Hypothesis testing for the variance of a population 
Someone is trying to introduce a new process in the production of a
  precision instrument for industrial use. The new process keeps the
  average weight but hopes to reduce the variability, which until now
  has been characterized by $\sigma^2 = 14.5$. 
Because the complete introduction of the new process has costs, a test
  has been done and 16 instruments have been produced with this new
  method. For $\alpha = 0.05$ and knowing that the sample variance $s^2
> = 6.8$, what is the decision to take? Suppose that the universe can be considered approximately normal.

I have the population variance, so I can use the normal distribution. My hypothesis is
$$H_0 : \sigma ^2 = 14.5$$
The test value:
$$X^2_0 = \frac{15*6.8^2}{14.5^2}) = 3.2989 $$
$$X^2_{\alpha,n-1} = X^2_{0.05,15} = 25.00$$
$X^2_0 > X^2_{\alpha,n-1}$ is false, so I fail to reject H_0?
 A: Because $\frac{(n-1)S^2}{\sigma^2} \sim \mathsf{Chisq}(\nu=n-1),$ a 95% confidence interval (CI) for $\sigma^2$ based on a sample variance $S^2$ is of the form
$\left(\frac{(n-1)S^2}{U},\, \frac{(n-1)S^2}{U}\right),$
where $L$ and $U$ cut probability $0.025$ from the lower and upper tails of $\mathsf{Chisq}(\nu=n-1),$ respectively. [This confidence interval is based on the assumption that weights are normal.]
Thus a 95% CI for $\sigma^2$ based on $S^2 = 6.8$ for a normal sample of size $n = 16$ is
$(3.71, 16.29).$  This CI represents non-rejectable values of $\sigma_0^2.$ [The computation is done below using R statistical software, but you can verify it using printed tables of the chi-squared distribution.]
15*6.8/qchisq(c(.975,.025), 15)
[1]  3.710657 16.288367

Because $\sigma_0^2 = 14.5$ lies in the CI so you cannot
reject $H_0: \sigma^2 = 14.5$ against $H_a: \sigma^2 \ne  14.5$ at the 5% level.
Can you find critical values for the test
statistic $Q = \frac{(n-1)S^2}{\sigma_0^2},$ beyond which the null hypothesis would be rejected (in either direction)?
Notes: There is some confusion in your question: (1) You do not state null and alternative hypotheses, so it is not clear whether you are doing a one- or two-sided test. (2) You seem to be mixing up sample standard deviation $S$ with $S^2$---you are squaring $S^2.$
If you want a one-sided test, you need a one-sided CI, which gives an upper bound $(-\infty, 14.05),$ which does not contain $14.5.$ 
15*6.8/qchisq(.05, 15)
[1] 14.04776

In this case, can you find a critical value for the test statistic $Q =\frac{(n-1)S^2}{\sigma_0^2},$ below which the null hypothesis
$H_0: \sigma^2 = 14.5$ would be rejected in favor of the one-sided
alternative $H_a: \mu < 14.5?$
A: 
"Suppose that the universe can be considered approximately normal."

If you are willing to make the assumption that the underlying data are IID normal then you can derive a confidence interval from the pivotal quantity $(n-1) S^2/\sigma^2 \sim \text{ChiSq}(n-1)$.  The other answer by BruceET shows you how to do this, and this will solve your immediate question.
However, if you want to learn sampling theory properly, I  strongly recommend you go beyond the immediate question here, and think about what you would need to do if you are not willing to make the (ridiculous) assumption that "the universe is approximately normal".  You can find a more general discussion of moment results in sampling theory, and a solution to this broader problem, in O'Neill (2014).  The main thing to note is that the variance of the sample variance depends on the kurtosis of the underlying population distribution.  If you assume a normal distribution then you are assuming knowledge of the kurtosis of the population, without reference to the actual data.  This means that you will often over or underestimate the variability of the sample variance and get bad results for confidence intervals for the true variance.
As a general rule, if you want to look at the true moments of sample moments, you need to add their orders to find out how many moments of the underlying population affect this.  Thus, determination of the variance (second order) of the sample variance (second order) requires knowledge up to the fourth moment of the underlying population distribution.  It is generally a bad idea to ignore the data and instead assume specific values for underlying moments that affect the problem at hand.  Thus, when formulating a confidence interval for the true variance of the population, it is generally a bad idea to assume normality of the population, since that assumes a specific value for the kurtosis, which affects the variance of the sample variance.
