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Chi squared test
1. CHI-SQUARED
TEST
Dr Lipilekha Patnaik
Professor, Community Medicine
Institute of Medical Sciences & SUM Hospital
Siksha ‘O’Anusandhan deemed to be University
Bhubaneswar, Odisha, India
Email: drlipilekha@yahoo.co.in
2. Pearson’s chi-squared test
• Chi- Square test is method of testing the significance of difference
between two proportions.
• It has the advantage that at can also be used when more than two
groups are to be compared.
• Commonly applied in biomedical research
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3. • Used to test categorical Data
Especially when the outcomes are in counts or frequency
• Nominal variables
Sex, Blood group
• Ordinal Variables
Birth order
Severity of disease (absent, mild, moderate, severe)
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4. Applications of chi-squared test
• This test is used as
• for finding difference in proportions
• for finding association between variables
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5. Example
• In a prevalence study of Hypertension, we found that
Hypertension No Hypertension
Non smokers 10 (10%) 90
Smokers 26 (26%) 74
• It is visible from the table that the proportion of HTN was higher
among smokers . The question that arises is whether HTN was really
higher among smokers or the difference was merely due to chance.
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7. Steps
• Null hypothesis
• Calculation of expected values
• Calculation of Chi-square value
• Calculation of degrees of freedom
• Conclusion
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8. Null hypothesis
• No significant difference in the proportions
• Alternate- significant difference in the proportions
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9. Calculation of expected values
• Expected value in any cell =
totalGrand
totalRowXtotalColumn
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Expected value of each cell should be more than 5 in
each cell.
10. Expected values in cells
Thus expected value in each cell is
• First cell: 36 X 100 / 200 = 18
• Second cell: 164 X 100 / 200 = 82
• Third cell: 36 X 100 / 200 = 18
• Fourth cell: 164 X 100 / 200 = 82
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11. Chi squared value
∑
−
=
E
EO 2
2 )(
χ
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df = (c -1) (r-1)
where c and r stand for number of columns and
number of rows respectively.
P < 0.05 – Significant difference between 2 proportions
P > 0.05 – No significant difference between 2
proportions
13. Calculation of degrees of freedom
df = (c -1) (r-1)
where c and r stand for number of columns and number of rows
respectively.
Thus df = (2 -1) X (2 – 1) = (1) X (1) = 1.
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15. Conclusion
• As the Chi-square value is more than the table value for 5%
level of significance at 1 df (3.84), we reject the null hypothesis
and state that there is a statistically significant difference of
proportions of hypertension between smokers and nonsmokers
(P < 0.05).
• Smoking is associated with hypertension.
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