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CHI-SQUARE TEST

BIOCHEMISTRY

mprasadnaidu
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CHI-SQUARE TEST

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  1. CHI-SQUARE TEST M.Prasad Naidu MSc Medical Biochemistry, Ph.D,.

  2. 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 ( χ)² and pounced as ‘kye’. Cont…

  3. 2 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…

  4. 3 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 > 70 kg s of weight.

  5. 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 .

  6. 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…

  7. 6 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 (χ)² d.f. (degrees of freedom) = χ²= ∑(O-E)²/E and d.f. is (c-1) (r-1)

  8. 7.Restractions in Application of (χ)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 < 5 in such cases Yates correction can be applied by subtracting ½ (χ)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…

  9. 8 In tables larger than 2 × 2 Yates correction can not be applied . The highest value of chi- square χ² obtainable by chance or worked out and given in (χ)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.

  10. 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 CT×RT/GT for (a) 15×35/100= 5.25 (b) 85×35/100= 29.75 (c) 15×65/100= 9.75 (d) 85×65/100= 55.25

  11. 10.Caliculation of Chi-square value χ²= ∑(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.

  12. 12.Method -2 Chi- square can be calculated in the following way also. (χ)² = ( ad-bc)²×N (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 > 6.64 at 0.01 with 1 d.f. shows significant difference b/n groups. So P is <0.01

  13. THANK YOU

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