CHI-SQUARE

x 2

Chi Square

• Symbolized by Greek x 2

• pronounced “Ki square”

• a Test of STATISTICAL SIGNIFICANCE for TABLE data

• “What are the ODDs the relationship in a TABLE using SAMPLE data found in the POPULATION

CHI SQUARE or x 2

A TEST of STATISTICAL SIGNIFICANCE

What do tests of statistical significance tell us?

• Are OBSERVED RESULTS• SIGNIFICANTLY

DIFFERENT than would be expected

• BY CHANCE

• Criteria p < .05

The Logic of Statistical Significance

• Use probability theory to determine whether OBSERVED results SIGNIFICANTLY different than expected BY CHANCE in RANDOM sample data

• By convention criteria p < .05 (1-in-20)

• p <.01 better

• p>.o5 OBSERVED results could be due

• to chance

HYPOTHESIS TESTING

Test of Significance:

Logic of chi-square

x 2 and the Logic of Formal Hypothesis Testing

• Assert RESEARCH HYPOTHESIS (H1) PRIOR to data collection:

• H1: There is a relationship between X & Y

• TEST the NULL HYPOTHESIS of NO RELATIONSHIP (Ho)

• H0: There is NO RELATIONSHIP X & Y

• GOAL: Reject NULL using x 2 (p<.05)

ASSESSING STATISTICAL SIGNIFICANCE

• Chi-square is one standard test of a relationship between 2 variables

• With it we ask the data distribution to tell us about the null hypothesis

• IS THE NULL TRUE?

CONCLUSIONS using x 2 to test“significance” of hypothesis

• Reject NULL HYPOTHESIS (H0) if find p <.05

• ....>conclude OBSERVED RESULT are

• STATISTICALLY SIGNIFICANT

• Cannot reject reject NULL (H0) if p>.05

• ...> conclude results NOT SIGNIFICANT

• ...>could be due to chance

Chi Square (x2) tells us• Whether the

OBSERVED results• we see in a TABLE• are SIGNIFICANTLY• different than would

be due to chance

• Values Expected by• CHANCE• Party• Dem Rep• Vote 50% 50%• Not 50% 50%• 100% 100%

Is relationship “significant”???

• Is pattern or relationship OBSERVED in sample also found in POPULATION??

Chi Square x2

• Compares OBSERVED values

• in a TABLE and• asked:• Are these significantly• different than would be

expected by chance? • * ( ) Expected p-value

• VOTE by PARTY *• Party• Deem Rep

• Vote 5 (5) 5 (5)• NOT 5 (5) 5(5)• 10 10• N=20

Basic Formula for Chi Square x2

• (O-E)2

• ----------

•E

• O = Observed cell value

• E = Expected by chance cell value

Value for x 2 depends upon

THREE factors: #1• 1. (O - E)2

• ________

• E

#1 Size of the observed relationship

– Age

– Young Old

– $lots10% 90%

– little 90% 10%

– 100% 100%

– %d=80%

– N=50

• Age

• Young Old

• $lots 50% 50%

• little 50% 50%

• 100% 100%

• %d= 0% • N=50

#2 THE DEGREE of FREEDOM associated with the TABLE

• d.f. degrees of freedom =

• (c-1) #columns - 1• (r-1) #rows -1• Must calculate d.f. or• (c-1) (r-1) to obtain• probability from x2

• table

• Example:• Party by Vote• Dem Rep• Vote• NOT• d.f. = (c-1) (r-1)• (2-1) (2-1)• d.f. = ?

Degrees of Freedom

• d.f. = (r-1) (c-1)• How many pieces of• independent

information can go into a table before rest of cell values are FIXED

• 2x2 table = 1 d.f.

• Example 2x2 Table• Dem Rep• Vote * x• NOT x • x x• How many cells filled

in before rest of values are fixed? *1

Degrees of Freedom cont.

