Statistics for the Social Sciences Psychology 340 Fall 2006 Relationships between variables.

Statistics for the Social Sciences

Psychology 340Fall 2006

Relationships between variables


Correlation

• Write down what (you think) a correlation is.

• Write down an example of a correlation

• Association between scores on two variables– Age and coordination skills in children, as kids get older their motor coordination tends to improve

– Price and quality, generally the more expensive something is the higher in quality it is


Correlation and Causality

• Correlational research design– Correlation as a kind of research design (observational designs)

– Correlation as a statistical procedure


Another thing to consider about correlation

• Correlations describe relationships between two variables, but DO NOT explain why the variables are related

Suppose that Dr. Steward finds that rates of spilled coffee and severity of plane turbulents are strongly positively correlated.

One might argue that turbulents cause coffee spills

One might argue that spilling coffee causes turbulents




Suppose that Dr. Cranium finds a positive correlation between head size and digit span (roughly the number of digits you can remember).

One might argue that bigger your head, the larger your digit span

1

2124

1537

One might argue that head size and digit span both increase with age (but head size and digit span aren’t directly related)




For many years instructors have noted that the reported fatality rate of

grandparents increases during midterm and final exam periods. One might argue that college exams cause grandparent death


Relationships between variables

• Properties of a correlation– Form (linear or non-linear)– Direction (positive or negative)– Strength (none, weak, strong, perfect)

• To examine this relationship you should:– Make a scatterplot - a picture of the relationship

– Compute the Correlation Coefficient - a numerical description of the relationship


Graphing Correlations

• Steps for making a scatterplot (scatter diagram)1. Draw axes and assign variables to them2. Determine range of values for each

variable and mark on axes3. Mark a dot for each person’s pair of

scores


Scatterplot

Y

X1

2

34

5

6

1 2 3 4 5 6

• Plots one variable against the other• Each point corresponds to a different individual

A 6 6

X Y


Scatterplot

Y

X1

2

34

5

6

1 2 3 4 5 6


A 6 6B 1 2

X Y


Scatterplot

Y

X1

2

34

5

6

1 2 3 4 5 6


A 6 6B 1 2C 5 6

X Y


Scatterplot

Y

X1

2

34

5

6

1 2 3 4 5 6


A 6 6B 1 2C 5 6

D 3 4

X Y


Scatterplot

Y

X1

2

34

5

6

1 2 3 4 5 6


A 6 6B 1 2C 5 6

D 3 4

E 3 2

X Y


Scatterplot

Y

X1

2

34

5

6

1 2 3 4 5 6

• Imagine a line through the data points


A 6 6B 1 2C 5 6

D 3 4

E 3 2

X Y

• Useful for “seeing” the relationship– Form, Direction, and Strength


Form

Non-linearLinear


NegativePositive

Direction

• X & Y vary in the same direction

• As X goes up, Y goes up

• Positive Pearson’s r

• X & Y vary in opposite directions

• As X goes up, Y goes down

• Negative Pearson’s r

Y

X

Y

X


Strength

• The strength of the relationship– Spread around the line (note the axis scales)

– Correlation coefficient will range from -1 to +1• Zero means “no relationship”• The farther the r is from zero, the stronger the relationship


Strength

r = 1.0“perfect positive corr.”r2 = 100%

r = -1.0“perfect negative corr.”r2 = 100%

r = 0.0“no relationship”r2 = 0.0

-1.0 0.0 +1.0

The farther from zero, the stronger the relationship


The Correlation Coefficient

• Formulas for the correlation coefficient:

r = XZ YZ∑N

€

r =SP

SSX SSY

€

SP = X − X ( ) Y −Y ( )∑

Used this one in PSY138 Common alternative


Computing Pearson’s r (using SP)

• Step 1: SP (Sum of the Products)

€

SP = X − X ( ) Y −Y ( )∑

mean3.64.0

6 61 25 6

3 4

3 2

X Y

€

X − X ( )

€

Y −Y ( )

€

X − X ( ) Y −Y ( )




€

SP = X − X ( ) Y −Y ( )∑

mean3.64.0

2.4

0.0

6 61 25 6

3 4

3 2

X Y

€

X − X ( )

€

Y −Y ( )

