Comparing Population Parameters (Z-test, t-tests and Chi-Square by olliegoblue23

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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, Data Coordinating Center
        College of Human Medicine
         Michigan State University
Is there an association between
 Drinking and Lung Cancer?


  Suppose a case-control study is
   conducted to test the above
           hypothesis?
       QUESTION: Is there a difference
       between the proportion of drinkers among
       cases and controls?



        Group 1                     Group 2
        Disease                    No Disease
P1= proportion of drinkers   P2= proportion of drinkrs
Elements of Testing hypothesis

   •   Null Hypothesis
   •   Alternative hypothesis
   •   Level of significance
   •   Test statistics
   •   P-value
   •   Conclusion
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
 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
                        Test Statistic
     How many standard deviations has our
      estimate deviated from the hypothesized
      value if the null hypothesis was true?

Z  (eP1  eP 2  0) /[(1/ n1  1/ n2)( p(1  p))]
where
p  (33  27) /(1700  2300)  60 / 4000  3/ 200  0.015
Z  [(33/1700)  (27 / 2300)  0)]/( (1/1700  1/ 2300)(0.015)(0.985)
Z  2.003
       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?
Chi-Square Test of Independence
 (based on a Contingency Table)

     (Observed  E xp ected )   2
 
  2

            Expected

df  (r  1)(c  1)
 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
    E11=1700(60)/4000=25.5                      E12=34.5
     E21=1674.5                                 E22=2265.5

                 (Observed  E xp ected )
          k 4                              2

   2
    obs                 Expected
                                                
          k 1


(33  25.5) (27  34.5)
                     2                2
                        
    25.5          34.5
(1667  1674.5) (2273  2265.5)
               2                2
                 
     1674.5            2265.5      4.0
   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
 Testing Hypothesis When Two
Population Means are Compared


       H0: 1= 2
       Ha: 1 2
     QUESTION: Is there an association
        between age and Lung Cancer?



      Group 1                  Group 2
      Disease                No Disease
Mean age of the cases   Mean age of the controls
   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
         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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