chi sqyre test.pptx

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CHI-SQUARE TEST

• M.Prasad Naidu
• MSc Medical Biochemistry, Ph.D,.

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CHI-SQUARE TEST

• It is a non parametric test , not based on any
  assumption or distribution of any variable.
• It is very useful in research and it is most
  commonly used when the data are given in
  frequencies .
• It can be used with any data which can be reduced
  to proportion or percentages.
• This test involves calculation of quantity called
  chi- square , which derived from Greek letter (
  ?)² and pounced as kye. Cont

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• It is an alternate test to find the significance
  of difference in two or more than two
  proportions.
• Chi square test is yet another useful test which
  can be applied to find the significance in the
  same type of data with the following advantages .
  
• cont

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• To compare the values of two binomial samples
  even if they are small such as incidence of
  diabetes in 20 obese persons with 20 non obese
  persons.
• To compare the frequencies of two multinomial
  samples such as number of diabetes and non
  diabetes in groups weighing 40-50, 50-60, 60-70,
  and gt 70 kg s of weight.

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4.TEST OF ASSOCIATION

• Test of association between events in binomial
  or multinomial samples .
• Two events often can be tested for their
  association such as cancer and smoking .
• Treatment and outcome of disease .
• Vaccination and immunity .
• Nutrition and intelligence .
• Cholesterol and heart disease .
• Weight and diabetes, B P and heart disease.
• To find they are independent of each other or
  dependent on each other i.e. associated .

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5.Caliculation of chi-square value

• Three essentials to apply chi - square test are
• A random sample
• Qualitative data
• Lowest frequency not less than 5
• Steps -
• Assumption of Null Hypothesis (HO)
• Prepare a contingency table and note down the
  observed frequencies or data (O)
• 
  cont

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• Determine the expected number (E) by multiplying
  CT RT /GT (column total ,row total and grand
  total )
• Find the difference between observed and expected
  frequencies in each cell (O-E)
• Calculate chi- square value for each cell with
• (O-E)²/E
• Sum up all chi square values to get the total
  chi-square value (?)² d.f. (degrees of freedom)
• ?² ?(O-E)²/E and d.f. is (c-1) (r-1)

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7.Restractions in Application of (?)²

• Chi- square test applied in a four fold table
  will not give reliable result , with one degree
  of freedom.
• If the observed value of any cell is lt 5 in such
  cases Yates correction can be applied by
  subtracting ½ (?)² ? (O-E-1/2)²/E
• Even with Yates correction the test may be
  misleading if any expected value is much below 5
  in such cases Yates correction can not be
  applied.
• 
  cont

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• In tables larger than 2 2 Yates correction can
  not be applied .
• The highest value of chi- square ?² obtainable by
  chance or worked out and given in (?)² table at
  different degrees of freedom under probabilities
  (P) such as 0.05, 0.01, 0.001 .
• If calculated value of chi-square of the sample
  is found to be higher than the expected value of
  the table at critical level of significance i.e.
  probability of 0.05 the H O of no difference
  between two proportions or the H O of
  independence of two characters is rejected.
• If the calculated value is lower the hypothesis
  of no difference is accepted.

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9.Exercise with out come results

• Groups Died Survived
  Total
• A (a) 10 (b)
  25 35
• B (c) 5 (d)
  60 65
• Total 15
  85 100
• Expected value can be computed with CTRT/GT for
  (a) 1535/100 5.25
• (b) 8535/100 29.75
• (c) 1565/100 9.75
  
• (d) 8565/100 55.25
  

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10.Caliculation of Chi-square value

• ?² ?(O-E)²/E d.f. ( c-1) (r-1)
• (A) (10-5.25)²/5.25 (4.75)²/5
  4.30
• (B) (25-29.75)²/29.75 (5)²/29.75 0
  .76
• (C) (5-9.75)²/9.75 (5)²/9.75
  2.31
• (D) (60-55.25)²/55.25 (5)²/55.25
  0.40
• 
  7.77
• The calculated chi- square value is 7.77 is more
  than
• Chi- square table value with 1 d.f. At 0.01,
  6.64 which significant. It can be said that
  there is significant difference between two
  groups.

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12.Method -2

• Chi- square can be calculated in the following
  way also.
• (?)² ( ad-bc)²N
• (a b c d ac b d) (600-125
  )²
• 
  ( 35 65 15 85)
• (475)² x100 225625 00
  /2900625
• (2900625) 7.77
  
• Here the calculated chi-square value7.77 is gt
  6.64 at
• 0.01 with 1 d.f. shows significant difference b/n
  groups. So P is lt0.01
  

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