Confidence interval for unknown $\mu$ and $\sigma^2$ I am stuck trying to solve this problem. I can only get to a certain point in my attempts. Any help is greatly appreciated.
I am given the following sample of size $10$ drawn from a $\mathcal N(\mu, \sigma^2)$-distributed population:
$$(100.4, 99.2, 98.4, 99.6, 103.6, 101.6, 103.2, 99.2, 97.6, 99.2)$$

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*Determine the $95 \%$ confidence interval for the unknown expected value $\mu$ and known variance $\sigma^2=4$

*Determine the $95 \%$ confidence interval for the unknow expected value $\mu$ and unknown variance $\sigma^2$

*Determine the $95\%$ confidence interval for the unkown variance $\sigma^2$ and unknown expected value $\mu$
What I have tried so far (with some resources from the web):

*

*$$\alpha=1-0.95=0.05\\ \Phi(c)=\frac{1}{2}\left(1-\frac{\alpha}{2}\right)=0.975$$
The mean of the sample is: $$\bar{X}=\frac{1002}{10}=100.2$$
Then I can use the following formula: $$\bar{X}-c \frac{\sigma}{\sqrt{n}}\le \mu \le \bar{X}+c \frac{\sigma}{\sqrt{n}}$$
I know $\bar{X}=100.2, \sigma=2,n=10$ but how can I find $c$?


*For question two, is it just the exact same calculation as 1. but now using the sample variance?


*What is the difference to question 2. ? Aren't they the same?
 A: In part 1, for $95\%$ confidence interval concerning normal distribution, the confidence coefficient $c$ is usually taken to be $1.96$. Of course, as pointed out by Amaan in the comment, a normal distribution table or software could be helpful.
I am not sure what is the exact difference between 2 and 3, but the main difference here is that $\sigma^2$ is unknown. In that case, you should use the sample variance $$s^2=\frac{1}{n-1}\sum_{i=1}^{n}{(x_i-\bar{x})^2}$$ to present an estimate of $\sigma^2$. In that case, you should know that $$\frac{\bar{X}-\mu}{\sqrt{s^2/n}}\sim t_{n-1}.$$ Thus a $95\%$ confidence interval for $\mu$ is given by $$\bar{X}-t_{n-1,0.975}\cdot\sqrt{\frac{s^2}{n}}<\mu<\bar{X}+t_{n-1,0.975}\cdot\sqrt{\frac{s^2}{n}}.$$ Here $t_{n-1,0.975}$ is the $97.5\%$ quantile of $t$-distribution with degree of freedom $n-1$. Note that $t$-distribution is also symmetric (Similar to normal distribution $t_{n-1,0.025}=-t_{n-1,0.975}$), so totally $0.025\cdot 2=0.05$ of probability is excluded.
On the other hand, note that $$\frac{(n-1)s^2}{\sigma^2}\sim\chi_{n-1}^2,$$ where $\chi_{n-1}^2$ is the chi-square distribution of degree of freedom $n-1$. Again, let $\chi_{n-1,0.025}^2$ and $\chi_{n-1,0.975}^2$ be the $2.5\%$ and $97.5\%$ quantiles of such distribution, resp. Then a $95\%$ CI is given as follows: $$\chi_{n-1,0.025}^2<\frac{(n-1)s^2}{\sigma^2}<\chi_{n-1,0.975}^2\implies\frac{(n-1)s^2}{\chi_{n-1,0.975}^2}<\sigma^2<\frac{(n-1)s^2}{\chi_{n-1,0.025}^2}.$$
For more about this, I recommend you to check the online lecture notes here.
