7
Non-Parametric Test-Chi-square
Parameters are population measures. The statistics t and F that we have discussed earlier take certain assumptions. They assume by normal distribution and homogeneity of the population. From these statistics, parameters can be estimated. But often, especially in experimental studies, we may have to deal with small number of samples; 10, 12, 15 or so. In such cases, we cannot expect representativeness of the population in the sample. In such cases, we use statistics that are known as non-parametric. We use these statistics also to test hypothesis, often in experimental conditions when the sample size is very small.
We have learnt about the parametric tests, namely t-test and F-test involving the assumptions based on the nature of the population distribution, and on the way the type of measurement scale is used to quantify the data or observations. These are the development of another kind of tests which do not make numerous and stringent assumptions about the nature of population distribution. Such kinds of tests are called ‘distribution-free’ or ‘non-parametric tests’. In this chapter, the nature and characteristics of non-parametric tests with special reference to Chi-square test is discussed.
But, a researcher always remembers that parametric procedures are more powerful than non-parametric procedures in detecting significant differences (if any) due to its nature as it levels all possible support from various mathematical models. Therefore, before an investigator resorts to non-parametric statistics or distribution-free statistics, he/she should consider whether any of the more powerful tools/tests can be used. We should prefer a parametric test if it is suitable. However, if a quick or rough estimate is required, an investigator may use non-parametric tests.
Nature of Non-Parametric Tests
Non-parametric tests are distribution-free statistical tests. These tests are not dependent upon the nature of population and measurement scales.
Non-parametric tests are used when
- The nature of the population, from which samples are drawn, is not known to be normal.
- The variables are repressed in nominal form, that is classified in categories and represented by frequency counts.
- The variables are repressed in ordinal form, that is ranked in order or repressed in numerical scores which have the strength of ranks.
Chi-Square (χ2) Test
In situations where the members of a random sample are classified into mutually exclusive categories, we wish to know whether the observed frequencies (i.e. number of subjects in different categories on the basis of our observation) in these categories are consistent with some hypothesis concerning the relative frequencies. This can further be explained with the following example:
Suppose we have taken a very limited opinion in a small sample of 90 students of class X regarding their plan for opting Arts, Science or Commerce at a later stage. The opinion of students and their frequencies are as follows:
| Arts | Science | Commerce |
|---|---|---|
35 |
28 |
27 |
As the three streams are equally popular among the students, we now assume that the difference in frequencies of these three exclusive categories is only due to chance. Presently, the chi-square test is the most suitable measure to test the agreement between these observed and expected results. Chi-square is denoted as χ2 and is defined by the following equation
where f0 is the observed or experimentally determined frequency and fe is the expected frequency of occurrence based on some hypothesis.
When the square of the difference between observed and expected frequencies is divided by the expected frequency, in each case, the sum of these quotients is chi-square.
Characteristics of Chi-Square
The Chi-Square test is applied only to discrete data which can be counted and so expressed in the form of frequencies or which can readily be transformed into frequencies. In other words, it does not require measured value. Therefore, if the data are not represented in intervals or ratio scales, then chi-square can be safely used. Chi-square is a non-parametric test and does not demand the normal distribution of traits. The corresponding parametric measure of the chi-square is the t-test.
Uses of Chi-Square
Chi-square test is used as a
- Test of goodness of fit
- Test of independence
Test of Goodness of Fit
Let us try to understand the meaning of this test with the help of an example. Suppose, 180 students of class X representing a random sample were asked to name the most interesting subject of the subjects, namely science, mathematics and English and if their responses are as 54 students liked science, 70 students responded to mathematics subject, while 56 students preferred English. Then these figures indicate that mathematics is the most popular subject for the present sample of students, followed by English and science. If we now assume that the above-mentioned three subjects are equally popular among the students (i.e. all the students of class X), then it is merely a chance that a larger number of students of the present sample preferred mathematics as their most interesting subject. If we accept this hypothesis, the sample of 180 students should be equally divided into these three categories. If someone is interested to investigate the discrepancy between the observed frequencies based on experimental results and the expected frequencies based on theoretical hypothesis, then this type of test is called ‘test of goodness of fit’. In other words the test of goodness of fit tests whether or not a table of observed frequencies ‘fits’ or is inconsistent with a corresponding set of theoretical set frequencies based on a given hypothesis. As mentioned earlier, chi-square is used in the test of goodness of fit. Larger the value of chi-square, the greater will be the probability of real divergence of observed frequencies from the expected one.
