Topic 17: Interaction Models

Interaction Models

• With several explanatory variables, we need to consider the possibility that the effect of one variable depends on the value of another variable

• Special cases

–One binary variable (Y/N) and one continuous variable

–Two continuous variables

One binary variable and one continuous variable

• X1 takes values 0 and 1 corresponding to two different groups

• X2 is a continuous variable• Model: Y = β0 + β1X1 + β2X2 + β3X1X2 + e

• When X1 = 0 : Y = β0 + β2X2 + e• When X1 = 1 : Y = (β0 + β1)+ (β2 + β3) X2 + e

One binary and one continuous

• β0 is the intercept for Group 1

• β0+ β1 is the intercept for Group 2

• Similar relationship for slopes (β2 and β3)

• H0: β1 = β3 = 0 tests the hypothesis that the

regression lines are the same

• H0: β1 = 0 tests equal intercepts

• H0: β3 = 0 tests equal slopes

KNNL Example p316

• Y is number of months for an insurance company to adopt an innovation

• X1 is the size of the firm (a continuous variable

• X2 is the type of firm (a qualitative or categorical variable)

The question

• X2 takes the value 0 if it is a mutual fund firm and 1 if it is a stock fund firm

• We ask whether or not stock firms adopt the innovation slower or faster than mutual firms

• We ask the question across all firms, regardless of size

Plot the data

symbol1 v=M i=sm70 c=black l=1;symbol2 v=S i=sm70 c=black l=3;

proc sort data=a1; by stock size;proc gplot data=a1; plot months*size=stock;run;

months

0

10

20

30

40

size

0 100 200 300 400

stock 0 1M M M S S S

M

M

MM

M

M

M

M

M

M

S

S SS

S S

S S

SS

Two symbols on plot

Interaction effects

• Interaction expresses the idea that the effect of one explanatory variable on the response depends on another explanatory variable

• In the KNNL example, this would mean that the slope of the line depends on the type of firm

Are both lines the same?

• From scatterplot, looks like different intercepts but can use the test statement for formal assessment

Data a1; set a1; sizestock=size*stock;Proc reg data=a1; model months=size stock sizestock; test stock, sizestock;run;

Output

Test 1 Results for Dependent Variable months

Source DFMean

Square F Value Pr > FNumerator 2 158.12584 14.34 0.0003

Denominator 16 11.02381    

Reject H0.There is a difference in the linear relationship across groups

Output•How are they different?

Parameter Estimates

Variable DFParameter

EstimateStandard

Error t Value Pr > |t|Intercept 1 33.83837 2.44065 13.86 <.0001

size 1 -0.10153 0.01305 -7.78 <.0001

stock 1 8.13125 3.65405 2.23 0.0408

sizestock 1 -0.00041714 0.01833 -0.02 0.9821

1. No difference in slopes assuming different intercepts

2. Potentially different intercepts assuming different slopes

Two parallel lines?

proc reg data=a1; model months=size stock;run;

OutputAnalysis of Variance

Source DFSum of

SquaresMean

Square F Value Pr > FModel 2 1504.4133 752.2066 72.50 <.0001

Error 17 176.38667 10.37569    

Corrected Total 19 1680.8000      

Root MSE 3.22113 R-Square 0.8951

Dependent Mean 19.40000 Adj R-Sq 0.8827

Coeff Var 16.60377    

Output

Int for stock firms is 33.87+8.05 = 41.92Common slope is –0.10

Parameter Estimates

Variable DFParameter

EstimateStandard

Error t Value Pr > |t|Intercept 1 33.87407 1.81386 18.68 <.0001

size 1 -0.10174 0.00889 -11.44 <.0001

stock 1 8.05547 1.45911 5.52 <.0001

Plot the two fitted lines

symbol1 v=M i=rl c=black l=1;symbol2 v=S i=rl c=black l=3;

proc gplot data=a1; plot months*size=stock;run;

months

0

10

20

30

40

size

0 100 200 300 400

stock 0 1M M M S S S

M

M

MM

M

M

M

M

M

M

S

S S

The plot

Two continuous variables

• Y = β0 + β1X1 + β2X2 + β3X1X2 + e

• Can be rewritten as follows

• Y = β0 + (β1 + β3X2)X1 + β2X2 + e

• Y = β0 + β1X1 + (β2 + β3X1) X2 + e

• The coefficient of one explanatory variable depends on the value of the other explanatory variable

Last slide

• We went over KNNL 8.2 – 8.7

• We used programs Topic17.sas to generate the output for today