CHI-SQUARE(X2) 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) Curve

CHI-SQUARE(X2) DISTRIBUTION 4. X2 relates to frequencies of occurrence of individuals (or events) in the categories of one or more variables ( So it is used with categorical variable ). 5. X2 test used to test the agreement between(the observed frequencies with certain characteristics) and (the expected frequencies under certain hypothesis).

Types of CHI-SQUARE(X2) DISTRIBUTION CHI-SQUARE(X2) test of Independence CHI-SQUARE(X2) test of Goodness of fit CHI-SQUARE(X2) test of homogeneity

CHI-SQUARE(X2) test of Independence It is the most common type. 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 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’

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 ∑(O-E)2 X2 =-------------------------------- E df=(r-1)(c-1) For 2 x 2 table, another formula to calculate X2 n(ad-bc)2 (a+c)(b+d)(a+b)(c+d)

Formula ∑ (O – E)2 X2 = E χ2 = The value of chi square O = The observed value E = The expected value ∑ (O – E)2 = all the values of (O – E) squared then added together

Steps in constructing X2 -test 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. Calculate the X2 value (calculated X2 c) 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 column total a+c b+d N=a+b+c+d

For r * c table X2 –test is not applicable if: X2 not applicable if the expected frequency of any cell is <1 If 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 159 152 311 173 177 350

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

ANSWER 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 Cell a =-------------=19.3 350 177 x 39 Cell b=--------------=19.7

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

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) 311 173 177 350

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

Chi-Square Table

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)-(25-159)}2 =------------------------------------- =3.22 39 x 311 x 173 x 177

EXERCISE In a clinical trial, 100 hypertensive patients were equally assigned on antihypertensive and a placebo. The outcome was classified as favourable and unfavourable. 43 showed favourable outcome, 34 were among the drug group. Construct the 2 x2 table and test the appropriate hypothesis at alpha= 0.05

EXRECISE 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 whether or not they exhibited symptoms of respiratory disorders. The results are shown in the following table:

EXERCISE Symptom High Exp Limited Exp. No Exp. Total Present 185 33 17 235 Absent 120 73 72 265 305 106 89 500

EXERCISE Do these data provide sufficient evidence at the alpha of 0.01 level of significance to indicate a relationship between level of exposure and the presence of respiratory disorders?