Chi-Square

Test of significance for proportions

FETP India

Competency to be gained from this lecture

Test the statistical significance of proportions using the relevant

Chi-square test

Key elements

• Principles of the Chi-square• Comparison of a proportion with an hypothesized value

• Chi-square for 2x2 tables• Chi-square for m x n tables• Testing dose-response with Chi-square

Analyzing quantitative and qualitative data

Quantitative data Qualitative

dataNormal Non-normal

Summary statistics

•Mean •Median •Proportions

Statistical tests

•T-test•F-test

•Non-parametric test

•Chi-square•Fisher exact test

Chi-square: Principle

• The Chi–square test examines whether a series of observed (O) numbers in various categories are consistent with the numbers expected (E) in those categories on some specific hypothesis (Null hypothesis)

O= Observed value E= Expected value

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χ 2 =∑ (O−E)2

E≥0

Principle

How the Chi-square works in practice

• X2 = 0 when every observed value is equal to the expected value

• As soon as an observed value differs from the expected value, the X

2 exceeds zero

• The value of the X2 is compared with a

tabulated value

• If the calculated value of X2 exceeds

the tabulated value under the column p = 0.05, the null hypothesis is rejected

Principle

Chi-square table:Percentage points of X

2 distribution

Principle

Use of Chi-square to compare a proportion with a hypothesized

value• The reported coverage for measles in a sub-center is 80%

• The chief medical officer of the district suspects that this coverage could be overestimated

• Validation survey with 80 children selected using simple random sampling 56/80 (70%) vaccinated

Hypothesized value

Chi-square calculations

VaccinatedNon-

vaccinated Total

Expected 64 16 80

Observed 56 24 80

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χ 2 =(56 −64)2

64+(24 −16)2

16=(−8)2

64+(8)2

16=6464

+6416

=1+ 4 =5

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χ 2 =∑ (O−E)2

E≥0

Interpretation of the Chi-square

• The calculated value of X2 (i.e.,

5) with 1 degree of freedom exceeds the table value (3.84) at 5% level

• Hence, the medical officer rejects the null hypothesis that the coverage is 80%

Hypothesized value

Use of Chi-square to compare proportions between two

samples• Cholera outbreak affecting a village• Cases clustered around a pond• Hypothesis generating interviews suggest that many case-patients washed their utensils in the pond

• The investigator compares those who washed their utensils with the others in terms of cholera incidence

2x2 tables

Incidence of diarrhea (cholera) among persons who

washed utensils in a pond and others, South 24 Parganas, West Bengal, India, 2006

2x2 tables

Chi-square to test the difference

in the two proportions• The proportion of persons affected by cholera in the exposed and unexposed groups differ

• Three steps to test whether this difference is significant:1. Calculate expected values2. Compare observed and expected to

calculate the Chi-square3. Compare the Chi-square with tabulated

value

2x2 tables

Step 1: Calculate the expected values (1/2)

•51% of the population became sick•If the cholera occurred at random, these

proportions apply to the two groups, exposed and unexposed 2x2 tables

Step 1: Calculate the expected values (2/2)

•51% of the 56 who washed utensils (=28) should have been sick•All other numbers can de deducted by subtraction (one degree of freedom)

2x2 tables

Step 2: Compare observed and expected values

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χ 2 =(50 −28)2

28+(6 −28)2

28+(8 −30)2

30+(49 −27)2

27=(22)2

28+(−22)2

28+(−22)2

30+(22)2

27=68.6

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χ 2 =∑ (O−E)2

E

2x2 tables

Step 3: Interpretation of the Chi-square

• The calculated value of X2 (i.e., 68.6)

with 1 degree of freedom exceeds the table value (3.84) at 5% level

• Hence, we reject the Null hypothesis that the attack rate of cholera is equal in the exposed and unexposed group

• This may suggest washing utensils in the pond is a source of infection if other elements of the investigation also support the hypothesis

2x2 tables

Simpler Chi-square formula

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χ 2 =N(ad−bc)2

(a+ c)(b+ d)(a+ b)(c+ d)

