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2 CHI-SQUARE(X ) DISTRIBUTION Chi-Square Test CHI-SQUARE(X2) DISTRIBUTION • PROPERTIES: 1.It is one of the most widely used distribution in statistical applications 2.This distribution may be derived from normal distribution 3.This distribution assumes values from ( zero to + infinity) CHI-SQUARE(X2) DISTRIBUTION 4. X2 relates to frequencies of occurrence of individuals (or events) in the categories of one or more variables. 5. X2 test used to test the agreement between the observed frequencies with certain characteristics and the expected frequencies under certain hypothesis. CHI-SQUARE(X2) DISTRIBUTION • CHI-SQUARE(X2) test of Goodness of fit • CHI-SQUARE(X2) test of homogeneity • CHI-SQUARE(X2) test of Independence CHI-SQUARE(X2) test of Independence • It is used to test the null hypothesis that two criteria of classification when applied to the same set of entities are independent (NO ASSOCIATION) CHI-SQUARE(X2) test of Independence • Generally , a single sample of size (n) can be drawn from a population, the frequency of occurrence of the entities are crossclassified on the basis of the two variables of interest( X &Y). The corresponding cells are formed by the intersections of the rows (r), and the columns (c). The table is called the ‘contingency table’ CHI-SQUARE(X2) test of Independence • Calculation of expected frequency is based on the Probability Theory • The hypotheses and conclusions are stated on in terms of the independence or lack of independence of the two variables. CHI-SQUARE(X2) test of Independence • X2=∑(O-E)2/E • df=(r-1)(c-1) For 2x2 table, another formula to calculate X2 n(ad-bc)2 X2 =-------------------------------(a+c)(b+d)(a+b)(c+d) Steps in constructing X2 -test 1. Hypotheses Ho: the 2 criteria are independent (no association) HA: The 2 criteria are not independent (There is association) 2. Construct the contingency table Steps in constructing X2 -test 3. Calculate the expected frequency for each cell By multiplying the corresponding marginal totals of that cell, and divide it by the sample size ∑E = ∑O for each row or column Steps in constructing X2 -test 4. Calculated the X2 value (calculated X2 c) X2=∑(O-E)2/E X2=∑(O-E)2/E For each cell we will calculate X2 value X2 value for all the cells of the contingency table will be added together to find X2 c Steps in constructing X2 -test 5. Define the critical value (tabulated X2) This depends on alpha level of significance and degree of freedom The value will be determined from X2 table df=(r-1)(c-1) r: no. of row c: no. of column Steps in constructing X2 -test 6. Conclusion If the X2 c is less than X2 tab we accept Ho. If the X2 c is more than X2 tab we reject Ho. Observed frequencies in a fourfold table Y1 Y2 Total row total X1 a b a+b X2 c d c+d Total column total a+c b+d N=a+b+c+d For r X c table X2 –test is not applicable if: 1. The expected frequency of any cell is <1 2. The expected frequencies of 20% of the cells is < 5 For 2 X 2 table X2 –test is not applicable if: The expected frequency of any cell is <5 EXERCISE • A group of 350 adults who participated in a health survey were asked whether or not they were on a diet. The response by sex are as follows EXERCISE male female Total On diet 14 25 39 Not on diet 159 152 311 Total 173 177 350 EXERCISE • At alpha =0.05 do these data suggest an association between sex and being on diet? ANSWER 1. Ho: Being on diet and sex are independent ( no association) HA: Being on diet and sex are not independent ( there is association) 2. Calculation of expected frequencies 173 x 39 Cell a =-------------=19.3 350 177 x 39 Cell b=--------------=19.7 350 2. Calculation of expected frequencies 173 x 311 Cell c =-------------=153.7 350 177 