€

X − X ( ) Y −Y ( )= 6 - 3.6

-2.6= 1 - 3.6

1.4= 5 - 3.6

-0.6= 3 - 3.6

-0.6= 3 - 3.6Quick check




€

SP = X − X ( ) Y −Y ( )∑

mean3.64.0

2.4-2.6

1.4

-0.6

-0.6

0.0 0.0

6 61 25 6

3 4

3 2

X Y

€

X − X ( )

€

Y −Y ( )

€

X − X ( ) Y −Y ( )2.0= 6 - 4.0-2.0= 2 - 4.0

2.0= 6 - 4.0

0.0= 4 - 4.0

-2.0= 2 - 4.0Quick check




€

SP = X − X ( ) Y −Y ( )∑

mean3.64.0

2.4-2.6

1.4

-0.6

-0.6

0.0

2.0-2.0

2.0

0.0

-2.0

0.0 14.0 SP

6 61 25 6

3 4

3 2

X Y

€

X − X ( )

€

Y −Y ( )

€

X − X ( ) Y −Y ( )4.8* =

5.2* =

2.8* =

0.0* =

1.2* =



• Step 2: SSX & SSY




mean3.64.0

2.4-2.6

1.4

-0.6

-0.6

0.0

2.0-2.0

2.0

0.0

-2.0

0.0 14.0

6 61 25 6

3 4

3 2

X Y

€

X − X ( )

€

Y −Y ( )

€

X − X ( ) Y −Y ( )4.85.2

2.8

0.0

1.2

€

X − X ( )2

5.76

15.20

SSX

2 =6.762 =

1.962 =

0.362 =

0.362 =




mean3.64.0

2.4-2.6

1.4

-0.6

-0.6

0.0

2.0-2.0

2.0

0.0

-2.0

0.0 14.0

6 61 25 6

3 4

3 2

X Y

€

X − X ( )

€

Y −Y ( )

€

X − X ( ) Y −Y ( )4.85.2

2.8

0.0

1.2

€

X − X ( )2

5.766.76

1.96

0.36

0.36

15.20

€

Y −Y ( )2

2 =4.02 =4.02 =4.02 =0.02 =4.0

16.0

SSY



• Step 3: compute r

€

r =SP

SSX SSY




mean3.64.0

2.4-2.6

1.4

-0.6

-0.6

0.0

2.0-2.0

2.0

0.0

-2.0

0.0 14.0

6 61 25 6

3 4

3 2

X Y

€

X − X ( )

€

Y −Y ( )

€

X − X ( ) Y −Y ( )4.85.2

2.8

0.0

1.2

€

X − X ( )2

5.766.76

1.96

0.36

0.36

15.20

€

Y −Y ( )2

4.04.0

4.0

0.0

4.0

16.0

SSYSSX

SP

€

r =SP

SSX SSY


Computing Pearson’s r


14.015.20 16.0

SSYSSX

SP

€

r =SP

SSX SSY




15.20 16.0

SSYSSX

r =14

SSXSSY




15.20

SSX

r =14

SSX * 16




€

r =14

15.2 *16



• Step 3: compute rr =

1415.2 * 16

=0.89

Y

X1

2

34

5

6

1 2 3 4 5 6

• Appears linear• Positive relationship• Fairly strong relationship• .89 is far from 0, near +1


The Correlation Coefficient

• Formulas for the correlation coefficient:

r = XZ YZ∑N

r =SPSSXSSY

SP = X−X( ) Y −Y( )∑

Used this one in PSY138 Common alternative



(using z-scores)

• Step 1: compute standard deviation for X and Y (note: keep track of sample or population)

6 61 25 6

3 4

3 2

X Y

• For this example we will assume the data is from a population



(using z-scores)


Mean 3.6

2.4-2.6

1.4

-0.6

-0.6

0.0

6 61 25 6

3 4

3 2

X Y

€

X − X ( )

€

X − X ( )2

5.766.76

1.96

0.36

0.36

15.20

SSXStd dev1.74

σ =SSX

N=

15.2

5= 1.74




(using z-scores)


Mean 3.6 4.0

2.4-2.6

1.4

-0.6

-0.6

2.0-2.0

2.0

0.0

-2.0

0.0

6 61 25 6

3 4

3 2

X YX −X( )

€

Y −Y ( )X −X( )2

5.766.76

1.96

0.36

0.36

15.20

€

Y −Y ( )2

4.04.0

4.0

0.0

4.0

16.0

SSYStd dev1.741.79


σ =SSY

N

=16.0

5= 1.79



(using z-scores)

