1. Exact probability test (Fisher’s Test) Oh Hun Kwon

2. Exact probability test (Fisher’s exact test) One of a class of exact tests, so called because the significance of the deviation from a null hypothesis can be calculated exactly, rather than relying on an approximation that becomes exact in the limit as the sample size grows to infinity, as with many statistical tests. Fisher's exact test is a statistical significance test used in the analysis of contingency tables where sample sizes are small. It is named after its inventor, R. A. Fisher.

3. Purpose and scope The test is useful for categorical data that result from classifying objects in two different ways; it is used to examine the significance of the association (contingency) between the two kinds of classification. For hand calculations, the test is only feasible in the case of a 2 × 2 contingency table. However the principle of the test can be extended to the general case of an m × n. Fisher's exact test assumes that the row and column totals are known in ad vance.

4. In our case NL vs AD  group difference (Thickness,FA,MD,Sulcal depth). Control age,gender if differentiate between two group significantly. –Age : independent two sample t-test –Gender : independent two sample t-test ?????  Chi square test, Fisher’s exact test NLADTotal Male(0) 5 1318 Female (1) 1 1112 Total6 2430 Calculating statistical power

5. Assumption Random sampling –The data must be randomly selected from, at least representative of, a larger sample. The data must be form a contingency table –The value must be the number of subjects actually observed, not percentages or rates. Small sample size –Total sample size is less than about 20~40. Independent observations –Each subject must be independently selected from the population. Independent samples –No pairing or matching.

6. Circumstance (example) Some soldiers are being trained as parachutists. One rather windy afternoon 55 practice jumps take place at two localities, dropping zone A and dropping zone B. Of 15 men who jump at dropping zone A, five suffer sprained ankles, and of 40 who jump at dropping zone B, two suffer this injury. The casualty rate at dropping zone A seems unduly high, so the medical officer in charge decides to investigate the disparity.

7. Null Hypothesis Is it a difference that might be expected by chance? If not it deserves deeper study. The null hypothesis is that there is no difference in the probability of injury generating the proportion of injured men at each dropping zone.

8. Possible table The number of possible tables with these marginal totals is eight, that is, th e smallest marginal total plus one.

9. Calculating Probability InjuredUninjurTotal Zone A a b a+b Zone B c d c+d Total a+c b+dn ※ Fisher showed that the probability of obtaining any such set of values was given by the hypergeometric distribution....

10. Hypergeometric distribution The hypergeometric distribution is a discrete probability distribution that describes the number of successes in a sequence of n draws from a finite population without replacement, just as the binomial distribution describes the number of successes for draws with replacement. A random variable X follows the hypergeometric distribution with parameters N, m and n if the probability is given by

11. Hypergeometric Distribution

12. Calculating P-value The observed set has a probabi lity of 0.0115427. The P value is the probability of getting the observed set, or one more extreme. A one tailed P value ⇒ 0.0115427 + 0.0009866 + 0.0000317 = 0.01256 The two tailed value ⇒ P = 0.025 for the conventional(double th e one tailed result) ⇒ P = 0.0136 for the mid P approach.

13. One tailed or Two tailed Observed table More extreme case than observed table

14. Conclusion In either case, the P value is less than the conventional 5% level The medical officer can conclude that there is a problem in dropping zone A. The way to present the results is: Injury rate in dropping zone A was 33%, in dropping zone B 5%; difference 28% (95% confidence interval 3.5 to 53.1% (from )), P = 0.0136 (Fisher's Exact test mid P).

15. Practice 9.1 Of 30 men employed in a small workshop 18 worked in one department and 12 in another department. In one year five of the 18 reported sick with septic hands, and of the 12 men in the other department one did so.  Is there a difference in the departments and how would you report this result?

16. Answer SickNot sick Total Dep A 5 1318 Dep B 1 1112 Total6 2430 P00.0016 P10.024 P20.1275 P30.3023 P40.3401 P50.1732 P60.0313

17. Answer P-value =0.003 +0.017 = 0.002 (one-tailed) P-value =0.003 +0.017 = 0.004 (two-tailed) Significantly different P is calculated from 2 x (0.5 x 0.173 + 0.031).