The theoretical hypothesis may be of two types, namely equal probability hypothesis and the normal probability hypothesis. It need not be mentioned that the equal distribution is not the same as normal distribution. In an earlier example, if we have equal probability hypothesis, the expected frequency will be 60 in each category. But if the hypothesis is of normal probability, we have to calculate the value of expected frequency with the help of characteristics of normal probability curve.
Thus chi-square is used in goodness of fit to test the null hypothesis: there is no significant difference (discrepancy) between observed frequencies and expected frequencies. This null hypothesis is tested in two ways. They are as follows.
- Given distribution versus equal probability hypothesis.
- Given distribution versus normal probability hypothesis.
These two types are separately illustrated with examples below.
Testing the Divergence of Observed Results from Those Expected on the Hypothesis of Equal Probability
EXAMPLE 7.1
The items on an attitude scale are answered by underlying one of the following phrases: strongly agree, agree, undecided, disagree and strongly disagree. The distribution of an item is marked by 100 subjects as shown in Table 7.1. Do these answers diverge significantly from the distribution to be expected, if there are no preferences in the group?
Solution:
As we see, the observed frequencies are given. Another information regarding the hypothesis is given as there are no preferences in the group. This clearly indicates that the hypothesis is of equal probability and the task is to test the significance of divergence between observed frequencies and expected frequencies based on this hypothesis. Therefore, the problem is whether the test of goodness of fit and chi-square will be computed to test significance of difference between the two frequencies.
As we know the formula for chi-square is
Since the hypothesis is of equal probability, therefore, the expected frequency of each category is equal to the total number of subjects in the sample divided by the number of categories, i.e. 
, and thus the table is as follows.
Table 7.1
The steps involved in the computation of χ2 are as follows:
Step 1: Calculation of (f0 − fe) for each category.
Step 2: (f0 − fe)2 for each category:
Step 3: Calculation of 
for each category.
Step 4: The summation of all these gives the chi-square
The basic purpose of computing chi-square is to test the null hypothesis: there is no significance of divergence between observed and expected frequencies. Higher the value of chi-square more will be the probability of rejecting the null hypothesis. Therefore, the obtained value of chi-square should be as high as it can be interpreted, as a real divergence between observed and expected frequencies results. At this point, we need the information that which value of chi-square is regarded as so high so that the hypothesis is rejected. Fortunately, we have statistical table of chi-square, where one can find out the significant value of chi-square to reject the null hypothesis at different levels. Generally, we reject our null hypothesis at any of the two levels of significance. They are 0.05 and 0.01; 0.05 or 0.01 are in fact the probability of committing error out of one. For example, if we decide the level of significance as 0.05 and reject the null hypothesis at this level of probability of committing error, then the null hypothesis is true, five times out of 100. To use the table of chi-square, we also need the degree of freedom (i.e how much we are free to choose). The degree of freedom depends on the number of categories and is calculated by the formula (r − 1) (c − 1), where r is the number of rows in the table and c is the number of columns in the table.
For the present problem the degree of freedom
Now we see in the table of chi-square at a degree of freedom 4, the calculated value of chi-square lies between the probability of 0.7 and 0.5 (see the table). For the same degree of freedom the value of chi-square from the table is 9.488 at p = 0.05. Therefore, the calculated value of chi-square could not reach the value of 9.488 which is a significant value of chi-square to reject the null hypothesis at the 0.05 level, and so the only possible interpretation is that no evidence is found in favour or against the proposition stated in the item of attitude scale.
Another problem is considered to clarify the various steps involved in computation and interpretation of chi-square when used to test the goodness of fit.
EXAMPLE 7.2
In 120 throws of a single dice, the following distribution of faces was obtained.
Do these results constitute a refutation of equal probability hypothesis?
Solution:
Since the hypothesis is of equal probability, all the faces of the dice have equal chances to come up and thus the expected frequency in each category is the same, i.e. 120/6 = 20 as given in Table 7.2.
Table 7.2
Faces of a Dice.
Steps for calculation of χ2.
