1. CHI-SQUARE(X2) DISTRIBUTION
Chi-Square Test

2. 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)

3. CHI-SQUARE(X2) Curve

4. 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).

5. 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

6. 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)

7. 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’

8. 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.

9. 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)

10. 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

11. 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

12. 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

13. 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

14. 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

15. 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.

16. Observed frequencies in a fourfold tableY1 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

17. 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

18. For 2 X 2 table X2 –test is not applicable if:
The expected frequency of any cell is <5

19. 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

20. EXERCISE male female Total On diet 14 25 39 Not on diet 159 152 311
173
177
350

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

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

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

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

25. 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

26. ANSWER 3. Calculate X2 : X2=∑(O-E)2/E
( )2 ( )2 ( )2 ( )2 = =
X2c =3.243

27. ANSWER
4. Find X2 tab
df= (r-1) (c-1)= (2-1)(2-1)=1
X20.95 df=1=3.841

28. Chi-Square Table

29. ANSWER
5. Conclusion
Since X2 c < X2 tab we accept Ho ( No association between sex and being on diet)

30. 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

31. 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

32. 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:

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

34. 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?