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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:

    Time Series Plot

    Scatter Plot

    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)) rho

    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!

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