Question

The standard deviation of the running time of a sample of 200 fuses was found to be

100 hours. Find a 99% confidence interval for the standard deviation of all such fuses.

Answer 1

The critical values for $\alpha = 0.01$ and $df =n-1=200-1=199$ degrees of freedom are $\chi_L^2 = \chi^2_{1-\alpha/2,n-1} = 151.3699$ and $\chi_U^2 = \chi^2_{\alpha/2,n-1} = 254.1352.$

The corresponding confidence interval is computed as shown below:

$$CI(Variance)=(\dfrac{(n-1)s^2}{\chi^2_{\alpha/2,n-1}},\dfrac{(n-1)s^2}{ \chi^2_{1-\alpha/2,n-1}}) ​$$

$$=(\dfrac{(200-1)(100)}{254.1352},\dfrac{(200-1)(100)}{ 151.3699}) ​$$

$$=(78.3048, 131.4660)$$

Now that we have the limits for the confidence interval, the limits for the 99% confidence interval for the population standard deviation are obtained by simply taking the squared root of the limits of the confidence interval for the variance, so then:

$$CI(Standard \ Deviation)$$

$$=( \sqrt{78.3048}, \sqrt{131.4660})=(8.8490,11.4659)$$

Therefore, based on the data provided, the 99% confidence interval for the population variance is $78.3048 < \sigma^2 < 131.4660,$ and the 99% confidence interval for the population standard deviation is $8.8490 < \sigma < 11.4659.$