Hi everyone!

Can anyone tell me how to calculate p-values from z-scores in R? Is this the correct way:

pvalue = pnorm(-abs(z))

Thanks!!!

Calculating P-Values From Z-Scores

13

Entering edit mode

10.7 years ago

Diana
▴
900

Hi everyone!

Can anyone tell me how to calculate p-values from z-scores in R? Is this the correct way:

pvalue = pnorm(-abs(z))

Thanks!!!

9

Entering edit mode

10.7 years ago

Manu Prestat
4.0k

your expression is good. Just don't forget, if relevant, to take into account the two-sided characteristic of the test, it would be then:

```
pvalue2sided=2*pnorm(-abs(z))
```

And, if you think you need to use the "apply" function, you might need to think about multitesting correction... but, I agree, this is another topic ;-)

1

Entering edit mode

10.7 years ago

Gjain
5.7k

Hi Diana,

I was explaining the same thing to my friend this morning.

Here is a very nice and useful link: Calculating p Values

- Calculating a Single p Value From a Normal Distribution
- Calculating a Single p Value From a t Distribution
- Calculating Many p Values From a t Distribution

I hope this helps.

1

Entering edit mode

7.5 years ago

Emre
▴
100

Based on the previous answers and comments, here is a function that considers both the one-sided case (two alternatives, observed scores are greater / z is positive: "+", observed scores are lower / z is negative: "-") and two sided case ("NULL").

```
convert.z.score<-function(z, one.sided=NULL) {
if(is.null(one.sided)) {
pval = pnorm(-abs(z));
pval = 2 * pval
} else if(one.sided=="-") {
pval = pnorm(z);
} else {
pval = pnorm(-z);
}
return(pval);
}
```

1

Entering edit mode

4.0 years ago

Kevin Blighe
83k

Yet another [late] answer, in the context of GWAS: SNP dataset and Z Score

Kevin

Similar Posts

Loading Similar Posts

Traffic: 1353 users visited in the last hour

Use of this site constitutes acceptance of our User Agreement and Privacy Policy.

Thanks! my test is one-sided as I am looking at enrichment only

In one-sided case, one is interested to find how extreme is the observation compared to random expectation. Therefore, it seems to be ill defined as it fails to take into account the sign of the z-scores: