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Distribution of p-value

  1. Sep 13, 2010 #1
    I read the following statement from wiki,but I don't know how to get this.

    "when a p-value is used as a test statistic for a simple null hypothesis, and the distribution of the test statistic is continuous, then the test statistic (p-value) is uniformly distributed between 0 and 1 if the null hypothesis is true."

    anyone can explain it more?
  2. jcsd
  3. Sep 14, 2010 #2


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    Hi chowpy, welcome to PF!

    Imagine that you have a data set A of one or more experimental observations. You also have a null hypothesis in mind (a possible distribution of results that data set A may or may not have come from). Say you're comparing the means of these two distributions (but it could be any parameter that you're comparing).

    The p-value is always defined as the expected frequency of obtaining your actual data set A from the null hypothesis. (If the p-value is incredibly low, we might decide that A came from another distribution, and therefore reject the null hypothesis; that's what hypothesis testing is all about.)

    If the null hypothesis is actually true, then we'd expect to get a p-value anywhere from 0% to 100%, distributed evenly. In other words, if the data set A (or a more extreme* data set) would only arise 20% of the time, then we'd expect a p-value of 0.20. *By more extreme I mean a data set with a mean farther away from the mean of the null hypothesis, in the example we're using.

    Does this answer your question?
  4. Sep 14, 2010 #3


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    Remember what it means for a random variable X to be uniformly distributed on (0,1)

    P(X <=a) = a for any a in (0,1)

    Let P denote the p-value as a random variable

    T stand for a generic Test statistic that has a continuous distribution.

    Pick an a in (0,1). Since T has a continuous distribution, there is a number ta that satisfies

    \Pr(T \le ta) = a

    Now, the events [tex] P \le a [/tex] and [tex] T \le ta [/tex] are equivalent, so that

    \Pr(P \le a) = \Pr(T \le ta) = a

    comparing this to the meaning of "uniformly distributed on (0,1) shows the result.
  5. Sep 16, 2010 #4
    Thanks Mapes and statdad~
    I understand it now~
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