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Suppose I want to turn statistical significance of 0.05 into a z-score of ~1.64

Can I use a normal distribution to convert these?

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This sounds like an odd thing to do... can you give more details about the circumstances that lead you to try to do this? –  Glen_b Mar 6 '13 at 8:23
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1 Answer

up vote 3 down vote accepted

Try the quantile function of the Normal distribution:

R> qnorm(c(0.05, 0.95))
[1] -1.64485  1.64485

And if you're wondering, the probability distribution function does the inverse:

R> pnorm( qnorm(c(0.05, 0.95)) )
[1] 0.05 0.95

Any decent numerical / statistical library or language will have these; I use R for convenience. Eg for C++ you have Boost Math.

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Thanks Dirik, can you explain how the distribution is a convertor for significance into a z-score. –  Matt Alcock Mar 6 '13 at 3:01
It's a bit of a tautology: the z-score $Z$ is the value such that the quantile is $\alpha$, here 0.05 (or $\alpha/2$ if you consider both tails). In that sense 1.64 corresponds to a 10% critical value ("z-score") with 5% in each tail. For 5% in both you need 1.96 corresponding to the two-sided 2.5% tails. –  Dirk Eddelbuettel Mar 6 '13 at 3:11
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