Sampling Distribution of r

Learning Objectives

1. State how the shape of the sampling distribution of $r$ deviates from normality
2. Transform $\mathrm{r}$ to $\mathrm{z}$'
3. Compute the standard error of $z$'
4. Calculate the probability of obtaining an $\mathrm{r}$ above a specified value

Assume that the correlation between quantitative and verbal SAT scores in a given population is $0.60$. In other words, $\rho=0.60$. If $12$ students were sampled randomly, the sample correlation, $r$, would not be exactly equal to $0.60$. Naturally different samples of $12$ students would yield different values of $\mathrm{r}$. The distribution of values of $r$ after repeated samples of $12$ students is the sampling distribution of $r$.

The shape of the sampling distribution of $r$ for the above example is shown in Figure 1. You can see that the sampling distribution is not symmetric: it is negatively skewed. The reason for the skew is that $r$ cannot take on values greater than $1.0$ and therefore the distribution cannot extend as far in the positive direction as it can in the negative direction. The greater the value of $\rho$, the more pronounced the skew.

Figure 1. The sampling distribution of $r$ for $N=12$ and $\rho=0.60$.

Figure 2 shows the sampling distribution for $\rho=0.90$. This distribution has a very short positive tail and a long negative tail.

Figure 2. The sampling distribution of $r$ for $N=12$ and $\rho=0.90$.

Referring back to the SAT example, suppose you wanted to know the probability that in a sample of $12$ students, the sample value of $r$ would be $0.75$ or higher. You might think that all you would need to know to compute this probability is the mean and standard error of the sampling distribution of $r$. However, since the sampling distribution is not normal, you would still not be able to solve the problem. Fortunately, the statistician Fisher developed a way to transform $\mathrm{r}$ to a variable that is normally distributed with a known standard error. The variable is called $z$' and the formula for the transformation is given below.

$z^{\prime}=0.5 \ln [(1+r) /(1-r)]$

The details of the formula are not important here since normally you will use either a table or calculator to do the transformation. What is important is that $z$' is normally distributed and has a standard error of

$\frac{1}{\sqrt{N-3}}$

where $N$ is the number of pairs of scores.

Let's return to the question of determining the probability of getting a sample correlation of $0.75$ or above in a sample of $12$ from a population with a correlation of $0.60$. The first step is to convert both $0.60$ and $0.75$ to their $z$' values, which are $0.693$ and $0.973$, respectively. The standard error of $z$' for $N=12$ is $0.333$ Therefore the question is reduced to the following: given a normal distribution with a mean of $0.693$ and a standard deviation of $0.333$, what is the probability of obtaining a value of $0.973$ or higher? The answer can be found directly from the applet "Calculate Area for a given $X$" to be $0.20$. Alternatively, you could use the formula:

$z=(x-\mu) / \sigma=(0.973-0.693) / 0.333=0.841$

and use a table to find that the area above $0.841$ is $0.20$.

Source: David M. Lane, https://onlinestatbook.com/2/sampling_distributions/samp_dist_r.html
This work is in the Public Domain.