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- Apr 14, 2013

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The variable $X$ is normally distributed with unknown expected value $\mu$ and unknown variance $\sigma^2$.

I want to determine the confidence interval for $\mu$ for $n=22; \ \overline{X}_n=7.2; \ S'=4; \ 1-\alpha=0.90$.

Is $S'$ the standard deviation? In some notes that I found in Google, it symbolized $s_n$ the standrard deviation of a sample and by $s_n'$ if we have the total population instead of a sample. If it is meant here like that, we follow the same steps to determine the confidence intervall in both case, or not?

If yes, I have done the following:

We have that $1-\alpha=0.90 \Rightarrow \alpha=0.10$.

So, we get $1-\frac{\alpha}{2}=1-\frac{0.10}{2}=1-0.05=0.95$.

Therefore we have $z =1.645$.

The median is equal to $\overline{X}_n=7.2$.

The half of the length of the confidence interval is equal to \begin{equation*}\frac{\sigma\cdot z}{\sqrt{n}}=\frac{S'\cdot z}{\sqrt{n}}=\frac{4\cdot 1.645}{\sqrt{22}}=1.40286\end{equation*}

The confidence interval is therefore equal to \begin{equation*}KI=\left [ \overline{x}-\frac{\sigma\cdot z}{\sqrt{n}};\overline{x}+\frac{\sigma\cdot z}{\sqrt{n}}\right ]=\left [ 7.2-1.40286;7.2+1.40286\right ]=\left [ 5.79714;8.60286\right ]\end{equation*}

Is everything correct?