Combining results from multiple distributions

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The discussion focuses on calculating the probability of an overlap interval (35-38.5) derived from three different test methods, each providing distinct 95% confidence intervals for a sample. It is established that the tests yield normally distributed outcomes, with means centered in their respective intervals. The consensus is that the combined probability for the overlap should exceed 95%, given that more consistent results typically increase confidence. A method was proposed that calculated the probability of results falling outside the overlap, yielding approximately 96% for the combined tests. This approach aligns with intuitive expectations, though caution is advised regarding reliance on intuition in statistical analysis.
mwoldinga
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Suppose I have a result where the outcome is that with 95% confidence interval of the sample is between 27 and 42. With a second method the test result of the same sample gives a 95% confidence interval between 27 and 48. And with a third method 95% confidence interval of the sample is between 35 and 38.5.
The overlap between the three intervals is in this case 35-38.5. How can I find out what the probability is for this interval, using all three test results?
Let's assume each of the three different test methods give an outcome that has a normal distribution, where the mean each time lies in the middle of each interval.
Of course the answer has to lie above 95%, since the probability will increase if you have more results that do not contradict the last one, but how can I calculate this?
Any suggestions?
Monica (The Netherlands)
 
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mwoldinga said:
Suppose I have a result where the outcome is that with 95% confidence interval ... The overlap between the three intervals is in this case 35-38.5. How can I find out what the probability is for this interval, using all three test results?...

Interpreting the CI percentages as exact probabilities, we could say for example there's a 0% chance the mean lies between 27 and 35 or 38.5 and 48, and 5% chance that the mean lies outside of 27 to 48 - I hope you agree that's nonsense.

...Let's assume each of the three different test methods give an outcome that has a normal distribution, where the mean each time lies in the middle of each interval.
Of course the answer has to lie above 95%, since the probability will increase if you have more results that do not contradict the last one, but how can I calculate this?
Any suggestions?

With a few additional assumptions we'd get one answer but then slightly different assumptions could give quite a different result.

It may be helpful to write down what facts are known about the 3 methods before translating those into assumptions.
 
I don't think it is very helpful if write down more facts about the methods. The tests are about dating bones. Each of the three test will give an outcome with a different confidence interval, using the same bone each time.
I calculated the probability where the boneresults were outside the overlap in all three cases. 1-this answer is then the probability that the boneresults are not outside the overlap in all three cases. This answer is about 96%. At least this method fits with my intuition, but intuition in statistics is a dangerous thing.
 
I'm taking a look at intuitionistic propositional logic (IPL). Basically it exclude Double Negation Elimination (DNE) from the set of axiom schemas replacing it with Ex falso quodlibet: ⊥ → p for any proposition p (including both atomic and composite propositions). In IPL, for instance, the Law of Excluded Middle (LEM) p ∨ ¬p is no longer a theorem. My question: aside from the logic formal perspective, is IPL supposed to model/address some specific "kind of world" ? Thanks.
I was reading a Bachelor thesis on Peano Arithmetic (PA). PA has the following axioms (not including the induction schema): $$\begin{align} & (A1) ~~~~ \forall x \neg (x + 1 = 0) \nonumber \\ & (A2) ~~~~ \forall xy (x + 1 =y + 1 \to x = y) \nonumber \\ & (A3) ~~~~ \forall x (x + 0 = x) \nonumber \\ & (A4) ~~~~ \forall xy (x + (y +1) = (x + y ) + 1) \nonumber \\ & (A5) ~~~~ \forall x (x \cdot 0 = 0) \nonumber \\ & (A6) ~~~~ \forall xy (x \cdot (y + 1) = (x \cdot y) + x) \nonumber...

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