0
$\begingroup$

This may be a basic question, so please pardon me: given a vector of univariate data, and an individual observation from the said univariate data, I would like to derive some probability of observing this individual observation, or a more extreme observation, from the empirical distribution of the said univariate data (aka. the p-value).

May I ask how can this be done in R?

$\endgroup$
1
$\begingroup$

First, let's start with $p$-value, because it seems you have a wrong understanding of $p$-values. They are not the "probability of observing this individual observation, or a more extreme observation, from the distribution". Quoting Wikipedia

$p$-value is a function of the observed sample results (a statistic) that is used for testing a statistical hypothesis. Before performing the test a threshold value is chosen, called the significance level of the test, traditionally 5% or 1% and denoted as $\alpha$. If the $p$-value is equal to or smaller than the significance level ($\alpha$), it suggests that the observed data are inconsistent with the assumption that the null hypothesis is true, and thus that hypothesis must be rejected and the alternative hypothesis is accepted as true. When the $p$-value is calculated correctly, such a test is guaranteed to control the Type I error rate to be no greater than $\alpha$.

An equivalent interpretation is that $p$-value is the probability of obtaining the observed sample results, or "more extreme" results, when the null hypothesis is actually true (where "more extreme" is dependent on the way the hypothesis is tested).

for learning more on $p$-values check this thread.

As about checking what is the probability of observing a certain value $x$ in your empirical distribution (i.e. data) - for this you simply count how many times $x$ occurred in your data.

data <- c(1,2,3,4,2,1,4,3,1,2,6,7,8,8,1,2)
x <- 5
mean(data == x) # P(data = x)
mean(data >= x) # P(data >= x)

however, let me say it again, this is not a $p$-value.

$\endgroup$
  • $\begingroup$ Thanks for your answer Tim. I see now where my confusion with p-values lie now, thanks for your clarification, and the solution you provided to my problem! $\endgroup$ – tohweizhong May 15 '15 at 7:41

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.