1. Community &family medicine
Dr. Mayasah A. Sadiq FICM-FM

2. DATA
QUALITATIVE
QUANTITATIVE
CHI SQUARE TEST
T-TEST

3. Dr. Mayasah A. Sadiq FICMS-FM
The Chi-Square test
Dr. Mayasah A. Sadiq FICMS-FM 2

4. Parametric and Nonparametric Tests
Introduce two non-parametric hypothesis tests using the chi-square statistic: the chi-square test for goodness of fit and the chi-square test for independence.

5. Parametric and Nonparametric Tests (cont.)
The term "non-parametric" refers to the fact that the chi‑square tests do not require assumptions about population parameters nor do they test hypotheses about population parameters.

6. Previous example of hypothesis test, such as the T test is a parametric test and includes assumptions about parameters and hypotheses about parameters.

7. Parametric and Nonparametric Tests (cont.)
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.

8. 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

9. Chi square distribution

10. 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)

11. 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.

12. The Chi-Square Test for Goodness-of-Fit (cont.)
The null hypothesis specifies the proportion of the population that should be in each category.
The proportions from the null hypothesis are used to compute expected frequencies that describe how the sample would appear if it were in perfect agreement with the null hypothesis.

13. 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.

15. An Example
Biased Coin?
Observed Expected
Frequency Frequency H T
sum

19. Degree of freedom for goodness of fit
K-1(k=no.of groups )
2-1=1
From the table for α 0.5= 3.841
Calc.Χ>tabulated
There is signif. Differ. In proportion.
There is association.

20. 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)

21. 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.

22. The Chi-Square Test for Independence (cont.)
Both chi-square tests use the same statistic. The calculation of the chi-square statistic requires two steps:
1. The null hypothesis is used to construct an idealized sample distribution of expected frequencies that describes how the sample would look if the data were in perfect agreement with the null hypothesis.

23. The Chi-Square Test for Independence (cont.)
For the goodness of fit test, the expected frequency for each category is obtained by
expected frequency = fe = pn
(p is the proportion from the null hypothesis and n is the size of the sample)
For the test for independence, the expected frequency for each cell in the matrix is obtained by
(row total)(column total)
expected frequency = fe = ───────────────── n

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

25. The Chi-Square Test for Independence (cont.)
2. A chi-square statistic is computed to measure the amount of discrepancy between the ideal sample (expected frequencies from H0) and the actual sample data (the observed frequencies = fo).
A large discrepancy results in a large value for chi-square and indicates that the data do not fit the null hypothesis and the hypothesis should be rejected.

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

27. Chi Square FORMULA:

28. 2nd application

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

30. 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

32. 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

33. 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

36. r x c contingency tables
SA A NO D SD Gr Gr Gr

38. Degree of freedom The d.f depends on subgroup and sample size.
( k-1 ) , or (r-1 ) (c-1)

39. degrees of freedom = (R –1)(C – 1)
R = number of rows
C = number of columns

40. For testing significance of patterns in qualitative data.
Chi-squared Test summary
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

41. Chi-Square (χ2) Any number squared is a positive number
Therefore, area under the curve starts at 0 and goes to infinity
To be statistically significant, needs to be in the upper 5% (α = .05)
Compares observed frequency to what we expected
9/14/2010

42. 2 Test of Independence Solution
H0: No Relationship
Ha: Relationship
 = .05
df = (2 - 1)(2 - 1) = 1
Critical Value(s):
Test Statistic:
Decision:
Conclusion:
 = .05 47

43. If your calculated chi-square value is greater than the critical value calculated, you“reject the null hypothesis.”
If your chi-square value is less than the critical value, you“fail to reject” the null hypothesis

44. 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;

45. YATES = ∑ -------------------------------------

46. 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.