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Suppose we have $2$ independent population parameters $p_1$ and $p_2$, such that the $90$% ( symmetric ) confidence intervals for $p_1$ and $p_2$ are given by $ (0.411, 0.498) $ and $(0.473, 0.567)$ respectively. Now, according to the original question ( copied verbatim ) , supposing that the $90 $% interval for $p_1$ lies entirely below the interval for $p_2$, how confident are we to conclude that $p_1 < p_2$?

Now, assuming that I have understood the problem correctly, I feel that this question does not make sense to me because $(0.411, 0.498)$ does not lie completely behind $(0.473,0.567)$. The question is, am I right in claiming so?

Next, assuming the problem is simply "How confident are we to conclude that $p_1 < p_2 $ ", given their $2$ $90$% symmetric confidence intervals as above, how would we go about approaching it, if it is even possible?

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When we say that an interval has confidence 95% we mean that in repeated sampling we would find that intervals constructed in the same way would cover the true value of the parameter about 95% of the times: the confidence is the probability of coverage in repeated experimentation.

Now, I do not think you can talk about "How confident are we to conclude that $p_1<p_2$" when no confidence interval is involved: $p_1 < p_2$ that will be either true or false. Perhaps from a Bayesian point of view you could talk about the posterior probability that $p_1 < p_2$ given that something happens, but that is an entirely different sort of statement than what goes under the name of confidence.

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    $\begingroup$ Yes, so what? My answer was that you cannot properly talk of "the confidence that $p_1 < p_2$". $\endgroup$ – F. Tusell Apr 19 '20 at 16:20
  • $\begingroup$ ok, I understand your point now. Thank you! $\endgroup$ – ONG SEE HAI Apr 19 '20 at 16:30

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