Step 1:
Step 2:
Step 4:
Degree of freedom |
(df) = (n − 1) |
|
= (6 − 1) = 5 |
Now we see from the table of chi-square that df = 5 lies between probabilities 0.02 and 0.05. The obtained value of chi-square is greater than the value (11.07) required for significance at the 0.05 level. Thus the null hypothesis, where there is no significance of divergence between observed and expected frequencies, is rejected and, therefore, the interpretation is that the result constitutes a refutation of the equal probability hypothesis.
Testing the Divergence of Observed Results from Those Expected on the Hypothesis of Normal Probability
The divergence between observed and expected frequencies may be tested by employing chi-square when the hypothesis is that the distribution of population is normal. The methods of computation are illustrated as follows.
EXAMPLE 7.3
Students of class IV were classified into three groups: above average, average and below average on the basis of their performance in a reading ability test.
| Above Average | Average | Below Average |
|---|---|---|
37 |
35 |
28 |
Does this distribution differ significantly from that to be expected if reading ability is normally distributed among the students?
Solution:
Since the problem is to test goodness of fit, we can employ chi-square. Therefore, the first task is to find out the expected frequencies. The hypothesis is not of equal probability, but of normal probability. Let us suppose that 90 cases lie within −3σ and +3σ of a normal curve and the curve is divided into three equal parts in terms of standard deviation unit, i.e. 2σ distance as shown in Fig. 7.1.
The curve is divided into three parts. Each part is of equal distance, i.e. 2σ; with the help of the statistical table we can find out the percentage of cases which lies within these three categories.
Using the NPC table, the percentage of cases which lies between M and 1σ = 34.13.
The percentage of cases which lies between + 1σ and − 1σ = 68.26.
Similarly the percentage of cases which lies between + 1σ and +3σ and the percentage of cases which lies between − 1σ and − 3σ = 50 − 34.26
Since the total number of subjects is 90, we calculate the exact number of individuals expected to lie in each category as follows:
Number of individuals lying within 
Number of individuals lying within 
Number of individuals lying within 
Thus we have calculated the expected frequencies for each category. The observed and expected frequencies are given in Table 7.3.
Table 7.3
Now we calculate the chi-square. All the steps are exactly the same as mentioned in the previous section.
Step 1:
Step 3:
χ2 = 11.33 + 11.37 + 13.18
χ2 = 35.88
df = (r − 1) (k − 1) = (2 − 1) (3 − 1) = 2.
From the table, we can see that the obtained value of chi-square is higher than the value written (χ2 = 9.21) at the 0.01 level. Therefore, we can reject the null hypothesis that there is no significant difference between observed and expected frequencies, and so we can safely conclude that students are not normally distributed with respect to their reading ability.
Evaluate Yourself
1. Define chi-square. Discuss at least two uses of the chi-square test.
2. Define the following terms.
- ‘Test of goodness of fit’
- ‘Test of independence’
3. 200 subjects were asked ‘should India make nuclear bomb’, 150 answered ‘no’, 40 answered ‘yes’ and the rest answered ‘do not know’. Do these results indicate any significant trend of opinion?
Test of Independence
There are situations where we wish to investigate the association or relationship of two attributes or traits which can be classified into two or more categories. For example, a group of people may be classified as male and female or boys and girls and at the same time they may be categorized with respect to their interest or attribute or performance into various categories. This can be represented in a doubleentry contingency table as given below (see Example 7.4). If we wish to investigate whether the sex and attitude or interest or performance is related to one another or independent from one another, not only for the sample but also for the entire population from which the sample subjects are drawn, we test it with the help of the chi-square test which is called the ‘test of independence’. It is illustrated with the help of the following example.
EXAMPLE 7.4
The following table represents boys and girls who choose each of the three possible answers to an item in an attitude scale.
Is the sex a variable associated with pattern of responses?
Solution:
As we see from the table, the group is classified with respect to two traits, namely sex and attitude, and in each cell of the table corresponding frequencies are given. The task is to find out the association between sex and attitude, therefore, the problem is of the test of independence. The chi-square is used to test the null hypothesis. There is no significance of association between sex and attitude. The formula for calculating chi-square remains the same as mentioned earlier.
The first step is to find out the expected frequencies for each cell. The expected frequency can be calculated by the following formula:
As shown in Table 7.4, the sum of observed frequencies of the first column is 95 and the sum of the first row is 123. The grand total of all the frequencies is 176.
Therefore, fe for the first cell 
Similarly, fe for the second cell of the first row 
In this way, we have calculated fe for each f0 and are given in the table within brackets.
