1. Topics: Significance Testing of Correlation Coefficients Inference about a population correlation coefficient: –Testing H 0 :  xy = 0 or some specific value –Testing H 0 :  xy = 0 for two or more correlations based on the same sample Inference about a difference between population correlation coefficients –Testing H 0 :  xy1 -  xy2 = 0 (or  xy1 =  xy2 )

2. Inference about a Correlation Coefficient Purpose: to determine whether two variables (X and Y) are linearly related in the population. H 0 :  xy = 0 H 0 :  xy =some specified value

3. Test of Correlation Coefficient H 0 :  xy = some specified value H 1 :  xy not= some specified value ( than some specified value) Transform sample and population correlation coefficients to Z r and Z  Calculate t z observed :distance of transformed r xy from the transformed population  xy in standard error points Test against z critical (determined from table for chosen level of significance)

4. Sampling Distribution of r xy

5. Example Study of relationship between achievement motivation and performance in school (grade point average). Theory and prior research suggests that the correlation between these two variables is positive and moderately high (.50) The observed correlation in this study was.75 based on N=63

6. Test of Correlation Coefficient: One Sample H 0 :  xy =.50 H 1 :  xy not=.50 Level of Significance:.05 Verify Assumptions –Independence of score pairs –Bivariate Normality –n >= 30

7. Assumptions Independence: where the pair of scores for any particular student is independent of the pair of scores of every other student. Bivariate Normality: For each value of X, the values of Y are normally distributed; for each value of Y the values of X are normally distributed; each variable normally distributed Sample Size: n >= 30

8. Bivariate Normal For each value of X the Y scores are normally distributed For each value of Y the X scores are normally distributed

9. Example Con’td Find Fisher Z transformation for r xy and  xy (from a Table I) –r =.75 so Z r =.973 –  =.50 so Z  =.549 Set up Z robserved : Z r -Z  /s Z to get distance of Z r from Z  in standard error points Computation formula for Z robserved : – (Z r -Z  ) (sqrt n-3) = –(.973 -.549)/7.75 = –(.424)(7.75) = 3.29

10. Example Con’td Find z critical (from table or memory) = 1.96 Decision Rule: –Reject H 0 if absolute value of z r observed >= 1.96 (3.29 is greater than 1.96) –Do not reject H 0 if absolute value of z robserved < 1.96 Conclusion: the relationship between achievement motivation and school performance (grade point average) is greater than the specified value of.50

11. The Simple Approach When H 0 :  xy = 0 H 0 :  xy = 0; H 1 :  xy > 0 Sample size = 102 r =.24 Compare r observed with r critical (.05,df=100) =.1638 (from Table G) Since.24 >.1638 can reject the null hypothesis and conclude that there is a positive correlation in the population--our best estimate of that correlation is.24

12. Example: Testing Two or More Correlation Coefficients Working example: Suppose the following measures were collected on 82 subjects: GPA,. Self-concept, and locus of control (internal-external)

13. Testing H 0 :  xy = 0 for Two or More Correlations Based on Same Sample (N=82)

14. Testing H 0 :  xy = 0 for Two or More Correlations Based on Same Sample H 0 : H 1 (non-directional): Level of significance:  =.01 (level of significance) Assumptions: Number of Variables: Number of dfs: Critical Value (from Table H) Decisions and conclusions:

15. Inference about Difference between Population Correlation Coefficients To determine whether or not the observed difference between two correlation coefficients (r 1 -r 2 ) may be due to chance or represents a difference in population coefficients

16. Example: Testing Difference between Two Correlation Coefficients To determine whether or not the observed difference between two correlation coefficients (r 1 -r 2 ) may be due to chance or represents a difference in population coefficients In the Overachievement Study, the correlation between SAT scores and GPA was.0214 for the sample of 40 subjects. However the correlation between SAT and GPA for men was -.2369 and for women was +.3760.

17. Example: Differences (con’t) H 0 : H 1 (non directional): Significance Level:  =.05 Check assumptions Convert sample r’s to Zrs (Table I): male Z rmale = -.239; female Z rfemale =.394 Compute standard error of the difference between correlations: s rmale-rfemale :.34 (via formula) Calulate z rmale- rfemale(observed) = Z rmale- - Z rfemale / s rmale-rfemale : -.633/.34 = -1.86 Find z rmale-rfemale(critical) : +/-1.96 (.05, two-tailed) Decision and Conclusion: