1. Assistant prof. Dr. Mayasah A. Sadiq FICMS-FM
The Chi-Square test
Assistant prof. Dr. Mayasah A. Sadiq FICMS-FM 2

2. DATA
QUALITATIVE
QUANTITATIVE
CHI SQUARE TEST
T-TEST

3. The most obvious difference between the chi‑square tests and the other hypothesis tests we have considered (T test) is the nature of the data.
For chi‑square, the data are frequencies rather than numerical scores.

4. For testing significance of patterns in qualitative data.
Chi-squared Tests
For testing significance of patterns in qualitative data.
Test statistic is based on counts that represent the number of items that fall in each category
Test statistics measures the agreement between actual counts(observed) and expected counts assuming the null hypothesis

5. Chi Square FORMULA:

6. Chi square distribution

7. Applications of Chi-square test:
Goodness-of-fit
The 2 x 2 chi-square test (contingency table, four fold table)
The a x b chi-square test (r x c chi-square test)

8. Steps of CHI hypothesis testing
1. Data :counts or proportion.
2. Assumption: random sample selected from a population.
3. HO :no sign. Difference in proportion
no significant association.
HA: sign. Difference in proportion
significant association.

9. 4. level of sign.
df 1st application=k-1(k is no. of groups)
df 2nd &3rd application=(column-1)(row-1)
IN 2nd application(conengency table)
Df=1, tab. Chi= always
Graph is one side (only +ve)

10. 5. apply appropriate test of significance

11. 6. Statistical decision & 7. Conclusion
Calculated chi <tabulated chi
P>0.05
Accept HO,(may be true)
If calculated chi> tabulated chi
P<0.05
Reject HO& accept HA.

12. The Chi-Square Test for Goodness-of-Fit
The chi-square test for goodness-of-fit uses frequency data from a sample to test hypotheses about the shape or proportions of a population.
The data, called observed frequencies, simply count how many individuals from the sample are in each category.

13. Example Eye colour in a sample of 40 Blue 12,brown 21,green 3,others 4
Eye colour in population
Brown 80%
Blue 10%
Green 2%
Others 8%
Is there any difference between proportion of sample to that of population .use α0.05

14. Observed counts(frequency)
Figure 18.1
Distribution of eye colors for a sample of n = 40 individuals. The same frequency distribution is shown as a bar graph, as a table, and with the frequencies written in a series of boxes.

16. Expected counts(frequency)
Expected blue10/100*40=4
Expected brown=80/100*40=32
Expexcted green=2/100*40=0.8
Expected others=8/100*40=3

17. 1. Data
Represents the eye colour of 40 person in the following distribution
Brown=21 person,blue=12 person,green=3,others=4

18. 2. Assumption
Sample is randomly selected from the population.

19. 3. Hypothesis
Null hypothesis: there is no significant difference in proportion of eye colour of sample to that of the population.
Alternative hypothesis: there is significant difference in proportion of eye colour of sample to that of the population.

20. 4. Level of significance; (α =0.05);
5% Chance factor effect area
95% Influencing factor effect area
d.f.(degree of freedom)=K-1; (K=Number of subgroups)
=4-1=3
D.f. for 0.5=7.81

21. 7.81 Accept Ho Influencing factor effect area 95% Reject Ho
Chance factor
effect area 5% 5%

22. Figure 18.4
For Example 18.1, the critical region begins at a chi-square value of 7.81.

23. 5. Apply a proper test of significance

24. Chi Square FORMULA:

26. Observed counts(frequency)
Figure 18.1
Distribution of eye colors for a sample of n = 40 individuals. The same frequency distribution is shown as a bar graph, as a table, and with the frequencies written in a series of boxes.

28. Expected counts(frequency)
Expected blue=10/100*40=4
Expected brown=80/100*40=32
Expexcted green=2/100*40=0.8
Expected others=8/100*40=3

29. =(12-4)² (21-32)² (3-0.8)² (4-3)²
=(64/4) + (121/32)+(4.8/0.8)+(1/3)
= =
Calculated chi =26.08

30. 7.81 Accept Ho Influencing factor effect area 95% Reject Ho
Chance factor
effect area 5% 5%

31. 6. Statistical decision
Calculated chi> tabulated chi
P<0.5

32. 7. Conclusion
We reject H0 &accept HA: there is significant difference in proportion of eye colour of sample to that of the population.

