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Comparing Population Parameters (Z-test, t-tests and Chi-Square test) Dr. M. H. Rahbar Professor of Biostatistics Department of Epidemiology Director,

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Presentation on theme: "Comparing Population Parameters (Z-test, t-tests and Chi-Square test) Dr. M. H. Rahbar Professor of Biostatistics Department of Epidemiology Director,"— Presentation transcript:

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 Is there an association between Drinking and Lung Cancer? Suppose a case-control study is conducted to test the above hypothesis?

3 QUESTION: Is there a difference between the proportion of drinkers among cases and controls?

4 Elements of Testing hypothesis Null Hypothesis Alternative hypothesis Level of significance Test statistics P-value Conclusion

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 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?

7 Test Statistic How many standard deviations has our estimate deviated from the hypothesized value if the null hypothesis was true?

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 Chi-Square Test of Independence (based on a Contingency Table)

10 In the following contingency table estimate the proportion of drinkers among those who develop Lung Cancer and those without the disease?

11 E11=1700(60)/4000=25.5 E12=34.5 E21=1674.5 E22=2265.5

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 Testing Hypothesis When Two Population Means are Compared H0:  1=  2 Ha:  1   2

14 QUESTION: Is there an association between age and Lung Cancer?

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 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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