• Step 2: compute z-scores

Mean 3.6 4.0

2.4-2.6

1.4

-0.6

-0.6

2.0-2.0

2.0

0.0

-2.0

6 61 25 6

3 4

3 2

X Y

€

X − X ( ) Y −Y( )X −X( )2

5.766.76

1.96

0.36

0.36

15.20

Y −Y( )2

4.04.0

4.0

0.0

4.0

16.0Std dev

ZX

1.741.79

1.38=2.4

1.74

X −X( )sX



(using z-scores)


Mean 3.6 4.0

2.4-2.6

1.4

-0.6

-0.6

2.0-2.0

2.0

0.0

-2.0

6 61 25 6

3 4

3 2

X Y

€

X − X ( ) Y −Y( )X −X( )2

5.766.76

1.96

0.36

0.36

15.20

Y −Y( )2

4.04.0

4.0

0.0

4.0

16.0Std dev

ZX

X −X( )sX

1.741.79

1.38-1.49

0.8

- 0.34

- 0.34

0.0 Quick check



(using z-scores)


Mean 3.6 4.0

2.4-2.6

1.4

-0.6

-0.6

2.0-2.0

2.0

0.0

-2.0

6 61 25 6

3 4

3 2

X YX −X( )

€

Y −Y ( )X −X( )2

5.766.76

1.96

0.36

0.36

15.20

€

Y −Y ( )2

4.04.0

4.0

0.0

4.0

16.0Std dev

ZX ZY

1.741.79

1.1

Y −Y( )sY

=2.0

1.791.38-1.49

0.8

- 0.34

- 0.34



(using z-scores)


Mean 3.6 4.0

2.4-2.6

1.4

-0.6

-0.6

2.0-2.0

2.0

0.0

-2.0

6 61 25 6

3 4

3 2

X YX −X( )

€

Y −Y ( )X −X( )2

5.766.76

1.96

0.36

0.36

15.20

€

Y −Y ( )2

4.04.0

4.0

0.0

4.0

16.0Std dev

ZX ZY

Y −Y( )sY

1.741.79

1.1-1.1

0.0

-1.1

1.1

0.0

1.38-1.49

0.8

- 0.34

- 0.34

Quick check



(using z-scores)


Mean 3.6 4.0

2.4-2.6

1.4

-0.6

-0.6

0.0

2.0-2.0

2.0

0.0

-2.0

0.0

6 61 25 6

3 4

3 2

X Y ZX ZY

5.766.76

1.96

0.36

0.36

15.20

€

Y −Y ( )2

4.04.0

4.0

0.0

4.0

16.0Std dev

ZX ZY

1.741.790.0

1.1-1.1

0.0

-1.1

1.1

0.0

1.52

X −X( ) X −X( )2

r =ZXZY∑N

Y −Y( )

1.38-1.49

0.8

- 0.34

- 0.34

* =



(using z-scores)


Mean 3.6 4.0

2.4-2.6

1.4

-0.6

-0.6

0.0

2.0-2.0

2.0

0.0

-2.0

0.0

6 61 25 6

3 4

3 2

X Y ZX ZY

5.766.76

1.96

0.36

0.36

15.20

€

Y −Y ( )2

4.04.0

4.0

0.0

4.0

16.0Std dev

ZX ZY

1.741.790.0

1.1-1.1

0.0

-1.1

1.1

0.0

1.521.64

0.88

0.0

0.37

X −X( ) X −X( )2

r =ZXZY∑N

=4.41

5

Y −Y( )

1.38-1.49

0.8

- 0.34

- 0.34

=0.89

4.41



(using z-scores)


Y

X1

2

34

5

6

1 2 3 4 5 6

• Appears linear• Positive relationship• Fairly strong relationship• .89 is far from 0, near +1

r =ZXZY∑N

=0.89


A few more things to consider about correlation

• Correlations are greatly affected by the range of scores in the data– Consider height and age relationship

• Extreme scores can have dramatic effects on correlations – A single extreme score can radically change r

• When considering "how good" a relationship is, we really should consider r2 (coefficient of determination), not just r.


Correlation in Research Articles

• Correlation matrix– A display of the correlations between more than two variables

Acculturation

• Why have a “-”?

• Why only half the table filled with numbers?


Next time

• Predicting a variable based on other variables

Statistics for the Social Sciences Psychology 340 Fall 2006 Relationships between variables.

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