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

Dr. M. H. RahbarProfessor 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?

G rou p 1D isease

P 1 = p rop ortion o f d rin kers

G rou p 2N o D isease

P 2 = p rop ortion o f d rin k rs

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, P1P2 or P1-P20

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?

( 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

P-value for a two tailed testP-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)

22 ( xp )

( 1)( 1)

Observed E ected

Expected

df r c

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

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

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= 2Ha: 1 2

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

G rou p 1D isease

M ean ag e o f th e cases

G rou p 2N o D isease

M ean ag e o f th e con tro ls

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