The variability in the amount of impurities present in a batch of chemicals used for a particular process depends on the length of time that the process is in operation. The standard deviation for the traditional process is 1.24. A new process has been developed to reduce the variability of impurities. Suppose a random sample of operation times is drawn under the new process. The results are given below:
$5.02$, $5.65$, $4.83$, $6.01$, $5.61$, $5.25$, $5.36$, $4.84$, $5.27$, $5.38$, $5.03$, $5.14$, $5.37$, $5.67$, $4.68$, $5.25$
(a) Find a 90% confidence interval for the population variance. What assumption do you need to make?
(b) Since the purpose of the research is to reduce the variability, the smaller the SD the better. Hence only the upper bound SD is of interest. Find a 90% confidence upper bound for the population standard deviation.
(c) Make a statistic inference whether the new process reduces the variability of impurities. Explain.
What I have tried so far:
(a) For the first part, I started by computing the point estimator, where I ended up getting $S = 0.35234$. From there, I used the pivotal quantity $\frac{(n-1)S^2}{\sigma^2}$~$\chi^2(n-1)$. In order to computer the confidence interval, I did the following: $$P(\chi_{0.95}^2 \le \frac{(n-1)S^2}{\sigma^2} \le \chi_{0.05}^2) = P(\frac{1}{\chi_{0.95}^2} \ge \frac{\sigma^2}{(n-1)S^2} \ge \frac{1}{\chi_{0.05}^2}) = P(\frac{(n-1)S^2}{\chi_{0.95}^2} \ge \sigma^2 \ge \frac{(n-1)S^2}{\chi_{0.05}^2}) $$
So the derived confidence interval should come out to be: $(\frac{(n-1)S^2}{\chi_{0.05}^2},\frac{(n-1)S^2}{\chi_{0.95}^2})$ where $n=16$, $S^2 = 0.12414$, $\chi_{0.05}^2 = 24.9958$, and $\chi_{0.95}^2 = 7.2609$. Plugging everything in, I ended up with $(0.074497, 0.25645)$. Here, I believe we have to assume that this is a random sample that follows a normal distribution.
(b) Here I wasn't too sure how to set this up. I thought that if I want the upper limit, I would want to set it up as follows: $$P(\chi_{0.9}^2 \le \frac{(n-1)S^2}{\sigma^2})$$ Again, solving this inequality for $\sigma$ (not $\sigma^2$ since we want SD, not variance), I get $$P(\sigma \le \sqrt{\frac{(n-1)S^2}{\chi_{0.9}^2}})$$ Using the same n and $S^2$, and using $\chi_{0.9}^2=8.54675$, I plug everything in to get $P(\sigma \le 0.46677)$
I'm not sure if I did all of these parts right, but any help would be appreciated. Thanks in advance.
Best Answer
Seems to me you didn't quite finish. I believe you want a 2-sided confidence interval.
If $L$ and $U$ cut 5% from lower and upper tails of $\mathsf{Chisq}(n-1),$ respectively, then from your first displayed equation a 90% CI for $\sigma^2$ is of the form $\left((n-1)S^2/U,\, (n-1)S^2/L\right).$