1. Text books: (1)Medical Statistics A commonsense approach A commonsense approach By By Michael J. Campbell & David Machin Michael J. Campbell & David Machin (2) An Introduction to Medical Statistics By By Martin Bland Martin Bland (3) Biostatistics: A Foundation for Analysis in the Health Sciences Health Sciences By By Wayne W. Daniel Wayne W. Daniel

2. THE CHI-SQUARE TEST BACKGROUND AND NEED OF THE TEST Data collected in the field of medicine is often qualitative. --- For example, the presence or absence of a symptom, classification of pregnancy as ‘high risk’ or ‘non- high risk’, the degree of severity of a disease (mild, moderate, severe)

3. The measure computed in each instance is a proportion, corresponding to the mean in the case of quantitative data such as height, weight, BMI, serum cholesterol. Comparison between two or more proportions, and the test of significance employed for such purposes is called the “Chi-square test”

4. ---- KARL PEARSON IN 1889, DEVISED AN INDEX OF DISPERSION OR TEST CRITERIOR DENOTED AS “CHI- SQUARE “. (X ). ---- KARL PEARSON IN 1889, DEVISED AN INDEX OF DISPERSION OR TEST CRITERIOR DENOTED AS “CHI- SQUARE “. (X 2 ). The formula for X –test is, The formula for X 2 –test is,

5. Chi- Square   ( o - e ) 2 e X = X 2 = Figure for Each Cell

6. reject H o if  2 >  2. ,df where df = (r-1)(c-1)  2 = ∑ (O - E) 2 E 3.E is the expected frequency ^ E =E =E =E = ^ (total of all cells) total of row in which the cell lies total of column in which the cell lies 1.The summation is over all cells of the contingency table consisting of r rows and c columns 2.O is the observed frequency 4.The degrees of freedom are df = (r-1)(c-1)

7. APPLICATION OF CHI-SQUARE TEST  TESTING INDEPENDCNE (OR ASSOCATION)  TESTING FOR HOMOGENEITY  TESTING OF GOODNESS-OF- FIT

8. Chi-square test Purpose To find out whether the association between two categorical variables are statistically significant Null Hypothesis There is no association between two variables

9. Chi-square test 1. 2. 3. 4. Test statistics

10. Chi-square test  Objective : Smoking is a risk factor for MI  Null Hypothesis: Smoking does not cause MI D (MI)D-(No MI)Total Smokers292150 Non-smokers163450 Total4555100

11. Chi-SquareChi-Square E O 29 E O 21 E O 16 E O 34 MINon-MI Smoker Non-Smoker

12. Chi-SquareChi-Square E O 29 E O 21 E O 16 E O 34 MINon-MI Smoker Non-smoker 50 5545100

13. Estimating the Expected Frequencies E =E =E =E = ^ (row total for this cell)(column total for this cell) n Classification 1 Classification 2 1 2 3 4c r 123 C1C1C1C1 C2C2C2C2 C3C3C3C3 C4C4C4C4 CcCcCcCc R1R1R1R1 R2R2R2R2 R3R3R3R3 RrRrRrRr E =E =E =E = ^ R2C3R2C3nnR2C3R2C3nnn

14. Chi-SquareChi-Square E O 29 E O 21 E O 16 E O 34 MINon-MI Smoker Non-smoker 50 5545100 50 X 45 100 22.5 = 22.5

15. Chi-SquareChi-Square E O 29 E O 21 E O 16 E O 34 AloneOthers Males Females 50 5545100 22.527.5 22.527.5

16. Chi-SquareChi-Square  Degrees of Freedom df = (R-1) (C-1)  Critical Value (Table A.6) = 3.84  Critical Value (Table A.6) = 3.84  X 2 = 6.84

18. Age Gender 45Total Male 60 (60)20 (30)40 (30)120 Female 40 (40)30 (20)10 (20)80 Total1005050200 Chi- square test Find out whether the sex is equally distributed among each age group

19. Test for Homogeneity (Similarity) To test similarity between frequency distribution or group. It is used in assessing the similarity between non-responders and responders in any survey Age (yrs)RespondersNon-respondersTotal <2076 (82)20 (14)96 20 – 29288 (289)50 (49)338 30-39312 (310)51 (53)363 40-49187 (185)30 (32)217 >5077 (73) 9 (13) 86 Total9401601100