Getting P-value While Using Variance Suppose we observe a random sample of five measurements: 10, 13, 15, 15, 17, from a normal distribution with unknown mean $\mu_1$ and unknown variance $\sigma_1^2$, A second random sample from another normal population with unknown mean $\mu_2$ and unknown variance $\sigma_2^2$ yields the measurements: 13, 7, 9, 11.
a) Test for evidence that $\sigma_1 > 1.0$. Complete the P-value for this test as accurately as possible. Draw a conclusion at $\alpha = 0.05$.
Here's what I've done so far:
Step#1: Calculate $\sigma_1$
$\sigma_1 = \sqrt \frac{(10-14)^2 + (13-14)^2 + (15-14)^2 + (15-14)^2 + (17-14)^2}{5} = \sqrt\frac{28}{5} = 2.366$
Step#2: Set up Hypothesis Test
$H_0: \sigma_1 = 2.366$
$H_a: \sigma_1 > 1.0$
How do I proceed from here? Thanks.
EDIT:
Also have this question, and would appreciate some insight. 
b) Use the pivotal method(and a pivotal statistic with F distribution) to derive a 95% confidence interval for $\frac{\sigma_2}{\sigma_1}$. Work it out for these data. And test the null hypothesis that $\sigma_2 = \sigma_1$ at the 5% level of significance.
 A: for part a), you've made a wrong null hypothesis test.  You want to test for evidence that $\sigma_1\gt{1}\to\sigma_1^2\gt{1}$.  Thus, your hypotheses should be:
$$H_0:\sigma_1^2=1$$
$$H_1:\sigma_1^2\gt1$$
We need $n_1$ which is 5.  We also need our sample variance, which you have the square root of. Thus, $S^2=5.6$ and now we calculate our chi-square statistic, which is 
$$\chi^2=\frac{(n-1)s^2}{\sigma_0^2}=\frac{(5-1)(5.6)}{1}=22.4$$
We know our $\alpha=.05$, so we need to find $\chi_{\alpha}^2(n-1)=\chi_{.05}^2(4)=9.488.$
Since $22.4\ge9.488$, we reject $H_0.$
By p-value, we can see that at $\alpha=.01, P(W\ge{13.28})=.01$, so our p-value is less than .01. Also, $P(W\ge{14.86})=.005$, so it would appear that our p-value for this problem is VERY small indeed!
for part b), we know that $s_1^2=5.6$ and $s_2^2=\frac{(13-10)^2+(7-10)^2+(9-10)^2+(11-10)^2}{4}=5.$.  ALso, let m be the number of measurements from the first random sample and let n be the number of measurements from the second random sample.  Considering the scope of deriving the confidence, I give it to you freely here:
$$.95CI\left(\frac{\sigma_2^2}{\sigma_1^2}\right)=\left[\frac{1}{F_{\frac{\alpha}{2}}(n-1)(m-1)}\frac{s_2^2}{s_1^2},F_{\frac{\alpha}{2}}(m-1)(n-1)\frac{s_2^2}{s_1^2}\right]=\left[\frac{1}{F_{.025}(4)(3)}\frac{5.6}{5},F_{.025}(3)(4)\frac{5.6}{5}\right]$$
Since I gave you the freebee, you should try and finish the hypothesis test for the ratio using what you know of the F-statistic and the information provided...
