frac{{5 + 8 + 3 + 8 + ... + 8 + 7}}{{26}}\\ = 6.577\end{array}\)

The standard deviation is computed asfollows:

\(\begin{array}{c}s = \sqrt {\frac{{\sum {{{\left( {x - \bar x} \right)}^2}} }}{{n - 1}}} \\ = \sqrt {\frac{{{{\left( {5 - 6.577} \right)}^2} + {{\left( {8 - 6.577} \right)}^2} + {{\left( {3 - 6.577} \right)}^2} + ... + {{\left( {7 - 6.577} \right)}^2}}}{{26 - 1}}} \\ = 1.880\end{array}\)

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:

\(\begin{array}{c}\sqrt {\frac{{\left( {n - 1} \right){s^2}}}{{\chi _R^2}}} < \sigma < \sqrt {\frac{{\left( {n - 1} \right){s^2}}}{{\chi _L^2}}} \\\sqrt {\frac{{\left( {26 - 1} \right){{1.880}^2}}}{{40.646}}} < \sigma < \sqrt {\frac{{\left( {26 - 1} \right){{1.880}^2}}}{{13.12}}} \\1.4744 < \sigma < 2.595\\1.5 < \sigma < 2.6\end{array}\)

Therefore, the 95% confidence interval estimate of the standard deviation of the population from which the sample was obtained is\(1.5\; < \sigma < 2.6\;\).