1
$\begingroup$

I have a question about finding a pvalue on bootstrapped data, which is very similar to here.

I have a numeric vector S of bootstrapped simulated data, for instance N values between 0 and 1, and one observed numeric data X, and I want to compute a pvalue which indicates if the observed value is more likely in the upper or lower tail (so, if I understand well, if the value X is significantly lower or higher than the null distribution).

I found the simple formula for the pvalue which computes the chances for x to be in the upper tail :

p = sum(S > X)/N and it's corrected one : p = (sum(S >X)+1)/(N+1). So if p > 0.05, X is most likely not in the upper tail. According to an article I found (here, on page 2), this pvalue can be seen as the proportion of boostrapped values which are more extreme than the observed value.

My questions (probably a bit silly) are :

  • is the equivalent formula for the lower tail p = sum(S<X)/N ? (edit : according to the article I'm reading, yes)
  • is there a formula for a two-sided pvalue ? because if I add the result of the two previous formula, I'll get 1 of course.
  • Like mentionned in the question linked above, is it correct to use the Z-table (assuming my boostrapped data are normally distributed) ?

Thanks in advance.

$\endgroup$
1
$\begingroup$

General: please read the answer of @whuber in the question that you are linking.

Ad 1: yes

Ad 2: I think it is more complicated.

If S follows a distribution which is symmetric around 0, then you can simply use (abs(S) > abs(X))/N

However, if I did not know anything about the distribution of S, then I would take the pairs quantiles of the experimental S distribution and ask whether X is within them. For example, if

X < quantile(S, 0.005) | X > quantile(S, 0.995)

then I would say that p < 0.01 in a two-tailed comparison.

Ad 3. I think the answer is no, since you are calculating your z scores based on mean which is calculated from the samples which undergo the test.

$\endgroup$
  • $\begingroup$ Thanks ! I had trouble to understand @whuber's answer. Now it's clear. $\endgroup$ – Micawber May 31 '18 at 7:52

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.