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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?

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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.

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  • $\begingroup$ @mdewwy but the interval you described includes the null $\mu$, so I can't reject the null hypothesis, which contradicts the p-value test. How is that possible? $\endgroup$ – gbi1977 Jan 6 at 13:52
  • $\begingroup$ @mdewwy Thanks, as I said this is the same result I got - I included the left rejection area of the two-sided interval. My question is why the 2 intervals are equivelant? just because they have the same $\alpha$ now? I'm not satisfied with this kind of answer... $\endgroup$ – gbi1977 Jan 6 at 13:59
  • $\begingroup$ What goes against my intuition, I guess, is why the proper one-sided inteval, which is also the non-rejection area, includes smaller possible values for $\mu$? Isn't the whole purpose of the non-rejection area to be "closer" to the null $\mu$? $\endgroup$ – gbi1977 Jan 6 at 14:35
  • $\begingroup$ The two intervals are not equivalent. One is a two-sided 90% the other one-sided 95% so both have 5% in one tail and one of them has a further 5% in the other tail $\endgroup$ – mdewey Jan 6 at 14:38
  • $\begingroup$ I think this is the understanding I lacked. It's counter-intuitive. $\endgroup$ – gbi1977 Jan 6 at 14:53

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