So, I have a population which is believed to have $H_0:\mu = 4620$ and I should test for $H_1: \mu < 4620$. Given a sample of $n=81, \bar{X}=4450, S=900$, the p-value is: $pv=\Pr\left(\frac{\bar{X}-4620}{900/\sqrt{81}}\leq\frac{4450-4620}{900/\sqrt{81}}\right)\\=\Pr\left(Z\leq-\frac{170}{100}\right)\\\approx0.0445<\alpha=0.05$
So I reject the null hypothesis.
Then, I'm being asked to find a confidence interval for $\alpha=0.1$, which is: $CI_{0.9}\left(\mu\right)=\left[\bar{X}\pm\frac{S}{\sqrt{n}}\cdot t_{80,0.95}\right]=\left[4283.588,4616.412\right]$
and to infer from it, again, that the null should be rejected with $\alpha=0.05$.
Now, clearly $\mu_{H_{0}}>\sup CI_{0.9}$ so we can reject with $\alpha=0.1$. Furthermore, if I take the one-sided confidence interval with $\alpha =0.05$ it has the same upper bound as the two-sided interval. It seems like all we have done is to expand the non-rejection region to the left tail of the two-sided interval.
But what I'm having trouble with is how to formulate the argument – why does it make sense? Is it enough to state that the one-sided interval overlaps the two-sided one? That the values in the left tail, for the two-sided case, "don't matter"?
Why is this transformation of the two-sided interval into a suitable one-sided interval valid, in this case? Is there a general method/explaination of why these intervals are equivelant, if at all?
Edit:
Found this answer Matching Confidence limits with One-Sided Hypothesis tests
which explains it the other way around – why the two-sided interval should have been constructed with $\alpha=0.9$. But even after drawing the bell curve and coloring the expanded acceptance region, I can't understand the logic behind "merging" the left rejection area to the interval. It fits the desired $\alpha$, that's fine, but why would anyone test a one-sided hypothesis using a two-sided interval in the first place?
Best Answer
The thing which is confusing you is where to put the lower bound. In the two-sided 90% interval you give the lower bound is finite (4283.588). If you want a 95% one-sided interval it has the same upper bound (as you say) but the upper bound is infinite. So it is from $-\infty$ to 4616.412.