91影视

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 ${蠂}_L^2=3.8157$and ${蠂}_R^2=21.92$.

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

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

Let x represents the sample observations.

The mean value is computed asfollows:

$$\begin{gathered} \stackrel{炉}{x}=\frac{鈭�x}{n} \\ =\frac{7+8+2+10+...+9}{12} \\ =7 \end{gathered}$$

The standard deviation is computed as follows:

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

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

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

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