## Introduction to Statistics

### V - Significance of Correlation Coefficient

- 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."

**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))

**e**^{2.86792}= (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

**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))

**e**^{1.78432}= (1 + ρ)/(1 - ρ)5.95550 = (1 + ρ)/(1 - ρ)

(1 - ρ) 5.95550 = 1 + ρ5.95550 - 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!