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

The shape of the sampling distribution of rr 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 rr 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 rr, the more pronounced the skew.

Figure 1: Sampling distribution of rr for N=12 N 12 and r=0.60 r 0.60 .

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

Figure 2: The sampling distribution of rr for N=12 N 12 and r=0.90 r 0.90

Referring back to the SAT example, suppose you wanted to know the probability that in a sample of 19 students, the sample value of rr 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 rr. 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 rr to a variable that is normally distributed with a known standard error. The variable is called z′ z′ and the formula for the transformation is given below. z′=0.5ln1+r1−r z′ 0.5 1 r 1 r The details of the formula are not important here since normally you will use either a table or a computer program to do the transformation. What is important is that z′ z′ is normally distributed and has a standard error of 1N−3 1 N 3 where NN 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 z′ z′ s. From a table, the values are 0.6931 and 0.9730 respectively. The standard error of z′ z′ for N=12 N 12 is 0.333. Therefore the question is reduced to the following: given a normal distributing with a mean of 0.6931 and a standard deviation of 0.333, what is the probability of obtaining a value of 0.9730 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−ms=0.9730−0.69310.333=0.8405 Z X m s 0.9730 0.6931 0.333 0.8405 and use a table to find that the area above 0.8405 is 0.20.

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This work is licensed by David Lane under a Creative Commons Attribution License (CC-BY 1.0), and is an Open Educational Resource.

Last edited by Charlet Reedstrom on Jun 19, 2003 12:00 am GMT-5.