1. Association between two categorical variables
Dr. Ahmed Samir Al-Naaimi
MBChB, MSc epid, PhD
Assistant Professor / Department of Community Medicine
Baghdad College of Medicine

2. Learning objectives
The student will analyze the relationship between two categorical variables with two or more categories.
Apply the Chi-square test for hypothesis testing.
Understand the concept of observed and expected frequencies.
Explore the link between Chi-square test and multiplication rule used for joint probability under the assumption of independence between 2 classification criteria.

3. Learning objectives
Relate the magnitude of difference between observed and expected frequency to resulting test statistic and the conclusion of statistical significance.
Evaluate the link between Z test and Chi-square test in the special condition of 2 x 2 contingency table.
List the conditions for a valid Chi-square test.

4. Introduction
The table shows the distribution of individuals according to 3 categories of Socioeconomic Index Level (SEIL).
SEIL N %
Low 50 25
Average
110 55
High 40 20
Total
200
100
In descriptive epidemiology, we learned how to tabulate a frequency distribution for a categorical variable. This table shows how individuals are distributed in each category of a variable.
For example, in a rural community, a random sample of 200 people were distributed according to the level of socioeconomic status.

5. Introduction
In the same sample the location of residence was also classified into 3 sectors: south, center and north. N %
South 44 22
Center 96 48
North 60 30
Total
200
100
In the same sample the location of residence was classified into 3 sectors: south, center and north.

6. Introduction
When we examine the relationship between two categorical variables, tabulated one against other.
This is a two way table or cross-tabulation.
Location
SEIL
South
Center
North
Total
Low 33 7 10 50
Average 9 81 20
110
High 2 8 30 40 44 96 60
200
Table show the residence area with a second variable, socioeconomic index level (SEIL).

7. Interpretation of a two by two table
There is an association between two categorical variables, if the distribution of a variable varies according to the value of the other.
The question we are interested in “Does the Socio- economic Index level (SEIL) varies by place of residence?
To answer this question we need to assess a cross- tabulation and calculate relative frequencies (percentages).

8. Interpretation of a two by two table
To answer the question of interest, what should we consider the relative frequencies of column or row totals?
Place of residence
SEIL
South
n %
Center
North
Low
Average
High
Total

9. Interpretation of a two by two table
If the distribution of SEIL is the same in each place of residence, the percentage of columns would be the same for each place of residence. It appears that the percentage of low SEIL differ between sites of residency, but the data are subject to sampling errors, so we need to assess whether these differences in the proportions of the sample reflect differences in populations.
To do this, we need a hypothesis test.

10. Expected frequencies
If the null hypothesis is true, there is no association between SEIL and area of residence, the percentages for each level of SEIL in each area, should be the same as the column of percentages in the total column. Or one can state the hypothesis as “the 2 methods of classification for people: SEIL and place of residence are independent”
For example: The percentage of people in low SEIL in the total sample is 25%. If the null hypothesis is true, we should expect that 25% of people in any place of residence are low SEIL , so the frequency of people in Center sector of residence with low SEIL, is 0.25 x 96 = 24.

11. Interpretation of a two ways table
Place of residence
SEIL
South
n %
Center
n %
North
Total
Low
Regular
High
Also, we should expect than 25% of people in the South have low SEIL. so the frequency (count) of people in South sector of residence with low SEIL is 0.25 x 44 = 11.

12. Expected frequencies
If there are no differences in the distribution of SEIL by places of residence, we should expect that the relative frequency of people with low SEIL is the same in each place of residence.
Note that the expected frequencies do not have to be integers.
Using the totals of columns and rows, we can calculate the expected frequency (count) in each cell.
E = (row total x column total) / grand total
Expected frequency = row total x column total /grand total

13. Expected frequencies
Under the null hypothesis of independence for 2 events, the joint probability is equal to the product of the probability of each event.
P (Low SEIL) = 50/200
P (South) = 44/200
P (Low SEIL and South) =
50/200 x 44/200
The frequency expected in (Low SEIL and South) is equal to the P (Low SEIL and South) multiplied by total sample size of 200.
Expected frequency (E) = 50/200 x 44/200 x 200
E = (row total x column total) / grand total
Location
SEIL
South
Center
North
Total
Low 33 50
Average
110
High 40 44 96 60
200
Expected frequency = row total x column total /grand total

14. Chi-square test
Expected frequencies are those that we should expect if the null hypothesis were true.
To test the null hypothesis, we must compare the expected frequencies with observed frequencies, using the following formula.
Where O = observed frequency, E = expected frequency, Ʃ = sum of all cells in the table. X2 statistic is the result of which is referred to tables of the X2 distribution for the value of p. It is known as the Chi square test.

15. Chi-Square test From the formula we can see that:
If there is a large or significant difference between the observed and expected values, the calculated (test statistic) 2 will be large, while if there is a small (or statistically insignificant) difference between the observed and expected values, the resulting 2 will be small also.

16. Chi-Square test
If the calculated (test statistic) 2 is large, then the sample data provides enough evidence to reject the null hypothesis (Ho) because the observed values are not what we expect under the null hypothesis.
If the calculated (test statistic) 2 is small in magnitude, then the sample data agrees with (accepts) the null hypothesis (Ho), which states that the observed values are similar to or not significantly different from those expected under the null hypothesis of independence.

