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To my understanding, I can invert a the normal test for $\ H_0: \theta \ = \theta_0 $ vs. $\ H_1: \theta > \theta_0 $

which means to get $\ H_1: \theta < \theta_0 $ and with that a (1-$\alpha $) confidence interval for $\theta $.

Is that then an exact confidence interval? If yes, why?

Would it make sense to assess its properties by the expected length? It that then a lower bound for $\theta $?

kjetil b halvorsen
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Bonsaibubble
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    Usually when people speak of "inverting a test" to get a confidence interval, they don't mean flipping the direction of the alternative. They mean identifying the set of values for the population parameter that would correspond to non-rejection of the null (i.e. "logical inversion" of the interval -- see the section on *consonance intervals* [here](https://projecteuclid.org/download/pdf_1/euclid.ss/1009213289)). Also see [Confidence intervals derived from 'inverted hypothesis test'](http://stats.stackexchange.com/questions/166478/confidence-intervals-derived-from-inverted-hypothesis-test) – Glen_b Jan 04 '17 at 02:34
  • On using inverted tests as *confidence intervals*, you might find this discussion on Andrew Gelman's blog "[*Why it doesn’t make sense in general to form confidence intervals by inverting hypothesis tests*](http://andrewgelman.com/2011/08/25/why_it_doesnt_m/)" of some value, in that it describes a circumstance where it doesn't make sense. – Glen_b Jan 04 '17 at 03:03

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