Determine the confidence level for the CI : $]14.6727, 37.38[$

Question

The lifetimes of eight computer batteries are 151, 153, 175, 134, 170, 172, 156, and 114 minutes. Assume that the durations are normally distributed. Peter, a student from statistically correctly determined a CI for the population standard deviation, obtaining $$CI: ]14.6727, 37.38[ $$

Determine the confidence level of the CI determined by Peter.

I did the following:

$\mu = 153.125$

$s= 20.8014$

$t_{\alpha/2,n-1}=t_{\alpha/2,7}$

$CI= \mu \pm \frac{s}{\sqrt{n}} t_{\alpha/2,7}$ (I used the -)

$\implies 14.6727= 153.125 - \frac{20.8014}{\sqrt{8}} t_{\alpha/2,7}$

...

$\implies t_{\alpha/2,7} = 18.82576$

When I check the table for a value that works for me, the $\frac{\alpha}{2}=0.05$ seems to work. But it is 1.895 instead of 18.82576

Answer 1

The t distribution has nothing to do with a CI for $\sigma.$

A 95% CI for normal variance $\sigma^2$ is of the form $\left(\frac{(n-1)S^2}{U},\,\frac{(n-1)S^2}{L} \right),$ where $L$ and $U$ cut probability $0.025$ from the lower and upper tails, respectively, of $\mathsf{Chisq}(\nu=n-1).$ Take square roots of endpoints to get a 95% CI for $\sigma.$

For your data, in R, we have the computations below, which give Peter's CI. So, what confidence level did he use?

x = c(151, 153, 175, 134, 170, 172, 156, 114) 
v = var(x);  v
[1] 432.6964
sqrt(7*v/qchisq(c(.95,.05),7))
[1] 14.67364 37.38317