Getting P-value While Using Variance

Question

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.

Answer 1

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...