• d.f. = (c-1)(r-1)• 2x2 table = 1• 2x3 table = ?• How many cell values

free to vary before rest of table is fixed?

• Party and Vote• Dem Rep• Vote * x• No * x• Un- x• decided• x x

Degrees of Freedom cont.

• d.f. = (c-1)(r-1)• 2x2 table = 1 d.f.• 2x3 table = 2 d.f.• 4x4 table = ???• 10x10 table = ???• information used to

calculate p=values or ODDS

#3 Factor used to calculate x2

• SAMPLE SIZE• LARGER samples• higher probability• x2 will be • STATISTICALLY• SIGNIFICANT • p<.05• N=5 N=50 N=500

“Significance” depends partly on SAMPLE SIZE

• Walks Quacks Like• Like a Duck• Duke Yes No• Yes 60% 40%• No 40 60

• (N) (25) (25)

• x2 =2.0;n.s. (p>.05)

• Walks Quacks Like• Like a Duck• Duke Yes No• Yes 60% 40%• No 40 60• 100% 100%

• (N) (50) (50)

• x2 = 4.00 ; p<.05

REMEMBER

• STATISTICAL SIGNIFICANCE does NOT = SUBSTANTIVE SIGNIFICANCE

• Need SAMPLE DATA collected via random sampling (N = 30+)

• For x2 need AT LEAST 5 observations per cell

• p< or = to .05• Used as tool in Formal Hypothesis test NOT• “bare foot empiricism i.e., run 5,000 tests

Conclusions will be tentative. Why?

• Using probability as standard of proof

• p<.05 means 95/100 times this relationship is REAL not RANDOM

• BUT always some chance conclusion

• is wrong

• How much using p<.05?

Probabilistic generalizations

• Will always be tentative

Type I and Type II Errors

• Decision about H0

• H0 is true H0 is false

• Reject H0 Type I No Error

• Do not reject

• H0 No Error Type II

Solution:

• Use “good” theory

• Use lowest possible p-value

• Replicate results.....Null results can be meaningful....why do these results mean?

Null results are important

• Must try to understand what they mean????

• for your• theory

Nothing replaces a good theory

• Can be guide to interpret results

• Sometimes need to be revised

TEST NULL

• REMEMBER:

• Assert H1 to be TRUE

• TEST the NULL

• H0 No relationship

• If H0 is < .05

• What do we conclude?

Correct Conclusion

• YES,“data supports hypothesis” or “statistically significant”

• NEVER “results PROOVE” H1

Example: TESTING NULLusing x2

• 2 Election variables -- Nature of Primary...>

• divisive or not• & Outcome of General

Election ...>win or lose• What would be your• Research hypothesis

• H1: ?????

TWO VARIABLES

• Type of Primary -- we’ll classify them as divisive or nondivisive

• Election Outcome -- we’ll classify as won or lost

Example cont.

• H1: If a candidate experiences a divisive primary race, s/he is more likely to lose the general election

• H0: There is no relationship between type of primary race and election outcome

• Test: calculate x 2 and p-value using d.f.

• and a x2 table (back of any stat book)

x2 = (0-E)2

E

• 0 E 0-E (0-E)2 (0-E)2/E

• ______ _______ Sum zero chi sq value

• x2 = ?; d.f. = ? p< or = ?

TYPE of Primary

• Divisive Nondivisive

• Won 8 ( ) 22 ( ) 30

• Lost 8 ( ) 6 ( ) 14

• 16 28 44

• x2 = ?? ; d.f. = ?? ; p< ??

TYPE of PRIMARY

• Calculate Measure of Association• • Cramer’s V = .295084 x2/mn• where m =(r-1) or• (c-1)• which ever is smaller • Other measures of association could use?

TYPE of PRIMARYConclusion

• Strength of relationship using

• Cramer’s V = .295

• Is this a significant relationship?

• x2 = d.f. = p<

• Overall conclusion????

• Next step: consider OTHER VARIABLES