2x2 tables

Application of simpler Chi-square formula to the cholera

example

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χ 2 =113(50x49 −6x8)2

(50 + 8)(6 + 49)(50 + 6)(8 + 49)=64

2x2 tables

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χ 2 =N(ad−bc)2

(a+ c)(b+ d)(a+ b)(c+ d)

Corrected Chi-square formula

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χ 2 =N{(ad−bc)−N/ 2}2

(a+ c)(b+ d)(a+ b)(c+ d)

Note that the corrected value will always be smaller than the uncorrected which tends to exaggerate

the significance of a difference2x2 tables

Example of a 4x2 table

Cataract

Present Absent Total

Hindu 10 90 100

Muslim 4 46 50

Christian 3 22 25

Others 1 9 10

Total 18 167 185

Degrees of freedom = (Nbr of rows-1) x (Nbr of columns-1)=(4-1)x(2-1)= 3x1=3 N x N tables

Calculation of the Chi-square for a 4x2 table

Cataract

Present Absent Total

Hindu 10 90 100

Muslim 4 46 50

Christian 3 22 25

Others 1 9 10

Total 18 167 185

•Overall prevalence of cataract = 9.6%•Apply 9.6% proportion to all groups to calculate expected values•Use generic formula

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χ 2 =∑ (O−E)2

EN x N tables

Interpretation of a Chi-square

for a m x n tableThe Chi-square tests the overall Null hypothesis that all frequencies are distributed at random

• If the Null hypothesis is rejected, it means the distribution is heterogeneous • It is not possible to:

✗“Attribute” the difference to a particular group✗ Regroup categories according to differences observed and test with a 2x2 table (i.e., post-hoc analysis)

✗ Test with multiple 2x2 tables (i.e., multiple comparisons)

N x N tables

Key rule about Chi-square

• Chi- square test should be applied on qualitative data set out in the form of frequencies

• Chi– square test should not be done on: Percentages Rates Ratios Mean values

N x N tables

Limitations to the use of Chi-square

• When sample size is small, other exact tests (e.g., Fisher exact test) are preferred and calculated with computer N < 30 Expected value < 5

• When several expected cell frequencies are less than one, it is better to amalgamate rows / columns

N x N tables

Testing a dose-response relationship

with a Chi-square• Overall m x n Chi-square

Tests the null hypothesis that the odds ratios do not differ

No particular conditions needed Overall test, easy to compute Rough conclusions

• Chi-square for trend

Dose-reponse

Exposure to injections with reusable needles and acute

hepatitis B, Thiruvananthapuram, Kerala,

India, 1992 Potential risk factors

Cases (N=160

)Controls (N=160)

Odds ratio

95% confidence interval

None 51 120 1.0 N/A 

Single injection 41 25 3.9 2.0-7.3

Multiple injections

29 7 9.8 3.8-26

Overall 3x2 Chi-square : 42, 2 degrees of freedom, p<0.00001

Heterogeneous

exposures categories

Dose-reponse

Testing a dose-response relationship

with a Chi-square• Overall m x n Chi-square• Chi-square for trend

Tests for a linear trend for the increase of the odds ratios with increased levels of exposure

Requires equal interval in the exposure categories

Can be calculated on a computer Refined conclusions

Dose-reponse

Odds of typhoid according to raw onion consumption,

Darjeeling, West Bengal, India, 2005-2006

Weekly raw onion servings

Cases (n=123)

Controls(n=123)

Odds ratio

# % # % OR 95% CI

1-2 19 23 28 34 1 N/A

3-4 25 31 27 61 1.4 0.6 – 3.3

5+ 38 46 8 5 7 2.5 – 21

Chi-square for trend: 16.8; P-value: 0.0004

Homogeneous

exposures categories

Dose-reponse

Take home messages

• The Chi-square compares expected and observed counts

• The Chi-square can compare an actual proportion with a theoretical one

• The Chi-square is the basic test to compare proportions in epidemiological 2x2 tables

• The Chi-square can be used also as a global test for m x n tables

• Specific Chi-squares can be used to test for dose-response effect