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# Chi sqyre test

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• 1. M.Prasad Naidu MSc Medical Biochemistry, Ph.D,.
• 2. &#xF09E; It is a non parametric test , not based on any assumption or distribution of any variable. &#xF09E; It is very useful in research and it is most commonly used when the data are given in frequencies . &#xF09E; It can be used with any data which can be reduced to proportion or percentages. &#xF09E; This test involves calculation of quantity called chi- square , which derived from Greek letter ( &#x3C7;)&#xB2; and pounced as &#x2018;kye&#x2019;. Cont&#x2026;
• 3. &#xF09E;It is an alternate test to find the significance of difference in two or more than two proportions. &#xF09E;Chi &#x2013;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 . &#xF09E; cont&#x2026;
• 4. &#xF09E;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. &#xF09E;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.
• 5. &#xF09E; Test of association between events in binomial or multinomial samples . &#xF09E; Two events often can be tested for their association such as cancer and smoking . &#xF09E; Treatment and outcome of disease . &#xF09E; Vaccination and immunity . &#xF09E; Nutrition and intelligence . &#xF09E; Cholesterol and heart disease . &#xF09E; Weight and diabetes, B P and heart disease. &#xF09E; To find they are independent of each other or dependent on each other i.e. associated .
• 6. &#xF09E; Three essentials to apply chi - square test are &#xF09E; A random sample &#xF09E; Qualitative data &#xF09E; Lowest frequency not less than 5 &#xF09E; Steps :- &#xF09E; Assumption of Null Hypothesis (HO) &#xF09E; Prepare a contingency table and note down the observed frequencies or data (O) &#xF09E; cont&#x2026;
• 7. &#xF09E; Determine the expected number (E) by multiplying CT &#xD7; RT /GT (column total ,row total and grand total ) &#xF09E; Find the difference between observed and expected frequencies in each cell (O-E) &#xF09E; Calculate chi- square value for each cell with &#xF09E; (O-E)&#xB2;/E &#xF09E; Sum up all chi &#x2013;square values to get the total chi-square value (&#x3C7;)&#xB2; d.f. (degrees of freedom) &#xF09E; = &#x3C7;&#xB2;= &#x2211;(O-E)&#xB2;/E and d.f. is (c-1) (r-1)
• 8. &#xF09E; Chi- square test applied in a four fold table will not give reliable result , with one degree of freedom. &#xF09E; If the observed value of any cell is &lt; 5 in such cases Yates correction can be applied by subtracting &#xBD; (&#x3C7;)&#xB2; = &#x2211; (O-E-1/2)&#xB2;/E &#xF09E; 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. &#xF09E; cont&#x2026;
• 9. &#xF09E; In tables larger than 2 &#xD7; 2 Yates correction can not be applied . &#xF09E; The highest value of chi- square &#x3C7;&#xB2; obtainable by chance or worked out and given in (&#x3C7;)&#xB2; table at different degrees of freedom under probabilities (P) such as 0.05, 0.01, 0.001 . &#xF09E; 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. &#xF09E; If the calculated value is lower the hypothesis of no difference is accepted.
• 10. &#xF09E; Groups Died Survived Total &#xF09E; A (a) 10 (b) 25 35 &#xF09E; B (c) 5 (d) 60 65 &#xF09E; Total 15 85 100 &#xF09E; Expected value can be computed with CT&#xD7;RT/GT for (a) 15&#xD7;35/100= 5.25 &#xF09E; (b) 85&#xD7;35/100= 29.75 &#xF09E; (c) 15&#xD7;65/100= 9.75 &#xF09E; (d) 85&#xD7;65/100= 55.25
• 11. &#xF09E; &#x3C7;&#xB2;= &#x2211;(O-E)&#xB2;/E d.f. =( c-1) (r-1) &#xF09E; &#xF09E; (A) =(10-5.25)&#xB2;/5.25 = (4.75)&#xB2;/5 = 4.30 &#xF09E; (B) = (25-29.75)&#xB2;/29.75 =(5)&#xB2;/29.75 = 0 .76 &#xF09E; (C) = (5-9.75)&#xB2;/9.75 = (5)&#xB2;/9.75 = 2.31 &#xF09E; (D) = (60-55.25)&#xB2;/55.25 = (5)&#xB2;/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. &#xF09E; Chi- square can be calculated in the following way also. &#xF09E; (&#x3C7;)&#xB2; = ( ad-bc)&#xB2;&#xD7;N (a b &#xD7; c d &#xD7;ac &#xD7;b d) = (600-125 )&#xB2; ( 35&#xD7; 65&#xD7; 15&#xD7; 85) = (475)&#xB2; 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