Null Hypothesis Testing with Pearson's r
When to Use
This is straightforward. Do this when you have a sample value of Pearson’s r and you want to test the null hypothesis that the value of r in the population is zero. For example, imagine that each of 50 subjects judges the intelligence level of a stimulus person on a 1 to 7 scale and also rates their mood on a 0 to 10 scale. Imagine also that for this sample of 50 subjects, Pearson’s r is +.23. We want to test the null hypothesis that Pearson’s r in the population is actually 0 and that this sample correlation of .23 represents nothing more than sampling error.
The Test Statistic
The test statistic in this case is t. However, this statistic is not usually computed or reported. Instead, we can proceed straight to a table (see below).
The p Value
p is the probability of a value of Pearson’s r as extreme as the one you got if the null hypothesis were true. Again, you do not have to figure out the exact p value. Instead, you can just figure out whether p is lower than your α level (.05). The table at right lists these critical r values for α levels of .10, .05, .02, and .01. Note that it also lists different critical values depending on the degrees of freedom, which is the number of cases minus two (df = N – 2). Again, notice from the table that as the sample size gets smaller, the r value needed to reject the null hypothesis gets larger.
You decide to reject the null hypothesis if your sample r value is more extreme than the critical r value. This means that the probability of getting an r value at least as extreme as yours, if the null hypothesis were true, is less than 5% (or whatever α is). So you decide that the null hypothesis is not true. You fail to reject the null hypothesis if your sample r value is less extreme than the critical t value. This means that the probability of getting an r value at least as extreme as yours, if the null hypothesis were true, is greater than 5%. So you decide that the null hypothesis could be true.
A group of 50 subjects rates the intelligence of a stimulus person and rates their moods. Pearson’s r for the sample is +.23. I can look in the table for 48 degrees of freedom. It is not there, but 50 is pretty close. For an α level of .05, the critical value is .27. So I fail to reject the null hypothesis; the correlation is not statistically significant. In other words, I cannot conclude that my sample correlation reflects anything other than sampling error.
Expressing the Result in Writing
Here is how such a result might be expressed in APA style: “The correlation between intelligence ratings and mood was not statistically significant, r(48) = .23, p > .05.
TABLE: CRITICAL VALUES OF r
Click the link below for a complete table of critical values of r: http://www.gifted.uconn.edu/siegle/research/Correlation/corrchrt.htm