Radon exposure in Egyptian tombs. Refer to the Radiation Protection Dosimetry (December 2010) study of radon exposure in tombs carved from limestone in the Egyptian Valley of Kings, Exercise 6.30 (p. 349). The radon levels in the inner chambers of a sample of 12 tombs were determined, yielding the following summary statistics: $\overline{x}=3643Bq/m^3$and $s=4487Bq/m^3$. Use this information to estimate, with 95% confidence, the true standard deviation of radon levels in tombs in the Valley of Kings. Interpret the resulting interval.

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

$\left(\frac{\left(n-1\right)s^2}{{\chi }_{0.05\frac{\alpha }{2}}^2}⩽{\sigma }^2⩽\frac{\left(n-1\right)s^2}{{\chi }_{0.95}^2}\right)$

Let the confidence level be 0.95.

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

Degrees of freedom is (n-1)=11

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

Now,

$\begin{array}{rcl}\left(\frac{\left(n-1\right)s^2}{{\chi }_{0.05\frac{\alpha }{2}}^2}⩽{\sigma }^2⩽\frac{\left(n-1\right)s^2}{{\chi }_{0.95}^2}\right)& =& \left(\frac{\left(12-1\right){\left(4487\right)}^2}{21.92}⩽{\sigma }^2⩽\frac{\left(12-1\right){\left(4487\right)}^2}{3.8157}\right)\\ & =& \left(\frac{221464859}{21.92}⩽{\sigma }^2⩽\frac{221464859}{3.8157}\right)\\ & =& \left(10103323.86⩽{\sigma }^2⩽58039666.91\right)\\ & & \end{array}$

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

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

$\left(\sqrt{10103323.86}⩽\sigma ⩽\sqrt{58039666.91}\right)=\left(3178.573⩽\sigma ⩽7618.376\right)$

Thus, the 95% confidence interval for $\sigma$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.