Help with this chi square stats problem? I'm presented with:

A rivet is to be inserted into a hole. If the standard deviation of
  hole diameter exceeds 0.02 mm, there is an unacceptably high
  probability that the rivet will not fit. A random sample of n = 15
  parts is selected, and the hole diameter is measured. The sample
  standard deviation of the hole diameter measurements is s = 0.016 mm.
Is there strong evidence to indicate that the standard deviation of
  hole diameter exceeds 0.02 mm? Calculate (a) lower bound and upper
  bound of P-value to draw conclusions. Round your answers to 1 decimal.
  Answers are exact.
(b) Construct a 95% lower confidence bound for σ. Round your answer to
  3 decimal places.
(c) Use the confidence bound in part (a) to test the hypothesis (if
  reject enter a value of 1, if not then enter a value of 2).

I'm really confused with chi square. I'm allowed to use the calculator (TI-84) or the table. I am completely confused even on how to procede. Can anyone help out?
 A: $$
S^2=\frac{1}{15-1}\sum_{i=1}^{15} (X_i-\bar X)^2, \text{ where } \bar X=\frac{X_1+\cdots+X_{15}}{15}.
$$
The probablity distribution of the sample variance $S^2$ is given by
$$
\frac{(15-1)S^2}{\sigma^2} \sim \chi^2_{15-1}
$$
where $\sigma$ is the population standard deviation.  From a table, find $A$ and $B$ such that
$$
\Pr(A <\chi^2_{15-1}) = \Pr(\chi^2_{15}<B)= 0.025.
$$
If follows that
$$
\Pr(A<\chi^2_{15}<B) = 0.95.
$$
So
$$
\Pr\left(A<\frac{(15-1)S^2}{\sigma^2}<B\right)=0.95.
$$
With a bit of algebra we can isolate $\sigma^2$:
$$
\Pr\left( \frac{(15-1)S^2}{B} < \sigma^2 < \frac{(15-1)S^2}{A} \right) = 0.95.
$$
That is a $95\%$ confidence interval for $\sigma^2$.  Plug in the value of $S^2$.
Does $0.02^2$ fall within the confidence interval?  If not, one would reject the null hypothesis that $\sigma^2=0.02^2$ when testing at the $5\%$ level.
However, in view of the way the question was phrased, it would make more sense to test the null hypothesis that $\sigma^2\le 0.02^2$ --- a one-sided null hypothesis.  Then one would reject the null hypothesis at the $5\%$ level if a suitable one-sided confidence interval fails to contain $0.02^2$.  For that you'd want $\Pr(C<\chi^2_{15})=0.05$ and you'd get $\Pr\left(\sigma^2 < \dfrac{(15-1)S^2}{C}\right) = 0.95$ and there you'd have a one-sided $95\%$ confidence interval for $\sigma^2$.
