In Exercises 9–16, assume that each sample is a simplerandom 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

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%.

The degrees of freedom are computed as follows:

$$\begin{gathered} df=n-1 \\ =26-1 \\ =25 \end{gathered}$$

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 25 degrees of freedom are ${\chi }_L^2=13.12$and ${\chi }_R^2=40.646$.

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:

$$\begin{gathered} \overline{x}=\frac{\sum x}{n} \\ =\frac{5+8+3+8+...+8+7}{26} \\ =6.577 \end{gathered}$$

The standard deviation is computed as follows:

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

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

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

Therefore, the 95% confidence interval estimate of the standard deviation of the population from which the sample was obtained is role="math" localid="1648111628961" $\mathbf{1}\mathbf{.}\mathbf{5}\mathbf{}\mathbf{<}\mathit{\sigma }\mathbf{<}\mathbf{2}\mathbf{.}\mathbf{6}\mathbf{}$
.