The document discusses statistical methods to test the association between drinking and lung cancer through case-control studies using z-tests, t-tests, and chi-square tests. It outlines the null and alternative hypotheses, significance levels, test statistics, and p-values, ultimately concluding that there is an association between drinking and lung cancer. Additionally, it elaborates on methods for calculating p-values and includes procedures for comparing population means.

1. Comparing Population Parameters
(Z-test, t-tests and Chi-Square test)

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?
G roup 1
Disease
P 1= proportion of drinkers
G roup 2
No Disease
P 2= proportion of drinkrs

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?
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. 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. 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)
2
2 ( xp )
( 1)( 1)
Observed E ected
Expected
df r c



  


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. E11=1700(60)/4000=25.5 E12=34.5
E21=1674.5 E22=2265.5
2
4
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






 

 




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?
G roup 1
Disease
M ean age of the cases
G roup 2
No Disease
M ean age of the controls

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