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CHI-SQUARE( X 2 ) DISTRIBUTION Chi-Square Test. CHI-SQUARE( X 2 ) DISTRIBUTION PROPERTIES: 1.It is one of the most widely used distribution in statistical.

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Presentation on theme: "CHI-SQUARE( X 2 ) DISTRIBUTION Chi-Square Test. CHI-SQUARE( X 2 ) DISTRIBUTION PROPERTIES: 1.It is one of the most widely used distribution in statistical."— Presentation transcript:

1 CHI-SQUARE( X 2 ) DISTRIBUTION Chi-Square Test

2 CHI-SQUARE( X 2 ) 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)

3 CHI-SQUARE( X 2 ) DISTRIBUTION 4. X 2 relates to frequencies of occurrence of individuals (or events) in the categories of one or more variables. 5. X 2 test used to test the agreement between the observed frequencies with certain characteristics and the expected frequencies under certain hypothesis.

4 CHI-SQUARE( X 2 ) DISTRIBUTION CHI-SQUARE( X 2 ) test of Goodness of fit CHI-SQUARE( X 2 ) test of homogeneity CHI-SQUARE( X 2 ) test of Independence

5 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)

6 CHI-SQUARE(X 2 ) test of Independence Generally, a single sample of size (n) can be drawn from a population, the frequency of occurrence of the entities are cross- classified 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’

7 CHI-SQUARE(X 2 ) 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.

8 CHI-SQUARE(X 2 ) test of Independence X 2 =∑(O-E) 2 /E df=(r-1)(c-1) For 2x2 table, another formula to calculate X 2 n(ad-bc) 2 X 2 =-------------------------------- (a+c)(b+d)(a+b)(c+d)

9 Steps in constructing X 2 -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

10 Steps in constructing X 2 -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

11 Steps in constructing X 2 -test 4. Calculated the X 2 value (calculated X 2 c) X 2 =∑(O-E) 2 /E For each cell we will calculate X 2 value X 2 value for all the cells of the contingency table will be added together to find X 2 c

12 Steps in constructing X 2 -test 5. Define the critical value (tabulated X 2 ) This depends on alpha level of significance and degree of freedom The value will be determined from X 2 table df=(r-1)(c-1) r: no. of row c: no. of column

13 Steps in constructing X 2 -test 6. Conclusion If the X 2 c is less than X 2 tab we accept Ho. If the X 2 c is more than X 2 tab we reject Ho.

14 Observed frequencies in a fourfold table Y1Y2Total row total X1aba+b X2cdc+d Total column total a+cb+dN=a+b+c+d

15 For r X c table X 2 –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

16 For 2 X 2 table X 2 –test is not applicable if: The expected frequency of any cell is <5

17 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

18 EXERCISE malefemaleTotal On diet142539 Not on diet159152311 Total173177350

19 EXERCISE At alpha =0.05 do these data suggest an association between sex and being on diet?

20 ANSWER 1.Ho: Being on diet and sex are independent ( no association) HA: Being on diet and sex are not independent ( there is association)

21 2. Calculation of expected frequencies 173 x 39 Cell a =-------------=19.3 350 177 x 39 Cell b=--------------=19.7 350

22 2. Calculation of expected frequencies 173 x 311 Cell c =-------------=153.7 350 177 x 311 Cell d=--------------=157.3 350

23 Observed and (expected) frequencies malefemaleTotal On diet14 (19.3) 25 (19.7) 39 Not on diet159 (153.7) 152 (157.3) 311 Total173177350

24 ANSWER 3. Calculate X 2 : X 2 =∑(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 X 2 c =3.243

25 ANSWER 4. Find X 2 tab df= (r-1) (c-1)= (2-1)(2-1)=1 X 2 0.95 df=1 =3.841

26 ANSWER 5. Conclusion Since X 2 c < X 2 tab we accept Ho ( No association between sex and being on diet)

27 Another solution Since this a 2x2 table we can use this formula: n(ad-bc) 2 X 2 =-------------------------------- (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

28 (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

29 Speech defect Socioeconomic Group UpperUpper middle Lower Middle LowerTotal Present8 (9.1) 24 (26.4) 32 (30.9) 27 (24.6) 91 Absent42 (40.9) 121 (118.6) 138 (139.1) 108 (110.4) 409 Total50145170135500

30 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)

31 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

32 ( 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:

33 Symptom Level of exposure HighLimitedNo exposure Total Present 185 (143.4) 33 (49.8) 17 (41.8) 235 Absent 120 (161.6) 73 (56.2) 72 (47.2) 265 Total 30510689500

34 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

35 X² =∑ (185 – 143.4)²/143.4 + (33 – 49.8)²/49.8 + (17-41.8)²/41.8 + (120- 161.6)²/161.6 + (73 -56.2)² /56.2 + (72- 47.2)²/47.2 = 33.47 Tab X² 0.99 = 9.21 Reject Ho Df = (3-1) (2-1) = 2

36 (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.

37 Out come Treatment DrugPlaceboTotal Favorable34943 Unfavorable164157 Total50 100

38 X² = n (ad – bc)²/(a+b)(c+d)(a+c)(b+d) = 100[(34x41) – (9x16)]²/(50) (50)(43)(57) =25.5

39 (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.

40 Hepatitis A. Baby MongoloidNon- Mongoloid Total +263460 -419361940 Total3019702000


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