The remaining steps are exactly the same as discussed earlier.
Step 1: Calculation of (f0 − fe)
6.61 |
0.09 |
−6.52 |
−6.61 |
0.09 |
6.25 |
Step 2: Calculation of 
43.69 |
0.0081 |
42.51 |
43.69 |
0.0081 |
42.51 |
Now we see from the table of chi-square the degree of freedom = 2. The calculated chi-square lies between probability 0.01 and 0.05. Hence, chi-square could not reach a value of significance at the 0.05 level, therefore, the calculated value of chi-square is not significant and we are not in a position to reject our null hypothesis which indicates that data do not show sex difference in attitude.
Evaluate Yourself
4. What is test of independence? Cite some examples from the field of educational research where chi-square can be used as a test of independence.
5. Responses of a group of boys and girls are shown in the following table.
Test the hypothesis that there is no association between sex and performance on the item.
Computation of Chi-Square When Variables are in 2 × 2 Fold Contingency Table
We can test the association between two variables when they are classified into various categories by employing chi-square as a test of independence. But when the two variables are classified only into two categories, we get a 2 × 2 fold contingency table as shown in the following table.
In the above table A, B, C and D are frequencies lying in four different cells. In this situation, we are not required to calculate the expected frequencies for the corresponding observed frequencies, and chi- square can be calculated directly by the following formula:
When N = A + B + C + D = Total number of subjects, then A, B, C and D are frequencies in four different cells of a 2 × 2 contingency table. It is illustrated with the following example.
EXAMPLE 7.5
The result of 180 students of class X is a mental ability test is summarized in the following table:
| Above the National Norm | Below the National Norm | |
|---|---|---|
Boys |
34 |
46 |
Girls |
56 |
44 |
Are the boys more intelligent than girls?
Solution:
Since the table is 2 × 2 fold, we use the formula
Now we see from the chi-square table whether the calculated value of chi-square is significant or not at dt = (r − 1) (c − 1) = (2 − 1) (2 − 1) = 1. The calculated value of chi-square lies between 0.01 and 0.05. Therefore, the value of chi-square is less than the value given in the table at the 0.05 level; therefore, the chi-square is not significant and thus the null hypothesis that there is no difference between boys and girls in terms of their mental ability is retained. Therefore, it can be safely concluded that boys are not more intelligent than girls.
Computation of Chi-Square When Frequencies are Small
As we have seen, chi-square can be used to test the divergence between observed and expected frequencies as well as to test the significance of relationship between two variables. These two tests are known as (i) test of goodness of fit and (ii) test of independence. In both of these two tests, sometime we have very small frequencies in one or more cells of a contingency table, especially when we apply chi-square to a problem with a degree of freedom 1 and when any of the frequencies is 5 or less. We make a correction in the formula. This correction is known as Yates’ correction for ‘continuity’. This correction reduces by 0.5 each observed frequency that is greater than the expected frequency and increases by the same amount each observed frequency that is less than expected. The detailed procedure is explained in the following example. The correction is needed because a computed chi-square being based on frequencies varies in discrete jump, whereas the chi-square table represents a continuous distribution of chi-square. If the frequencies in the 2 × 2 table are small and we fail to make Yates’ correction, the value of chi-square is considerably increased and an erroneous result may ensue.
EXAMPLE 7.6
A student answered 16 true—false questions correctly out of a test of 20 questions. Using the chi-square test, determine whether this subject was merely guessing.
Solution:
If the student was merely guessing, the probability of correct and wrong responses will be equal as given in the table.
| Right | Wrong | |
|---|---|---|
f0 |
16 |
4 |
fe |
10 |
10 |
Since one of the frequencies is less than 5, we use Yates’ correction. As mentioned earlier, 16 is considered as 15.5, while 4 is considered as 4.5.
|
|
Right |
Wrong |
Step 1: |
After correction (f0 − fe) |
15.5 − 10 = 5.5 |
4.5 − 10 = −5.5 |
The remaining procedure is the same as mentioned earlier. |
|||
Step 2: |
(f0 − fe)2 |
30.25 |
30.25 |
Step 3: |
|
3.025 |
3.025 |
Using the table of chi-square, the calculated value of chi-square lies between the probability of 0.02 and 0.01 which is higher than the table value of chi-square at the 0.05 level. Therefore, we conclude that the responses of the student are not based on merely guessing.
