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

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%.

The degrees of freedom are computed as follows:

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

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 ${\chi }_R^2=21.92$.

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

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

Let x represents the sample observations.

The mean value is computed asfollows:

$$\begin{gathered} \overline{x}=\frac{\sum x}{n} \\ =\frac{7+8+2+10+...+9}{12} \\ =7 \end{gathered}$$

The standard deviation is computed as follows:

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

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

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

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