Comparison parameters

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Comparison parameters

  1. 1. Comparing Population Parameters (Z-test, t-tests and Chi-Square test) Dr. M. H. Rahbar Professor of Biostatistics Department of Epidemiology Director, Data Coordinating Center College of Human Medicine Michigan State University
  2. 2. Is there an association between Drinking and Lung Cancer? Suppose a case-control study is conducted to test the above hypothesis?
  3. 3. QUESTION: Is there a difference between the proportion of drinkers among cases and controls? G r o u p 1 D is e a s e P 1 = p r o p o r t io n o f d r in k e r s G r o u p 2 N o D is e a s e P 2 = p r o p o r t io n o f d r in k r s
  4. 4. Elements of Testing hypothesis • Null Hypothesis • Alternative hypothesis • Level of significance • Test statistics • P-value • Conclusion
  5. 5. Case Control Study of Drinking and Lung Cancer Null Hypothesis: There is no association between Drinking and Lung cancer, P1=P2 or P1-P2=0 Alternative Hypothesis: There is some kind of association between Drinking and Lung cancer, P1P2 or P1-P20
  6. 6. Based on the data in the following contingency table we estimate the proportion of drinkers among those who develop Lung Cancer and those without the disease? Lung Cancer Total Case Control Drinker Yes A=33 B=27 60 No C=1667 D= 2273 3940 eP1=33/1700 eP2=27/2300
  7. 7. Test Statistic How many standard deviations has our estimate deviated from the hypothesized value if the null hypothesis was true? ( 1 2 0)/[(1/ 1 1/ 2)( (1 ))] (33 27)/(1700 2300) 60/ 4000 3/ 200 0.015 [(33/1700) (27/ 2300) 0)]/( (1/1700 1/ 2300)(0.015)(0.985) 2.003 Z eP eP n n p p where p Z Z = − − + − = + + = = = = − − + =
  8. 8. P-value for a two tailed test P-value= 2 P[Z > 2.003] = 2(.024)=0.048 How does this p-value compared with =0.05? Since p-value=0.048 < =0.05, reject the null hypothesis H0 in favor of the alternative hypothesis Ha. Conclusion: There is an association between drinking and lung cancer. Is this relationship causal?
  9. 9. Chi-Square Test of Independence (based on a Contingency Table) 2 2 ( xp ) ( 1)( 1) Observed E ected Expected df r c χ − = = − − ∑
  10. 10. In the following contingency table estimate the proportion of drinkers among those who develop Lung Cancer and those without the disease? Lung Cancer Total Case Control Drinker Yes O11=33 O12=27 R1=60 No O21=1667 O22= 2273 R2=3940 Total C1 = 1700 C2 = 2300 n = 4000 E11=1700(60)/4000=25.5 E12=34.5 E21=1674.5 E22=2265.5
  11. 11. E11=1700(60)/4000=25.5 E12=34.5 E21=1674.5 E22=2265.5 24 1 2 2 2 2 2 ( xp ) (33 25.5) (27 34.5) 25.5 34.5 (1667 1674.5) (2273 2265.5) 1674.5 2265.5 4.0 k k obs Observed E ected Expected χ = = − + = − − + − − + = ∑
  12. 12. How do we calculate P-value? • SPSS, Epi-Info statistical packages could be used to calculate the p-value for various tests including the Chi-Square Test • If p-value is less than 0.05, then reject the null hypothesis that rows and column variables are independent
  13. 13. Testing Hypothesis When Two Population Means are Compared H0: 1= 2 Ha: 1 2
  14. 14. QUESTION: Is there an association between age and Lung Cancer? G r o u p 1 D is e a s e M e a n a g e o f t h e c a s e s G r o u p 2 N o D is e a s e M e a n a g e o f t h e c o n t r o ls
  15. 15. Use Two-sample t-test when both samples are independent • H0: 1 = 2 vs Ha: 1  2 • H0: 1 - 2 = 0 vs Ha: 1 - 2  0 • t= difference in sample means – hypothesized diff. SE of the Difference in Means • Statistical packages provide p-values and degrees of freedom • Conclusion: If p-value is less than 0.05, then reject the equality of the means
  16. 16. Paired t-test for Matched case control study • H0: 1 = 2 vs Ha: 1  2 • H0: 1 - 2 = 0 vs Ha: 1 - 2  0 • Paired t-test= Mean of the differences –0 SE of the Differences in Means • Statistical packages provide p-values for paired t-test • Conclusion: If p-value is less than 0.05, then reject the equality of the means
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