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Z score to p values

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lou
lou on 7 Dec 2016
I have a large matrix 1xN containing z values. I would like to know how to turn these z scores to p values using normcdf function?
How to obtain p values both for one-tailed and two-tailed p values using normcdf?
Many thanks in advance!

Accepted Answer

Star Strider
Star Strider on 7 Dec 2016
If I remember correctly, the probability of a one-tailed test is twice the probability of a two-tailed test, so:
p_one = 2*normcdf(z_vector);
p_two = normcdf(z_vector);
  11 Comments
Star Strider
Star Strider on 30 Mar 2017
It depends upon the hypothesis you are testing.
Noah
Noah on 23 Jul 2021
Note that the accepted answer is backwards, unless you mean something strange by your hypothesis. The probability of one-tailed test is HALF the probability of a two-tailed test. The area under a bell curve on one side is half the area on both sides.
p_oneTailed = normcdf(z_vector);
p_twoTailed = 2*normcdf(z_vector);

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More Answers (1)

Ziwei Liu
Ziwei Liu on 18 Aug 2023
Two-tailed p value should actually be 2 * (1 - normcdf(z)).
normcdf(z) gives the area under curve on the left side of z. This is not p value. One-tailed p value should be the area on the right side, which is (1 - normcdf(z)).Two-tailed p value should be the double of that.
You can use the arrayfun function to compute p value for each entry in your z score matrix. i.e. p = arrayfun(@(x) 2*(1-normcdf(x)), ZScoreMatrix).
  1 Comment
Germano Gallicchio
Germano Gallicchio on 28 Feb 2024
For this to work with negative z scores, you also need to take the absolute value of z:
z = [-2.58 -1.96 -1.65 0 1.65 1.96 2.58]; % vector of z scores
p = 2 * (1 - normcdf(abs(z))); % vector of associated pvalues
disp([z' p'])
-2.5800 0.0099 -1.9600 0.0500 -1.6500 0.0989 0 1.0000 1.6500 0.0989 1.9600 0.0500 2.5800 0.0099

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