2. 1

3. 2 Two-samples tests, X 2 Dr. Mona Hassan Ahmed Prof. of Biostatistics HIPH, Alexandria University

4. 3 Z-test (two independent proportions) P1= proportion in the first group P2= proportion in the second group n1= first sample size n2= second sample size

5. 4 Critical z = 1.96 at 5% level of significance 2.58 at 1% level of significance

6. 5 Researchers wished to know if urban and rural adult residents of a developing country differ with respect to prevalence of a certain eye disease. A survey revealed the following information Total Eye disease Residence NoYes 30027624Rural 50048515Urban Test at 5% level of significance, the difference in the prevalence of eye disease in the 2 groups Example

7. 6 P1 = 24/300 = 0.08 p2 = 15/500 = 0.03 2.87 > Z* The difference is statistically significant Answer

8. 7 t-Test (two independent means) = mean of the first group = mean of the second group S 2 p = pooled variance

9. 8 Critical t from table is detected at degree of freedom = n 1 + n 2 - 2 level of significance 1% or 5%

10. 9 Sample of size 25 was selected from healthy population, their mean SBP =125 mm Hg with SD of 10 mm Hg. Another sample of size 17 was selected from the population of diabetics, their mean SBP was 132 mmHg, with SD of 12 mm Hg. Test whether there is a significant difference in mean SBP of diabetics and healthy individual at 1% level of significance Example

11. 10 S 1 = 12 S 2 =11 State H0 H 0 :  1 =  2 State H1 H 1 :  1   2 Choose α α = 0.01 Answer

12. 11 Critical t at df = 40 & 1% level of significance = 2.58 Decision: Since the computed t is smaller than critical t so there is no significant difference between mean SBP of healthy and diabetic samples at 1 %. Answer

13. 12 Degrees of freedom Probability (p value) 0.100.050.01 16.31412.70663.657 52.0152.5714.032 101.8132.2283.169 171.7402.1102.898 201.7252.0862.845 241.7112.0642.797 25 1.7082.0602.787  1.6451.9602.576

14. 13 Paired t- test (t- difference) Uses: To compare the means of two paired samples. Example, mean SBP before and after intake of drug.

15. 14 di = difference (after-before) Sd = standard deviation of difference n = sample size Critical t from table at df = n-1

16. 15 The following data represents the reading of SBP before and after administration of certain drug. Test whether the drug has an effect on SBP at 1% level of significance. Example

17. 16 SBP (After) SBP (Before) Serial No. 1802001 1651602 1751903 185 4 1702105 1601756

18. 17 di 2 di After-Before BP After BP Before Serial No. 400-201802001 2551651602 225-151751903 00185 4 1600-401702105 225-151601756 2475-85 Total ∑ di2∑di Answer

19. 18 Answer

20. 19 Critical t at df = 6-1 = 5 and 1% level of significance = 4.032 Decision: Since t is < critical t so there is no significant difference between mean SBP before and after administration of drug at 1% Level. Answer

21. 20 Degrees of freedom Probability (p value) 0.100.050.01 16.31412.70663.657 52.0152.5714.032 101.8132.2283.169 171.7402.1102.898 201.7252.0862.845 241.7112.0642.797 25 1.7082.0602.787  1.6451.9602.576

22. 21 Chi-Square test It tests the association between variables... The data is qualitative. It is performed mainly on frequencies. It determines whether the observed frequencies differ significantly from expected frequencies.

23. 22 Where E = expected frequency O = observed frequency

24. 23  Critical X 2 at df = (R-1) ( C -1) Where R = raw C = column  I f 2 x 2 table X 2* = 3.84 at 5 % level of significance X 2* = 6.63 at 1 % level of significance

25. 24

26. 25 In a study to determine the effect of heredity in a certain disease, a sample of cases and controls was taken: Total Disease Family history ControlCases 20012080Positive 300160140Negative 500280220Total Using 5% level of significance, test whether family history has an effect on disease Example

27. 26 X 2 = (80-88) 2 /88 + (120-112) 2 /112 + (140-132) 2 /132 + (160-168) 2 /168 = 2.165 < 3.84 Association between the disease and family history is not significant Total Disease Family history ControlCases 200 120 112 80 88 positive O E 300160 168 140 132 Negative O E 500280220Total Answer

28. 27 The odds ratio was developed to quantify exposure – disease relations using case- control data Once you have selected cases and controls  ascertain exposure Then, cross-tabulate data to form a 2-by-2 table of counts

29. 28 2-by-2 Crosstab Notation Disease +Disease -Total Exposed +ABA+B Exposed -CDC+D TotalA+CB+DA+B+C+D Disease status A+C = no. of cases B+D = no. of non-cases Exposure status A+B = no. of exposed individuals C+D = no. of non-exposed individuals

30. 29 Disease +Disease - Exposed +AB Exposed -CD Cross-product ratio

31. 30 Exposure variable = Smoking Disease variable = Hypertension D+D- E+3071 E-122 Total3193 Example

32. 31 Interpretation of the Odds Ratio Odds ratios are relative risk estimates Relative risk are risk multipliers The odds ratio of 9.3 implies 9.3× risk with exposure

33. 32 No association OR < 1 OR = 1 OR > 1 Positive association Higher risk Negative association Lower risk (Protective)

34. 33 In the previous example OR = 9.3 95% CI is 1.20 – 72.14

35. 34 Multiple Levels of Exposure Smoking levelCasesControls Heavy smokers213274 Moderate smokers61147 Light smokers1482 Non-smokers8115 Total296618

36. 35 Multiple Levels of Exposure k levels of exposure  break up data into (k – 1) 2  2 tables Compare each exposure level to non-exposed e.g., heavy smokers vs. non-smokers CasesControls Heavy smokers213274 Non-smokers8115

37. 36 Multiple Levels of Exposure Smoking levelCasesControls Heavy smokers213274OR 3 =(213)(115)/(274)(8)=11.2 Moderate smokers61147OR 2 =(61)(115)/(147)(8) = 6.0 Light smokers1482OR 1 =(14)(115)/(82)(8) = 2.5 Non-smokers8115 Total605115 Notice the trend in OR (dose-response relationship)

38. 37 Small Sample Size Formula For the Odds Ratio It is recommend to add ½ to each cell before calculating the odds ratio when some cells are zeros D+D- E+3171 E-022 Total3193 OR Small Sample = (A+0.5)(D+0.5) (B+0.5)(C+0.5) OR Small Sample = (31+0.5)(22+0.5) =19.8 (71+0.5)(0+0.5)

39. 38