1. Hypothesis Testing and Sample Size Calculation Po Chyou, Ph. D. Director, BBC

2. Hypothesis Testing on Population mean(s) Population median(s)  Population proportion(s) Population variance(s) Population correlation(s)  Association based on contingency table(s) Coefficients based on regression model Odds ratio Relative risk Trend analysis Survival distribution(s) / curve(s) Goodness of fit

3. Hypothesis Testing 1.Definition of a Hypothesis An assumption made for the sake of argument 2.Establishing Hypothesis Null hypothesis - H 0 Alternative hypothesis - H a 3. Testing Hypotheses Is H 0 true or not?

4. Hypothesis Testing 4.Type I and Type II Errors Type I error: we reject H 0 but H 0 is true α = Pr(reject H 0 / H 0 is true) = Pr(Type I error) = Level of significance in hypothesis testing Type II error: we accept H 0 but H 0 is false  = Pr(accept H 0 / H 0 is false) = Pr(Type II error)

5. Hypothesis Testing 5.Steps of Hypothesis Testing - Step 1Formulate the null hypothesis H 0 in statistical terms - Step 2Formulate the alternative hypothesis H a in statistical terms - Step 3Set the level of significance α and the sample size n - Step 4Select the appropriate statistic and the rejection region R - Step 5Collect the data and calculate the statistic

6. Hypothesis Testing 5.Steps of Hypothesis Testing (continued) - Step 6If the calculated statistic falls in the rejection region R, reject H 0 in favor of H a ; if the calculated statistic falls outside R, do not reject H 0

7. Hypothesis Testing A random sample of 400 persons included 240 smokers and 160 non- smokers. Of the smokers, 192 had CHD, while only 32 non-smokers had CHD. Could a health insurance company claim the proportion of smokers having CHD differs from the proportion of non-smokers having CHD? 6. An Example

8. Hypothesis Testing Example (continued) Let P 1 = the true proportion of smokers having CHD P 2 = the true proportion of non-smokers having CHD - Step 1 H 0 : P 1 = P 2 - Step 2 H a : P 1  P 2 - Step 3 α =.05, n = 400

9. Hypothesis Testing Example (continued) - Step 4statistic =  = P 1 - P 2 where P 1 = x 1, P 2 = x 2 and P = x 1 + x 2 n 1 n 2 n 1 + n 2 P(1-P) (1/n 1 + 1/n 2 )

10. Hypothesis Testing Example (continued) P 1 = x 1 n1n1 = 192 =.80 240 P 2 = x 2 n2 n2 = 32 =.20 160 P = x 1 + x 2 n 1 + n 2 = 192 + 32 = 224 = 0.56 240 + 160 400  = P 1 - P 2 P(1-P) (1/n 1 + 1/n 2 ) =.80 -.20 =.60 = 11.84 > 1.96 (.56) (1-.56) (1/240 + 1/160).05066 - Step 5

11. Hypothesis Testing Example (continued) - Step 6 Reject H 0 and conclude that smokers had significantly higher proportion of CHD than that of non-smokers. [P-value <.0000001]

12. Hypothesis Testing 7. Contingency Table Analysis The Chi-square distribution (  2 )

13. Hypothesis Testing Equation for chi-square for a contingency table For i = 1, 2 and j =1, 2  2 = (O 11 - E 11 ) 2 + (O 12 - E 12 ) 2 + (O 21 - E 21 ) 2 + (O 22 - E 22 ) 2 E 11 E 12 E 21 E 22  2 =  (O ij - E ij ) 2 i, j E ij

14. Hypothesis Testing Equation for chi-square for a contingency table (cont.) E 11 = n 1 m 1 E 12 = n 1 - n 1 m 1 = n 1 m 2 nnn E 21 = n 2 m 1 E 22 = n 2 - n 2 m 1 = n 2 m 2 nnn

15. Hypothesis Testing Example : Same as before - Step 1 H 0 : there is no association between smoker status and CHD - Step 2 H a : there is an association between smoker status and CHD - Step 3  =.05, n = 400 - Step 4statistic =  2 = (O 11 - E 11 ) 2 + (O 12 - E 12 ) 2 + (O 21 - E 21 ) 2 + (O 22 - E 22 ) 2 E 11 E 12 E 21 E 22

16. Hypothesis Testing Example (continued) : Same as before

17. - Step 5

18. Hypothesis Testing Example (continued) : Same as before E 11 = n 1 m 1 = 240 * 224 = 134.4 n 400 E 12 = n 1 - n 1 m 1 = 240 - 134.4 = 105.6 n E 21 = n 2 m 1 = 160 * 224 = 89.6 n 400 E 22 = n 2 - n 2 m 1 = 160 - 89.6 = 70.4 n - Step 5 (continued) Expectation Counts

19. Hypothesis Testing Example (continued) : Same as before - Step 5 (continued)  2 = (O 11 - E 11 ) 2 + (O 12 - E 12 ) 2 + (O 21 - E 21 ) 2 + (O 22 - E 22 ) 2 E 11 E 12 E 21 E 22 = (192 - 134.4) 2 + (48 - 105.6) 2 + (32 - 89.6) 2 + (128 - 70.4) 2 134.4 105.6 89.6 70.4 = 24.68 + 31.42 + 37.03 + 47.13 = 140.26 > 3.841

20. Hypothesis Testing Example (continued) : Same as before - Step 6 Reject H 0 and conclude that there is an association between smoker status and CHD. [P-value <.0000001]

21. Sample Size Estimation and Statistical Power Calculation Definition of Power Recall :  = Pr (accept H 0 / H 0 is false) = Pr (Type II error) Power = 1 -  = Pr(reject H 0 / H 0 is false)

22. Sample Size Estimation for Intervention on Tick Bites Among Campers Assumptions 1.Given that the proportion (P CON ) of tick bites among campers in the control group is constant. 2.Given that the proportion (P INT ) of tick bites among campers in the intervention group is reduced by 50% compared to that of the control group after intervention has been implemented. 3.Given that a one- or two- tailed test is of interest with 80% power and a type-I error of 5%.

23. Sample Size Estimation for Intervention on Tick Bites Among Campers Summary Table 1

24. Statistical Power Calculation for Intervention on Obesity of Women in MESA Assumptions 1.Given that the proportion (P CON ) of women who are obese at baseline (i.e., the control group) is constant. There are a total of 840 women in the control group. Based on our preliminary data analysis results, approximately 50% of these 840 women at baseline are obese (BMI >= 27.3). 2.Given that the proportion (P INT ) of women who are obese in the intervention group is reduced by 5% or more compared to that of the control group after intervention has been implemented. There are a total of 680 women who had been newly recruited. Based on our preliminary data analysis results, 50% of these 680 newly recruited women are obese. Assume that 60% of these women will agree to participate, we will have 200 women to be targeted for intervention.

25. Statistical Power Calculation for Intervention on Obesity of Women in MESA (continued) 3.Given that a one-tailed test is of interest with a type-I error of 5%, then the estimated statistical powers are shown in Table 1 for detecting a difference of 5% or more in the proportion of obesity between the control group and the intervention group. Assumptions Table 1

26. Reference “Statistical Power Analysis for the Behavioral Sciences” Jacob Cohen Academic Press, 1977

27. Take Home Message: You’ve got questions : Data ? S T A TI S T I C S ?... Contact Biostatistics and consult with an experienced biostatistician –Po Chyou, Director, Senior Biostatistician (ext. 9-4776) –Dixie Schroeder, Secretary (ext. 1-7266) OR Do it at your own risk

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