17. Chi-Square distribution
The values of test statistic in Chi-square distribution is between zero and + ∞. No negative values are present since they are squared values.
The Chi-square distribution has one tail only (positively skewed distribution).
The higher the df the
more flattened is the
curve.
Hypothesis testing is
always one tailed

18. Chi-Square test
The X2 distribution is obtained from the sum of the squares of many standard Normal variables. The number of independent variables commonly used in this sum is the “degrees of freedom”, df = (r-1) x (c-1), where r is the count of rows in the table and c is the count of columns.
The tabulated X2 for 2x2 table with df=1 and alpha error = 0.05 is equal to (Z1-alpha/2)2 = (1.96)2 = 3.84.
This procedure is similar to that we used in other presentations, where we referrer results of Z in Normal distribution tables or t results in t distribution table.

19. Chi-Square test Place of residence SEIL South O E Center O E North
Total n
Low 50
Regular
110
High 40
200
Expected frequency = row total x column total /grand total.
Example: the expected frequency in the first cell of the table (the left upper) = (50 x 44) / 200 = 11, while the observed frequency is 33.

20. Chi-Square test SEIL Place of residence Observed Expected O - E (O-E)2
(O-E)2/E
Low
South 33 11 22
484 44
Center 9 24
- 15
225
9.38
North 2 15
- 13
169
11.27
Regular 7
24.2
-17.2
295.8
12.2 81
52.8
28.2
795.2
15.1 8
- 25
625
18.9
High 10
8.8
1.2
1.44
0.2 20
19.2
0.8
0.64
0.03 30 12 18
324 27
Total
138.1
Knowing the value of X2 and degrees of freedom, we can obtain the probability of obtaining the observed or more extreme if the null hypothesis were true. We see the tables of the distribution of X2 and the line of 4 degrees of freedom we seek the value obtained (138.1) in the columns at 4 degrees of freedom and the value of

21. Steps for hypothesis testing
1. State the statistical hypothesis
Ho: There is no association between SEIL and residence location
HA: There is an association
2. Fill in the observed frequencies for contingency table.
3. Calculate expected frequencies.
4. Calculate the test statistic (Chi-square)
5. Calculate the degrees of freedom (df) = (r-1) x (c-1)
= (3-1) x (3-1) = 2 x 2 = 4
6. Get the tabulated 2 (decision rule) for the specified df.

22. Steps for hypothesis testing
6. The tabulated X2 (decision rule) for df=4 is 9.5
7. Compare the test statistic (calculated X2) and decision rule. Since is > 9.5, then reject the Ho in favor of HA.
8. Conclusion: there is a statistically significant association between SEIL and residence location.

23. Chi-Square test in 2 x 2 tables
When both variables are binary (dichotomous), the cross-tabulation table becomes a 2 x 2.
The 2 test can be applied in the same way as for a larger number of categories table.
This special condition for 2 is very common in medical literature. It will give the same result as that of Z test used for the difference between 2 proportions studied earlier in the biostatistics module. Remember that the decision rule for 2 at df=1 is which is the square value of Z at alpha 0.05 = 1.96.

24. Example (2 x 2 table)
There was a study of the bacteriological efficacy of clarithromycin Vs penicillin, in acute pharyngo-tonsillitis in children by Streptococcus Beta Haemolytic Group A. The results are shown below
Drug
Cure
Not cure
Total
Clarithromycin 91 9
100
Penicillin 82 18
173 27
200

25. Example (2 x 2) table Statistical hypothesis
Ho: There is no association between type of treatment and cure. While in case of Z test we would say “There is no difference in bacteriological efficacy (response rate) between the two treatments, against Streptococcus Beta Hemolytic Group A.
HA: There is an association between type of treatment and patient’s response to treatment.
Drug
Cure
O E
Not cure
Total
Clarithromycin
100
Penicillin
173 27
200

26. Example (2 x 2) table df = (r-1) x (c-1) = (2-1) x (2-1) = 1 x 1 = 1
Calculate expected frequencies
Calculate the test statistic (2) for each cell in the table and its sum = 3.47
Get the decision rule 2 at df=1 which is 3.841
Drug
Effect
Observed
Expected
O - E
(O-E)2
(O-E)2/E
Clarithromycin
Cure 91
86.5
4.5
20.25
0.234
Not cure 9
13.5
- 4.5
1.5
Penicillin 82 18
Total
3.47

27. Example (2 x 2) table
Compare the test statistic (3.47) and decision rule (3.841), since the test statistic is larger, we accept the Ho.
Conclusion: There is no statistically significant association between the type of treatment and the patients response to treatment
Try to solve this example by Z test and compare the results obtained by both methods.

28. A quick formula for 2 x 2 tables
2 can be calculated without the need for expected frequencies in the special case of 2 x 2 table. Use the observed frequencies in a table and marginal totals. If we labeled the cells and marginal totals as follow:
Exposure
Result
Yes No
Total a b
a + b c d
c + d
a + c
b + d N
When the sample size is small, we should disminsh the difference between the observed and expected values, in each cell of the table.
This is obtained by a modification in the previous formulae: X2 =(|ad – bc|) – N/2)2 x N /(a+b) (c+d) (a+c) (b+d)
Vertical bars to the sides |ad-bc |, show that we need take the absolute values from ad-bc. So, if this value is negative, we do not take into account the sign and take the positive value.
N/2 is called the continuous correction.
The results between the formulae with and without continuous correction are slightly different, because to the continuous correction.
2=[(ad – bc)2 x N ]/[(a+b) (c+d) (a+c) (b+d)]

29. Validity of Chi-Square tests
Chi square tests are based on the assumption that the test statistic follows approximately the 2 distribution.
This is reasonable for large samples but for the small one we should use the following guidelines:
a) For 2 x 2 tables
If the total sample size is> 40, then 2 can be used.
If n is between 20 and 40, and the smallest expected value is > 5, 2 can be used. Otherwise, use the Fisher exact significance test.
b) For r x c tables: The 2 test is valid if not more than 20% of expected values is less than 5 and none is less than 1.