What does Pearson's Correlation Coefficient Tell Me?

Question

I run a cor.test(data) in R and get the following output:

t = 1.3596, df = 315, p-value = 0.1749
alternative hypothesis: true correlation is not equal to 0
95 percent confidence interval:
 -0.0340665  0.1849804
sample estimates:
       cor 
0.07637844 

Now, while I see that cor > 0, there is some correlation, how exactly do I determine if there is some significant correlation? How am I supposed to interpret the p-value, t-value, cor-value, etc. ?

Answer 1

Your correlation is not statistically different from $0$.

To go further, what follows is What Pearson's Correlation Coefficient implemented in cor.test(data) tells you.

Your $2$-dimension dataset (where $2$d stands for the two variables between which the correlation coefficient is computable) is built by sampling from the population data. This sampling process, in other situations, could have led to other correlation coefficients. From all the possible samples, derives all the possible correlation coefficients, $\rho$, and those are assumed to follow a normal distribution under some conditions, $\mathcal{N}(\bar{\rho},\sigma)$, identified with an expected value, $\bar{\rho}$, and a standard deviation, $\sigma$.

Among all the possible $\rho$s, you have "found" one of them, let call it $\hat{\rho}$. Note that as $n\rightarrow\infty$, $\hat{\sigma}\rightarrow 0$ and $\hat{\rho}\rightarrow\bar{\rho}$. In other words, as $n\rightarrow\infty$ the number of possible samples becomes $1$, i.e. the population data.

Briefly, Theory "says" that $(\rho-\bar{\rho})/\sigma$ follows a student distribution (incidentally itself reasonably well approximated by a normal distribution of mean $0$ and a standard deviation of $1$, $\mathcal{N}(0,1)$, as the sample increases in size).

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In your case

1. $\hat{\rho}=0.07637844$
2. $\hat{\sigma}=0.05617714$ (deduced)
3. The null hypothesis is $\bar{\rho}=0$
4. $t=(\hat{\rho}-0)/\hat{\sigma}=1.3596$

Then, the student distribution parameterized with $315$ degrees of freedom tells that

• The probability of having $t<1.3596$ is $0.1749$ or $17.49\%$

Which means that at a $5\%$ significance level (traditionally denoted by $\alpha$) one fails to reject the null under which the true correlation coefficient is $0$. This also means that up to a significance level of $17.49\%$ one fails to reject the null. Put differently, as already said above, your correlation is not statistically different from $0$.

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Then at a confidence level of $1-\alpha=0.95=95\%$, the student distribution with $315$ degrees of freedom tells us that

• with a probability of $5\%$ one has $t<-1.96752...$ or $t>1.96752...$. Equivalently that with a probability of $95\%$, one has $-1.96752<t<1.96752$

And this is used to obtain the confidence interval of your correlation coefficient. One has $95\%$ chance of having an occurrence of the random variable $\rho$ between $-0.0340665$ and $0.1849804$. Note that

$-0.0340665 = 0.07637844 - 1.96752 \times 0.05617 $

and

$\ \ \ 0.1849804 = 0.07637844 + 1.96752 \times 0.05617$

Logically (i.e. we have concluded that your correlation is not statistically different from $0$) $0$ belongs to $[-0.0340665, 0.1849804]$.