A sample of $n \in \mathbb{N}$ screws is taken from a large batch of screws that are produced. The length of a bolt is approximated as normally distributed with variance $4$ and the lengths of the screws are distributed as independent and identical.
(a) $n = 10$ screws were taken and their length (in mm) was measured, with the following result: $$10 \ \ \ 8 \ \ \ 9 \ \ \ 10 \ \ \ 11 \ \ \ 11 \ \ \ 9 \ \ \ 12 \ \ \ 12 \ \ \ 8$$ Calculate the sample mean and determine a two-sided confidence interval for the above sample for the mean length of the confidence probability $0.95$. Also state the quantile used and its (approximate) value.
(b) How many screws had to be tested at least, thus a confidence interval for the mean length with a confidence level of $0.95$ has at most the length $2$ or length 1? Explain.
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I have done the following :
(a) The sample mean is equal to \begin{equation*}\overline{x}_{10}=\frac{1}{10}\left (10+8+9 + 10 + 11 + 11 + 9 + 12 + 12 +8\right )=\frac{1}{10}\cdot 100=10\end{equation*}
For the two-sided confidence interval do we use the formula \begin{equation*}\left [M(x)-\frac{\sigma}{\sqrt{n}}q_{1-\frac{a}{2}},M(x)+\frac{\sigma}{\sqrt{n}}q_{1-\frac{a}{2}}\right ] \end{equation*} where $M(x)$ is the mean, i.e. $M(x)=10$ and $\sigma=\sqrt{4}=2$ ? The value of $q$ is related to the confidence level, right?