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Comment by tprice7 | Hacker News Reader

And in anticipation of the rebuttal that the probability is still 0 because it's never exactly normal, what I really meant originally but didn't write out explicitly for brevity and because I assumed it would be implicit: when I talk about giving an upper bound on p(null | data) from p(data | null), what I really mean is giving an upper bound on p(null | data, normal) from p(data | null, normal) where normal is the assumption that the distribution of whatever parameter we are looking at is normally distributed and null is the event that the difference in means between the two groups we are looking at it is less than some predetermined positive threshold. Or, for a 1-sample test, that the mean of a single group is within that threshold of some default value.

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If you write out the actual calculation you will see normality (which was just one example of an assumption) is actually part of the null model being tested. It is not something different or outside of it.

That is just a trivial semantics issue, and yes I am familiar with the calculation.

Where do you think the flaw is specifically? Say we are doing a 1-sample test.

0. (setup) Suppose we have a real number mu and a positive epsilon. Define the interval I as [mu – epsilon, mu + epsilon]. For each “candidate mean” within this interval, we have a corresponding t-statistic. Let the statistic t_0 be the inf of all these t-statistics. Let T be the event that t_0 is at least as big as the observed value.

1. You can use Student’s t-distribution to compute an upper bound for the probability of T under the assumptions that the observations are iid normal and the mean lies in I. I will call this probability p(T | null, normal, iid), where “null” is the event that the mean exists and is in I. It makes no difference that it is more typical to lump these assumptions together as “null” because in math you can define things however you want as long as you are consistent.

2. We have that p(null | T, normal, iid) = p(T | null, normal, iid) * p(null | normal, iid) / p(T | normal, iid).

3. Therefore, if we have an upper bound for x = p(null | normal, iid) / p(T | normal, iid) then we can get an upper bound for p(null | T, normal, iid). That is my main claim.

Which of the above statements do you object to?

>"You can use Student’s t-distribution to compute an upper bound for the probability of T under the assumptions"

I'm not sure what you are arguing anymore. I am saying you will never test a parameter value in isolation, it is always part of a model with other assumptions. There is simply no such thing as testing a parameter value alone. To define a likelihood you need more than simply a parameter...

You seemed to be disagreeing with that, but are now acknowledging the presence of the other assumptions.

“I'm not sure what you are arguing anymore.”

It’s the claim I make in 3, and then the secondary claim that making our upper bound on p(null | T, normal, iid) small for significant p-values (i.e. p(T | null, normal, id)) could be used as a criterion for whether our threshold for statistical significance is small enough.

“You seemed to be disagreeing with that”

I’m not sure what I said that gave that impression. I didn’t mention anything about the normal / iid assumptions initially not because I thought we weren’t making these assumptions but because I didn’t think these details were essential to my point.

"Let the statistic t_0 be the inf of all these t-statistics. Let T be the event that t_0 is at least as big as the observed value." Oops, I meant the inf of their absolute values, and T is the event that t_0 is at least as big in absolute value as the observed value.

Also, every probability mentioned should also include in the list of conditions that the number of samples observed matches our experiment.

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