Sampling Distribution of r

Site: Saylor Academy
Course: MA121: Introduction to Statistics
Book: Sampling Distribution of r
Printed by: Guest user
Date: Tuesday, April 16, 2024, 5:00 PM

Description

Now, we'll talk about how the shape of the sampling distribution of Pearson correlation deviates from normality and then discusses how to transform r to a normally distributed quantity. Then, we will discuss how to calculate the probability of obtaining an r above a specified value.

Sampling Distribution of Pearson's 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
Public Domain Mark This work is in the Public Domain.

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Questions

Question 1 out of 4.
What is the shape of the sampling distribution of r?

Normal

Bimodal

Skewed


Question 2 out of 4.
What is the corresponding z' for r = -.65?



Question 3 out of 4.
Which of these r values is the most different from its corresponding z'?

r = 0

r = .6

r = .2

r = -.8


Question 4 out of 4.
The population has a correlation of .60. What is the probability that you will get a sample correlation of at least r = .50 if you sample 19 pairs?

Answers

  1. Skewed- Unless r = 0, the sampling distribution is skewed. The reason for the skew is that r cannot take on values greater than 1.0 or less than -1.0, and therefore the distribution cannot extend as far in one direction as it can in the other.

  2. -.775 Use a calculator or table to transform r to z'. You get -.775.

  3. r = -.8 Although you could plug all of these values into the r to z' calculator, you don't need to do that. You know the r with the biggest absolute value has the most skewed sampling distribution, so it is the most different from its corresponding z'.

  4. .718 Population correlation = .6 to z' = .693, SE = 1/sqrt(N-3) = 1/sqrt(19-3) = .25, r = .50 to z' = .549, Use the "Calculate area for a given X" applet. Plug in .693 for the mean and .25 for the SD, and then calculate area above .549. You get .718.