x 311 Cell d=--------------=157.3 350 Observed and (expected) frequencies male female Total On diet 14 (19.3) 25 (19.7) 39 Not on diet 159 (153.7) 152 (157.3) 311 Total 173 177 350 ANSWER 3. Calculate X2 : X2=∑(O-E)2/E (14-19.3)2 (25-19.7)2 (159-153.7)2 (152-157.3)2 =-----------+-----------+--------------+------------19.3 19.7 153.7 157.3 =1.455+1.426+0.183+0.17 X2c =3.243 ANSWER 4. Find X2 tab df= (r-1) (c-1)= (2-1)(2-1)=1 X20.95 df=1=3.841 ANSWER 5. Conclusion Since X2 c < X2 tab we accept Ho ( No association between sex and being on diet) Another solution • Since this a 2x2 table we can use this formula: n(ad-bc)2 X2 =-------------------------------(a+c)(b+d)(a+b)(c+d) 350{(14 x 152)-(25 x 159)}2 =------------------------------------- =3.22 39 x 311 x 173 x 177 (Example) Five hundred elementary school children were cross classified by socioeconomic group and the presence or absence of a certain speech defect. The result were as follows Speech defect Socioeconomic Group Upper Present 8 (9.1) Upper Lower Lower Total middle Middle 24 32 27 91 (26.4) (30.9) (24.6) Absent 42 (40.9) 121 138 108 (118.6) (139.1) (110.4) Total 50 145 170 135 409 500 • Are these data compatible with the hypothesis that the speech defect is unrelated to socioeconomic status? • 1) Ho :Speech defect and SE group are independent ( no Association) • HA: Speech defect and SE group are not independent ( Association exist) • 2)Calculate the expected frequencies • 3)Calculate the X2 value ( calculated value) • X² = ∑ (0 –E)² / E • • X² = ∑ (8 – 9.1)² /9.1 + (24 – 26.4)²/26.4 + (32 – 30.9)² /30.9 + (27-24.6)² /34.6 + (121 – 118.6)²/118.6 + (138 139.1)²/139.1 + (108 – 110.4)²/110.4 X²=0.5 • • Tab X² DF = (2-1) (4-1) =3 → X²0.95 = 7.815 (Example 2) • Five hundred employees of a factory that manufacture a product suspected of being associated with respiratory disorders were cross classified by level of exposure to the product and weather or not they exhibited symptoms of respiratory disorders. The results are shown in following table: Symptom Present Absent Total Level of exposure High Limited 185 (143.4) 120 (161.6) 305 33 (49.8) 73 (56.2) 106 No Total exposure 17 (41.8) 72 (47.2) 89 235 265 500 • Do these data provide sufficient evidence, at the 0.01 level of significance to indicate a relationship between level of exposure and the presence of respiratory and the presence of respiratory disorder ? • 1) Ho : The presence of respiratory symptoms and the level of exposure are independent. • HA : The two criteria are not independent • 2)Calculate the expected frequencies • 3) Calculate the X2 • X² =∑ (185 – 143.4)²/143.4 + (33 – 49.8)²/49.8 + (17-41.8)²/41.8 + (120161.6)²/161.6 + (73 -56.2)² /56.2 + (7247.2)²/47.2 = 33.47 Tab X² 0.99 = 9.21 Reject Ho • Df = (3-1) (2-1) = 2 (Example 3) • In a clinical trial involving a potential hypothesis drug, patients are assigned at random either to receive the active drug or placebo. The trial is double blind, that is neither the patient nor the examining physician knows with of the 2 treatment the patient is receiving. Patients response to treatment is categorized as favorable or unfavorable on the basis of degree and duration of response in BP. There are 50 patients assigned to each group. Treatment Out come Drug Placebo Total Favorable 34 9 43 Unfavorable 16 41 57 Total 50 50 100 • X² = n (ad – bc)²/(a+b)(c+d)(a+c)(b+d) • = 100[(34x41) – (9x16)]²/(50) (50)(43)(57) • =25.5 (Example 4) • A study found that mongolism in babies is associated with hepatitis A injection of the mother during pregnancy. Suppose a study of 2000 randomly selected mothers to be yielded the following table after the births of their babies. Hepatitis A. Baby Mongoloid NonMongoloid Total + 26 34 60 - 4 1936 1940 Total 30 1970 2000