This page's content is no longer actively maintained, but the material has been kept on-line for historical purposes.
The page may contain broken links or outdated information, and parts may not function in current web browsers.

## Introduction to Statistics

### V - Significance of Correlation Coefficient

1. How to test to see if your result gives you a "high degree" of confidence that there is a relationship between x and y.

What if you had the following data:  Statistical Test for Significance of r:
"Fisher's Z-transformation"

1/2 ln((1+r)/(1-r)) ±1.96 * Square root of (1/(n-3)) = 1/2 ln ((1+ρ)/(1-ρ))

The purpose of this test is to see if r would still be different than 0 if you had infinite data. The test gives upper and lower bounds for the "real" r = ρ. If the 0 is included in the range between ρ low end and ρ high end, then the statistical test says, "You can't claim your result is Statistically significant at the 95% confidence level." If 0 is not in the range between ρ low end and ρ high end, then the test says "You do have a Statistically significant relationship between the two variables at the 95% confidence level."

2. Example using r = 0.6 and n = 10:

1/2 ln ((1+0.6)/(1-0.6)) ± 1.96 * Square root of (1/(n-3)) = 1/2 ln ((1+ρ)/(1-ρ))

.69315 ± .074081 = 1/2 ln ( (1 + ρ)/(1 - ρ) )

High End Case for ρ

(1.43396) = 1/2 ln ((1 + ρ)/(1 - ρ))

2.86792 = ln ((1 + ρ)/(1 - ρ))

e2.86792 = (1 + ρ)/(1 - ρ)

17.60037 = (1 + ρ/1 - ρ)

(1 - ρ)17.60037 = 1 + ρ

17.60037 - 17.60037 ρ = 1 + ρ

16.60037/18.60037 = ρ

ρ high end = .89248

Low End Case for ρ

-0.047766 = 1/2 ln ((1 + ρ)/(1 - ρ)) -0.09532 = ln ((1 + ρ)/(1 - ρ))

e-.009532 = ((1 + ρ)/(1 - ρ))

0.90908 - 0.90908 ρ = 1 + ρ

-0.90918/1.90908 = ρ

ρ low end = -0.04762

3. Example using r = 0.6 and n = 100:

You improve your chances of finding a statistically significant relationship if you gather more data. So let's suppose that we gather 90 more data points for the same case as Example 1 and that the "r" is still 0.6.

Is this proof of a statistically significant relationship?

Use the Fisher Test:

n = 100
r = 0.6

1/2 ln ((1+0.6)/(1-0.6) ) ± 1.96 * Square root of (1/(n-3))=1/2 ln ((1+ρ)/(1-ρ))

.69315 ± .19901 = 1/2 ln ((1 + ρ)/(1 - ρ))

High End Case for ρ

.89216 = 1/2 ln ((1 + ρ)/(1 - ρ))

1.78432 = ln ((1 + ρ)/(1 - ρ))

e1.78432 = (1 + ρ)/(1 - ρ)

5.95550 = (1 + ρ)/(1 - ρ)

(1 - ρ) 5.95550 = 1 + ρ

5.95550 - 5.95550 ρ = 1 + ρ

4.95550/6.95550 = ρ

ρ high end = .71246

Low End Case for ρ

Work it out for yourself and the result should be this:

ρ low end = +0.45750 Zero is outside the interval, so this is a statistically significant relationship!