14
votes

I have data that contain 54 samples for each condition (x and y). I have computed the correlation the following way:

> dat <- read.table("http://dpaste.com/1064360/plain/",header=TRUE)
> cor(dat$x,dat$y)
[1] 0.2870823

Is there a native way to produce SE of correlation in R's cor() functions above and p-value from T-test?

As explained in this web (page 14.6)

2
Perhaps you're looking for ?cor.test instead.A5C1D2H2I1M1N2O1R2T1

2 Answers

22
votes

I think that what you're looking for is simply the cor.test() function, which will return everything you're looking for except for the standard error of correlation. However, as you can see, the formula for that is very straightforward, and if you use cor.test, you have all the inputs required to calculate it.

Using the data from the example (so you can compare it yourself with the results on page 14.6):

> cor.test(mydf$X, mydf$Y)

    Pearson's product-moment correlation

data:  mydf$X and mydf$Y
t = -5.0867, df = 10, p-value = 0.0004731
alternative hypothesis: true correlation is not equal to 0
95 percent confidence interval:
 -0.9568189 -0.5371871
sample estimates:
       cor 
-0.8492663 

If you wanted to, you could also create a function like the following to include the standard error of the correlation coefficient.

For convenience, here's the equation:

enter image description here

r = the correlation estimate and n - 2 = degrees of freedom, both of which are readily available in the output above. Thus, a simple function could be:

cor.test.plus <- function(x) {
  list(x, 
       Standard.Error = unname(sqrt((1 - x$estimate^2)/x$parameter)))
}

And use it as follows:

cor.test.plus(cor.test(mydf$X, mydf$Y))

Here, "mydf" is defined as:

mydf <- structure(list(Neighborhood = c("Fair Oaks", "Strandwood", "Walnut Acres",
  "Discov. Bay", "Belshaw", "Kennedy", "Cassell", "Miner", "Sedgewick", 
  "Sakamoto", "Toyon", "Lietz"), X = c(50L, 11L, 2L, 19L, 26L, 
  73L, 81L, 51L, 11L, 2L, 19L, 25L), Y = c(22.1, 35.9, 57.9, 22.2, 
  42.4, 5.8, 3.6, 21.4, 55.2, 33.3, 32.4, 38.4)), .Names = c("Neighborhood", 
  "X", "Y"), class = "data.frame", row.names = c(NA, -12L))
3
votes

Can't you simply take the test statistic from the return value? Of course the test statistic is the estimate/se so you can calc se from just dividing the estimate by the tstat:

Using mydf in the answer above:

r = cor.test(mydf$X, mydf$Y)
tstat = r$statistic
estimate = r$estimate
estimate; tstat

       cor 
-0.8492663 
        t 
-5.086732