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Text books : Medical Statistics A commonsense approach By Michael J. Campbell & David Machin (2) An Introduction to Medical Statistics By Martin Bland (3) Biostatistics: A Foundation for Analysis in the Health Sciences

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Presentation Transcript
slide1
Text books:
  • Medical Statistics

A commonsense approach

By

Michael J. Campbell & David Machin

(2) An Introduction to Medical Statistics

By

Martin Bland

(3) Biostatistics: A Foundation for Analysis in the

Health Sciences

By

Wayne W. Daniel

the chi square test

THE CHI-SQUARE TEST

BACKGROUND AND NEED OF THE TEST

Data collected in the field of medicine is often qualitative.

--- For example, the presence or absence of a symptom, classification of pregnancy as ‘high risk’ or ‘non-high risk’, the degree of severity of a disease (mild, moderate, severe)

slide3
The measure computed in each instance is a proportion, corresponding to the mean in the case of quantitative data such as height, weight, BMI, serum cholesterol.

Comparison between two or more proportions, and the test of significance employed for such purposes is called the “Chi-square test”

slide4
---- KARL PEARSON IN 1889, DEVISED AN INDEX OF DISPERSION OR TEST CRITERIOR DENOTED AS “CHI-SQUARE “. (X 2 ).

The formula for X 2 –test is,

chi square

[

]

S

( o - e) 2

e

Chi- Square

X 2 =

Figure for Each Cell

slide6

2 = ∑

^

total of column in

which the cell lies

3. E is the expected frequency

(O - E)2

E

total of row in

which the cell lies

reject Ho if 2 > 2.,df

where df = (r-1)(c-1)

^

E =

(total of all cells)

1. The summation is over all cells of the contingency table consisting of r rows and c columns

2. O is the observed frequency

4. The degrees of freedom are df = (r-1)(c-1)

application of chi square test
APPLICATION OF CHI-SQUARE TEST
  • TESTING INDEPENDCNE (OR ASSOCATION)
  • TESTING FOR HOMOGENEITY
  • TESTING OF GOODNESS-OF-FIT
chi square test
Chi-square test

Purpose

To find out whether the association between two categorical variables are statistically significant

Null Hypothesis

There is no association between two variables

chi square test1
Chi-square test

Test statistics

1.

2.

3.

4.

chi square test2
Chi-square test
  • Objective : Smoking is a risk factor for MI
  • Null Hypothesis:Smoking does not cause MI
chi square1

29

21

O

O

E

E

16

34

O

O

E

E

Chi-Square

MI

Non-MI

Smoker

Non-Smoker

chi square2

29

21

O

O

E

E

16

34

O

O

E

E

Chi-Square

MI

Non-MI

50

Smoker

50

Non-smoker

45

55

100

slide13

^

E =

R2C3

n

Classification 1

1

2

3

4

c

Classification 2

1

R1

2

R2

3

R3

Estimating the Expected Frequencies

r

Rr

^

(row total for this cell)•(column total for this cell)

n

E =

C1

C2

C3

C4

Cc

chi square3

21

O

E

16

34

O

O

E

E

Chi-Square

MI

Non-MI

50

29

Smoker

50 X 45

100

O

22.5 =

22.5

E

50

Non-smoker

45

55

100

chi square4

21

O

E

16

34

O

O

E

E

Chi-Square

Alone

Others

50

29

Males

O

22.5

27.5

E

50

Females

22.5

27.5

45

55

100

chi square5
Chi-Square
  • Degrees of Freedomdf = (R-1) (C-1)
  • Critical Value (Table A.6) = 3.84
  • X2 = 6.84
slide18

Age

Gender <30 30-45 >45 Total

Male 60 (60) 20 (30) 40 (30) 120

Female 40 (40) 30 (20) 10 (20) 80

Total 100 50 50 200

Chi- square test

Find out whether the sex is equally distributed among each age group

test for homogeneity similarity
Test for Homogeneity (Similarity)

To test similarity between frequency distribution or group. It is used in assessing the similarity between non-responders and responders in any survey