91影视

The$100\left(1-伪\right)\%$confidence interval is given below:

$\left(\frac{\left(n-1\right)s^2}{{蠂}_{0.05\frac{伪}{2}}^2}猢�{蟽}^2猢�\frac{\left(n-1\right)s^2}{{蠂}_{0.95}^2}\right)$

Let the confidence level be 0.95.

$\begin{array}{rcl}1-伪& =& 0.95\\ 伪& =& 1-0.95\\ & =& 0.05\\ \frac{伪}{2}& =& 0.025\\ & & \end{array}$

Degrees of freedom is (n-1)=11

From table, the value of ${蠂}_{0.025}^2=21.92$and${蠂}_{0.0975}^2=3.8157$

Now,

$\begin{array}{rcl}\left(\frac{\left(n-1\right)s^2}{{蠂}_{0.05\frac{伪}{2}}^2}猢�{蟽}^2猢�\frac{\left(n-1\right)s^2}{{蠂}_{0.95}^2}\right)& =& \left(\frac{\left(12-1\right){\left(4487\right)}^2}{21.92}猢�{蟽}^2猢�\frac{\left(12-1\right){\left(4487\right)}^2}{3.8157}\right)\\ & =& \left(\frac{221464859}{21.92}猢�{蟽}^2猢�\frac{221464859}{3.8157}\right)\\ & =& \left(10103323.86猢�{蟽}^2猢�58039666.91\right)\\ & & \end{array}$

Hence, the 95% confidence interval for${蟽}^2$ is$\left(10103323.86,58039666.91\right)$

Now, take the square root on the lower and upper limit. That is,

$\left(\sqrt{10103323.86}猢�蟽猢�\sqrt{58039666.91}\right)=\left(3178.573猢�蟽猢�7618.376\right)$

Thus, the 95% confidence interval for $蟽$is $\left(3178.573,7618.376\right)$

The true standard deviation of radon levels in tombs in the Valley of kings lies between 3178.573 $Bq/m^3$ and 7618.376 $Bq/m^3$at 95% confidence interval.