Chisquare Test of Association or Independence 1 Chisquare

Chi-square : Test of Association or Independence 1

Chi-square (χ2) • A chi square (χ2 ) distribution is a probability distribution. • The chi-square is useful in making statistical inferences about categorical data in which the categories are two and above. 2

χ2 • Qualitative, or categorical, data are frequently collected in medical investigations. • For example, variables assessed might include üsex, üblood group, üclassification of disease, üpresence or absence of a certain disease, or üwhether the patient survived or not. 3

Chi-square… • Definition A statistic which measures the discrepancy between K observed frequencies O 1, O 2, . Ok and the corresponding expected frequencies e 1, e 2, . . . , ek. • Chi square = χ2 = Σ{ (Oi - ei)2 } / ei 4

Chi-square… • The sampling distribution of the chi-square statistic is known as the chi square distribution. • As in t distributions, there is a different χ2 distribution for each different value of degrees of freedom, but all of them share the following characteristics. 5

Characteristics 1. Every χ2 distribution extends indefinitely to the right from 0. 2. Every χ2 distribution has only one (right ) tail. 3. As df increases, the χ2 curves get more bell shaped and approach the normal curve in appearance (but remember that a chi square curve starts at 0, not at - ∞ ) 6

The Chi Square Distribution • The chi square distribution is asymmetric and its values are always positive. Degrees of freedom are based on the table and are calculated as (rows-1)X(columns-1). 7

Assumptions • The observations must be independent • No more than 20% of the excepted frequencies are less than 5. • The values in the contingency table are counts, not percent /proportions • No cell has 0 or 1 observed value 8

Procedures of Hypothesis Testing 1. State the hypothesis ØThe null hypothesis, H 0: There is no association between the two variables versus Ø Alternative hypothesis, H 1: There is association between the two variables 9

Procedures… 2. Fix the level of significance and computing the critical (tabulated) value: the critical value can be read from the chisquare distribution table at α level of significance and df = (r-1)(c-1) where; r= no. of rows and c= no. of columns Hence, χ2 tab = χ2 (α, df) 10

Procedures… 3. Compute the test statistic: χ2 calc = Σ{ (Oi - ei)2 } / ei 4. Decision rule: ØIf χ2 calc > χ2 tab then the decision is “reject the null hypothesis” and conclude, there is association between the two categorical variables BUT ØIf χ2 calc < χ2 tab then the decision is “accept the null hypothesis” 11

Procedures… • We can use the P-value to test the hypothesis This can be shown symbolically as v. P-value= P(χ2 ≥ χ2 calc) v. To determine the P-value look at the χ2 distribution table with df degrees of freedom and find where the χ2 calc falls on this table 12

Interpreting Chi Square • The chi square test tells us only if the variables are independent or not. • It does not tell us the pattern or nature of the relationship. 13

Example • To study the association of smoking and symptoms of asthma. The study involved 150 individuals and the result is given in the following table: Symptoms of Asthma Ever Smoke Total Yes No Yes 20 30 50 No 22 78 100 Total 42 108 150 14

Example… • The question is, is there association between smoking cigarettes and symptoms of asthma at 0. 05 level of significance? Solution 1. Hypothesis Ø H 0: there is no association between smoking and symptoms of Asthma ØH 1: there is association between smoking and symptoms of Asthma 15

Example… 2. Critical Value (χ2α, df): α=0. 05 df= (2 -1)(21) =1 Then χ20. 05, 1 = 3. 841 3. Test statistic: χ2 calc = Σ{ (Oi - ei)2 } / ei Ø 1 st compute the excepted values for each observed values, for each I observed value (oi) the expected I value (ei) is computed as; ei= (Ritotal x Citotal)/N • Let us do it in the given table 16

Example… Symptom s of Asthma Ever Smoke Yes Obser Expected ved Total No Obse Expected rved Yes 20 42*50/150= 30 14 108*50/150 = 36 No 22 42*100/150 78 = 28 108*100/150 100 = 72 Total 42 108 50 17

Example… Calculate • χ2 calc = Σ{ (Oi - ei)2 } / ei • • = (20 -14)2/14 + (22 -28) 2/28 + (30 -36) 2/36 + (78 -72) 2/72 = 2. 57+1. 28+1+0. 5= 5. 38 18

Example… 4. Decision rule: Since χ2 calc > χ2 tab i. e. 5. 38>3. 841, the null hypothesis is rejected. 5. Conclusion: Thus, there is association between smoking and symptoms of asthma. 19

Example… • P-value: P-value = P(χ2 ≥ χ2 calc, df) • P-value = P(χ2 ≥ 5. 38, 1) read on the table • P-value = 0. 05>P>0. 02 (in between), • since the P-value is less than the significant level of 0. 05, the null hypothesis is rejected. • What if the level of significance is 0. 01? ? ? 20

Example: • Question: Are the homicide rate and volume of gun sales related for a sample of 25 cities? HOMICIDE RATE GUN SALES Low High Totals High 8 5 13 Low 4 8 12 Totals 12 13 N = 25 • The bivariate table showing the relationship between homicide rate (columns) and gun sales (rows). This 2 x 2 table has 4 cells. 21

Practice it 22