Given the following set of values:
10,11,14,95,73,30,29,9,97,94,70
How do I calculate a 99% confidence interval for the sample mean? I am assuming that the variance is 10
Well, the idea I have is to assume that the distribution is normal, but after that i'm not completely sure what to do next. In particular, I am unable to find the z-score that corresponds to the z-score in the formula for the CI, i'm not sure what to input in R to find the CI. The formula for the confidence interval is:
$x-z_{\frac{a}{2}} \frac{\sigma^2}{sqrt(n)}$ and $x+z_{\frac{a}{2}} \frac{\sigma^2}{sqrt(n)}$ where x denotes the mean. Here the significance level (a) is 0.01.
It is likely that you are asked to find a confidence interval for the (unknown) population mean, not the sample mean. The sample mean is not a parameter of interest, it can be calculated from the sample.
A conventional setup for the problem is that you have a sample $\mathbf X=(X_1,X_2,\ldots,X_n)$ of size $n=11$ from a $N(\mu,\sigma_0^2)$ population (by assumption) with $\sigma_0^2=10$. You have to find a confidence interval for the mean $\mu$.
A suitable pivotal quantity here is $$Q(\mathbf X,\mu)=\frac{\sqrt{n}(\overline X-\mu)}{\sigma_0}\sim N(0,1)$$
, where $\overline X=\frac{1}{n}\sum\limits_{i=1}^n X_i$ is the sample mean.
So if $z_{\alpha/2}$ be such that $P(Z>z_{\alpha/2})=\alpha/2$ where $Z\sim N(0,1)$, you have $$P_{\mu}(-z_{\alpha/2}\le Q\le z_{\alpha/2})=1-\alpha\quad\,\forall\,\mu$$
You have to use the above to arrive at the form $$P_{\mu}(c_1\,\le \mu\le\,\, c_2)=1-\alpha\quad\,\forall\,\mu$$
A $100(1-\alpha)\%$ confidence interval for $\mu$ is then $[c_1,c_2]$.