2 CHI-SQUARE(X2) DISTRIBUTION PROPERTIES:1.It is one of the most widely used distribution in statistical applications2.This distribution may be derived from normal distribution3.This distribution assumes values from( zero to + infinity)
3 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.
4 CHI-SQUARE(X2) DISTRIBUTION CHI-SQUARE(X2) test of Goodness of fitCHI-SQUARE(X2) test of homogeneityCHI-SQUARE(X2) test of Independence
5 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)
6 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 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(X2) test of Independence Calculation of expected frequency is based on the Probability TheoryThe hypotheses and conclusions are stated on in terms of the independence or lack of independence of the two variables.
8 CHI-SQUARE(X2) test of Independence X2=∑(O-E)2/Edf=(r-1)(c-1)For 2x2 table, another formula to calculate X2n(ad-bc)2X2 =(a+c)(b+d)(a+b)(c+d)
9 Steps in constructing X2 -test HypothesesHo: 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 X2 -test 3. Calculate the expected frequency for each cellBy 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 X2 -test 4. Calculated the X2 value (calculated X2 c)X2=∑(O-E)2/E X2=∑(O-E)2/EFor each cell we will calculate X2 valueX2 value for all the cells of the contingency table will be added together to find X2 c
12 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 tabledf=(r-1)(c-1)r: no. of rowc: no. of column
13 Steps in constructing X2 -test 6. ConclusionIf the X2 c is less than X2 tab we accept Ho.If the X2 c is more than X2 tab we reject Ho.
14 Observed frequencies in a fourfold table Y1Y2Totalrow totalX1aba+bX2cdc+dcolumn totala+cb+dN=a+b+c+d
15 For r X c table X2 –test is not applicable if: The expected frequency of any cell is <1The expected frequencies of 20% of the cells is < 5
16 For 2 X 2 table X2 –test is not applicable if: The expected frequency of any cell is <5
17 EXERCISEA 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 male female Total On diet 14 25 39 Not on diet 159 152 311 173177350
19 EXERCISEAt alpha =0.05 do these data suggest an association between sex and being on diet?
20 ANSWER 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 Cell a = =19.3350177 x 39Cell b= =19.7
22 2. Calculation of expected frequencies Cell c = =153.7350177 x 311Cell d= =157.3
23 Observed and (expected) frequencies malefemaleTotalOn diet14(19.3)25(19.7)39Not on diet159(153.7)152(157.3)311173177350
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 frequencies3)Calculate the X2 value ( calculated value)
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 Level of exposure Symptom Present Absent High Limited No exposure TotalPresent185(143.4)33(49.8)17(41.8)235Absent120(161.6)73(56.2)72(47.2)26530510689500
34 Do these data provide sufficient evidence, at the 0 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 independent2)Calculate the expected frequencies3) Calculate the X2
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 Treatment Out come Drug Placebo Total Favorable 34 9 43 Unfavorable 16 415750100
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 Baby Hepatitis A. what is the conclusion. Mongoloid Non- Mongoloid Total+263460-4193619403019702000what is the conclusion.