33. Applications of Chi-square test:
Goodness-of-fit
The 2 x 2 chi-square test (contingency table, four fold table)
The a x b chi-square test (r x c chi-square test)

34. The Chi-Square Test for Independence
The second chi-square test, the chi-square test for independence, can be used and interpreted in two different ways:
1. Testing hypotheses about the relationship between two variables in a population, or(2×2)
2. Testing hypotheses about differences between proportions for two or more populations.(a×b)

35. The Chi-Square Test for Independence (cont.)
The data, called observed frequencies, simply show how many individuals from the sample are in each cell of the matrix.
The null hypothesis for this test states that there is no relationship between the two variables; that is, the two variables are independent.

36. 2× 2 chi square (contingency table )

37. The Chi-Square Test for Independence (cont.)
The calculation of chi-square is the same for all chi-square tests:

38. Chi Square FORMULA:

39. 2nd application

40. Example A total 1500 workers on 2 operators(A&B)
Were classified as deaf & non-deaf according to the following table.is there association between deafness & type of operator .let α 0.05

41. Result
not deaf.
deaf
total
Operator A
100
900
1000 B 60
440
500
total
160
1340
1500

42. Operator
Result A B
deaf
not deaf.
total
100
900
1000 60
440
500
160
1340
1500
Total number of items=1500
Total number of defective items=160

44. Operator
Result A B
def
not def.
total
100
900
1000 60
440
500
160
1340
1500
Expected deaf from Operator A
= 1000 * 160/1500 = 106.7
(expected not deaf= =893.3)
Expected deaf from Operator B
= 500 * 160/1500 = 53.3

45. Result
Operator A B
def
not def.
total
100
900
1000 60
440
500
160
1340
1500
Operator
Expected A B
def
not def.
total
106.7
893.3
53.3
446.7

46. 1. Data
Represent 1500 workers,1000 on operator A 100 of them were deaf while 500 on operator B 60 of them were deaf

47. 2. Assumption
Sample is randomly selected from the population.

48. 3. Hypothesis
HO: there is no significant association between type of operator & deafness.
HA:there is significant association between type of operator & deafness.

49. 4. Level of significance; (α = 0.05);
5% Chance factor effect area
95% Influencing factor effect area
d.f.(degree of freedom)=(r-1)(c-1) =(2-1)(2-1)=1
D.f. 1 for 0.05=3.841

50. 3.841 Accept Ho Influencing factor effect area 95% Reject Ho
Chance factor
effect area 5% 5%

51. 5. Apply a proper test of significance

53. =( )² ( )² ( )²
+( )² =
446.7
= o
=1.41

54. 3.841 Accept Ho Influencing factor effect area 95% Reject Ho
Chance factor
effect area 5% 5%

55. 6. Statistical decision
Calculated chi< tabulated chi
P>0.5

56. 7. Conclusion We accept H0 HO may be true
There is no significant association between type of operator & deafness.

57. Applications of Chi-square test:
Goodness-of-fit
The 2 x 2 chi-square test (contingency table, four fold table)
The a x b chi-square test (r x c chi-square test)

58. aₓ b
SA A NO D SD Gr Gr Gr

60. Degree of freedom The d.f depends on colunms number & rows number.
or (r-1 ) (c-1)
i.e. if 3 row ,4 colunms
Df=(3-1)(4-1)
df =6
If 3 rows,3 colunms
Df=(3-1)(3-1)
Df=4

61. Yates Correction
When we apply 2x2 chi-square test and one of the expected cells was <5
Or when we apply axb chi-square test and one of the expected cells was <2,
Or when the grand total is <40
we have to apply Yates' correction formula;

62. YATES = ∑ -------------------------------------

63. Note
When 2x2 chi-square test have a zero cell (one of the four cells is zero) we can not apply chi-square test because we have what is called a complete dependence criteria.
But for axb chi-square test and one of the cells is zero when can not apply the test unless we do proper categorization to get rid of the zero cell.

64. 12 3  Or 5 6 7 11 12 3  5 6 7 13