 # Statistics and Confidence Intervals. Given the following set of values: 10,11,14,95,73,30,29,9,97,94,70 foyerir 2022-09-11 Answered
Statistics and Confidence Intervals
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:
$x-{z}_{\frac{a}{2}}\frac{{\sigma }^{2}}{sqrt\left(n\right)}$ and $x+{z}_{\frac{a}{2}}\frac{{\sigma }^{2}}{sqrt\left(n\right)}$ where x denotes the mean. Here the significance level (a) is 0.01.
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Step 1
A conventional setup for the problem is that you have a sample $\mathbf{X}=\left({X}_{1},{X}_{2},\dots ,{X}_{n}\right)$ of size $n=11$ from a $N\left(\mu ,{\sigma }_{0}^{2}\right)$ 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
$Q\left(\mathbf{X},\mu \right)=\frac{\sqrt{n}\left(\overline{X}-\mu \right)}{{\sigma }_{0}}\sim N\left(0,1\right)$
, where $\overline{X}=\frac{1}{n}\sum _{i=1}^{n}{X}_{i}$ is the sample mean.
So if ${z}_{\alpha /2}$ be such that $P\left(Z>{z}_{\alpha /2}\right)=\alpha /2$ where $Z\sim N\left(0,1\right)$, you have
${P}_{\mu }\left(-{z}_{\alpha /2}\le Q\le {z}_{\alpha /2}\right)=1-\alpha \phantom{\rule{1em}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\mathrm{\forall }\phantom{\rule{thinmathspace}{0ex}}\mu$
You have to use the above to arrive at the form
${P}_{\mu }\left({c}_{1}\phantom{\rule{thinmathspace}{0ex}}\le \mu \le \phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}{c}_{2}\right)=1-\alpha \phantom{\rule{1em}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\mathrm{\forall }\phantom{\rule{thinmathspace}{0ex}}\mu$
$100\left(1-\alpha \right)\mathrm{%}$ confidence interval for $\mu$ is then $\left[{c}_{1},{c}_{2}\right]$.

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