Let $Y_1
Confidence interval for a sample from the norm dist
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0The order statistics don't even matter in this case because you are considering their mean. – 2011-11-14
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0Any progress on [this](http://math.stackexchange.com/q/33993/6179)? – 2011-11-16
1 Answers
The expression $P(\bar{X} - c\sigma < \mu < \bar{X} + c\sigma)$ is not actually an equation until you put "$=\text{something}$" after it. Presumably you want it to be equal to some number between $0$ and $1$. Let's call that number $p$.
Notice that $\bar{X} = (Y_1+Y_2)/2$, and $\bar{X} \sim \mathcal{N}(\mu, \sigma^2/2)$. So $$ \frac{\bar{X} - \mu}{\sigma/\sqrt{2}} \sim \mathcal{N}(0,1). $$ Now find $c$ such that $P(-c
Then notice that $$ -c < \frac{\bar{X}-\mu}{\sigma/\sqrt{2}} < c \text{ if and only if }\bar{X}-c\frac{\sigma}{\sqrt{2}} < \mu < \bar{X}+c\frac{\sigma}{\sqrt{2}}. $$
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0Let's say p = 0.5, what would you get for c? I get c to be 0.67 to satisfy this: P(−c
– 2011-11-14 -
0The value of $c$ for which $\Pr(-c
is the same as the value for which $\Pr(Z , and if I can trust the software package I'm using, that is 0.7733726. – 2011-11-15