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Chapter 10

Correlation

Positive and Negative Correlation

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Strength of Correlation

Correlations actually vary with respect to their strength. Scatter plot

= scores on any two variables, X and Y

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Curvilinear Relationships

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The Correlation Coefficient

Correlation coefficients range between -1.00 and +1.00 as follows: -1.00 perfect negative correlation -.60 strong negative correlation -.30 moderate negative correlation -.10 weak negative correlation .00 no correlation +.10 weak positive correlation +.30 moderate positive correlation +.60 strong positive correlation +1.00 perfect positive correlation

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Pearson’s Correlation Coefficient

o For example, we might be interested in examining the relationship between one’s attitude towards legalization of prostitution (X) and their attitudes towards legalization of marijuana (Y)

Prostitution(x)

Marijuana(Y)

A 1 2

B 6 5

C 4 3

D 3 3

E 2 1

F 7 4

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Pearson’s Correlation CoefficientSummary Table

Child X Y X2 Y2 XY

A 1 2

B 6 5

C 4 3

D 3 3

E 2 1

F 7 4

Total

Calculating the Correlation Coefficient

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Using the results from the summary table, calculate the correlation coefficient.

ΣX = 23 ΣY = 23 ΣX2 = 115 ΣY2 = 64 ΣXY = 83

 

 

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Testing the Significance of Pearson’s r Pearson's r gives us a precise measure of

the strength and direction of the correlation in the sample being studied.

If we have taken a random sample from a specified population, we may still seek to determine whether the obtained association between X and Y exists in the population and is not due merely to sampling error.

To test the significance of a measure of correlation, we usually set up the null hypothesis that no correlation exists in the population.

Can use either a t test or a simplified method using r to assess significance

Correlation Steps

Step 1: Create a summary table Step 2: Find the values of ΣX, ΣY,

ΣX2, ΣY2, ΣXY, and the mean of X and Y.

Step 3: Insert values from step 2 into the correlation formula.

Step 4: Find the degrees of freedom, alpha, and critical r

Step 5: Compare computed r with critical value of r using Table F 13

Importance of Graphing

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Review

Correlation Strength Direction

Test of significance Curvilinear correlation Importance of graphing

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Partial Correlation

Usually, researchers examine more than two variables at a time.

Must consider if a correlation between two measures holds up when controlling for a third variable.

Requires a correlation matrix Useful statistic for finding spurious

variables

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Correlation Matrix

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Attitude toward

School (X)

Grades (Y)

Employment (Z)

Attitude toward School (X)

1.00 ------- -------

Grades (Y) .89 1.00 -------

Employment (Z)

-.59 -.41 1.00

How Years on Force (Z) affects correlation

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Physical Fitness (X)

Salary (Y) Years on Force (Z)

Physical Fitness (X)

1.00 --- ---

Salary (Y) -.44 1.00 ---

Years on Force (Z)

-.68 .82 1.00

Rxy = -.44 Rxz = -.68Ryz = .82

 

Formula

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Correlations:Rxy = -.44 Rxz = -.68Ryz = .82

Rxy.z = -.44 – (-.68)(.82)√1-(-.68)2 *√1-(.82)2

Rxy.z = +.28

The partial correlation of physical fitness score (X) and salary (Y) while holding constant years on the force (Z) is calculated as follows:

 

Testing for significance

When testing for significance, we use t scores for partial correlations and not Table F.

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Chi Square & Strength of Association

Knowing that the result is significant is not enough

Only use this when examining two variables and the correlation is significant!!

Need to know how strong the association between the two is

Phi coefficient Cramer’s V correlation coefficient

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Phi Coefficient

A researcher is examining participation in a GED program and whether or not the individual once released from prison was arrested within a 2 year time frame. The researcher found there was a statistically significant difference and found the following results:

x2 = 7.44 N = 100

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Cramer’s V

A researcher is examining those who participate in a GED program, work skills program, and those who do not and whether or not the individual, once released from prison, was arrested within a 2 year time frame. The researcher found there was a statistically significant difference and found the following results:

x2 = 8.42 N = 120

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