The question is: given that $H_0: \mu=34, H_a:\mu<34$ gives p-value $p$, find the largest confidence level, $c$, that does not include $34$.

The answer does this:$1-2p=c$ But I do t understand this. I have gotten this far:

$p=P(x<\bar x)$ if $\mu=34$ and $\sigma \approx S_x/\sqrt(n)$ so $2p= P(x<\bar x \cup x>68-\bar x)$ thus

$$ 1-2p=P(\bar x<x<68-\bar x)$$ when $\mu=34, \sigma \approx S_x/\sqrt(n)$

Now, $$P(\bar x-t^*\frac{S_x}{\sqrt(n)}<\mu<\bar x+t^*\frac{S_x}{\sqrt(n)})=c$$ when $$\mu = \bar x$$

How do I relate the two; why is $c=1-2p$?


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