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 (H 1 ) PRIOR to data collection: H 1 : There is a relationship between X & Y TEST the NULL HYPOTHESIS of NO RELATIONSHIP (H o ) H 0 : 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 (H 0 ) if find p < >conclude OBSERVED RESULT are STATISTICALLY SIGNIFICANT Cannot reject reject NULL (H 0 ) if p>.05...> conclude results NOT SIGNIFICANT...>could be due to chance

Chi Square (x 2 ) 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 x 2 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) N=20

Basic Formula for Chi Square x 2 (O-E) 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 x 2 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 x 2 SAMPLE SIZE LARGER samples higher probability x 2 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 (N) ( 25 ) ( 25 ) x 2 =2.0;n.s. (p>.05) Walks Quacks Like Like a Duck Duke Yes No Yes 60% 40% No % 100% (N) (50) (50) x 2 = 4.00 ; p<.05

REMEMBER STATISTICAL SIGNIFICANCE does NOT = SUBSTANTIVE SIGNIFICANCE Need SAMPLE DATA collected via random sampling (N = 30+) For x 2 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 H 0 H 0 is true H 0 is false Reject H 0 Type I No Error Do not reject H 0 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 H 1 to be TRUE TEST the NULL H 0 No relationship If H 0 is <.05 What do we conclude?

Correct Conclusion YES,“data supports hypothesis” or “statistically significant” NEVER “results PROOVE” H 1

Example: TESTING NULL using x 2 2 Election variables -- Nature of Primary...> divisive or not & Outcome of General Election...>win or lose What would be your Research hypothesis H 1 : ?????

TWO VARIABLES Type of Primary -- we’ll classify them as divisive or nondivisive Election Outcome -- we’ll classify as won or lost

Example cont. H 1 : If a candidate experiences a divisive primary race, s/he is more likely to lose the general election H 0 : There is no relationship between type of primary race and election outcome Test: calculate x 2 and p-value using d.f. and a x 2 table (back of any stat book)

x 2 = (0-E) 2 E 0 E 0-E (0-E) 2 (0-E) 2 /E _____________ Sum zero chi sq value x 2 = ?; d.f. = ? p< or = ?

TYPE of Primary Divisive Nondivisive Won 8 ( ) 22 ( ) 30 Lost 8 ( ) 6 ( ) x 2 = ?? ; d.f. = ?? ; p< ??

TYPE of PRIMARY Calculate Measure of Association Cramer’s V = x 2 /mn where m =(r-1) or (c-1) which ever is smaller Other measures of association could use?

TYPE of PRIMARY Conclusion Strength of relationship using Cramer’s V =.295 Is this a significant relationship? x 2 = d.f. = p< Overall conclusion???? Next step: consider OTHER VARIABLES