# Confidence interval from p value

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

$$1-2p=P(\bar x when $$\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 $$c=1-2p$$?