Inverting a test in order to get a (1-$\alpha $)-confidence interval for $\theta $

Asked Dec 03 '16 at 16:09

Active Nov 05 '18 at 11:12

Viewed 2,467 times

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 $?

edited Nov 05 '18 at 11:12

asked Dec 03 '16 at 16:09

Bonsaibubble

2

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). Also see 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" 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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