When we have a problem of test of independence and the two variables are classified into two categories, we use the following formula to compute chi-square.
But when any of the entry in such a 2 × 2 contingency table is 5 or less than 5, we again use Yates’ correction, and the formula is
The formula reads as χ2 is equal to N multiplied by the square of modulus (|AD − BC|) of AD − BC divided by the total of each combination. Here modulus indicates that the sign of difference between the two is not to be considered.
Let us illustrate this with the help of an example.
EXAMPLE 7.7
In a debate organized by a political awareness form, 10 male participants out of 15 and five females out of nine expressed their views in favour of the parliamentary form of government for India. Are the views of male and female participants independent?
Solution:
The problem is of test of independence and is summarized in the following table.
For dr = 1, chi-square lies between p = 0.95 and p = 0.90. Therefore, the chi-square is not significant. It means that male and female participants have independent views regarding the form of government suited for India.
Evaluate Yourself
6. What is Yates’ correction for continuity? Calculate chi-square without using Yates’ correction for the data in Example 7.6 and compare the result with the chi-square calculated earlier. What inference can you draw from these two values?
7. An attitude scale to measure attitude towards secularism was administered on a random sample of rural and urban people of India. The results are presented in the following table.
| Above Median | Below Median | |
|---|---|---|
Urban |
15 |
21 |
Rural |
26 |
19 |
Do the data indicate that rural people have strong favourable attitude towards secularism as compared with urban people?
8. Compute X2 for the following contingency table and test the independence of the two variables.
9. A group of 100 students are classified into five categories: excellent, good, satisfactory, bad and very bad, on the basis of their performance in an achievement test as given below:
Does this distribution differ significantly from that to be expected, if they are normally distributed with respect to achievement of the students in the population?
10. In the following distribution, apply chi-square test and interpret the result.
| Cl | Observed Frequency | Expected Frequency |
|---|---|---|
140–160 |
1 |
0 |
130–140 |
4 |
3 |
120–130 |
10 |
12 |
110–120 |
32 |
35 |
100–110 |
80 |
74 |
90–1 00 |
107 |
109 |
80–90 |
105 |
110 |
70–80 |
87 |
79 |
60–70 |
36 |
40 |
50–60 |
16 |
14 |
40–50 |
2 |
3 |
30–40 |
0 |
1 |
Summary
- Non-parametric tests are distribution-free statistical tests. These tests are dependent upon the nature of population and measurement scales.
- The chi-square is a non-parametric test.
It is defined as
where f0 is the observed frequency and fe is the expected frequency.
- Chi-square is used
- to test the divergence between observed results and expected results based on some theoretical hypothesis. This type of test is called as ‘test of goodness or fit’. Test of goodness of fit may be of two types: (a) when the hypothesis is of equal distribution and (b) when the hypothesis is of normal distribution
- to test the association between two variables that are classified into various categories. This type of test is called the test of independence or the test of association.
- Chi-square is computed by the above-mentioned formula in both the cases, namely test of goodness of fit and test of association or independence. When the two variables are classified into only two categories, the frequencies are represented in a 2 × 2 contingency table. In this case, chi-square can be computed by the formula
where N is the total number of subjects and A, B, C and D are frequencies in four different cells of the contingency table.
- When we have small frequencies (less than or equal to 5) in more cases, a correction is made in the formula which is called Yates’ correction for continuity. For a 2 × 2 contigency table when we introduce Yates’ correction the formula for χ2 is as follows.
- Chi-square test is used as a (i) test of goodness of fit and (ii) test of independence.
Key Words
- Non-Parametric Tests: These tests are distribution-free tests and are not dependent upon the nature of population and measurement scale.
- Characteristics of Chi-Square Test: It is applied only for discrete data which can be counted and so expressed in the form of frequencies.
- Test of Goodness of Fit: Whether or not a table of observed frequencies fits or is inconsistent with a corresponding set of theoretical frequencies based on a given hypothesis.
- Test of Independence: There are situations when we wish to investigate the association or relationship of two attributes or traits which can be classified into two or more categories.
References
1. Garrett, H.E. and Wood Worth R.S. (1986), Statistics in Psychology and Education (Indian edition). Vikils, Feffer and Simsons Pvt. Ltd: Bombay, p. 52.
2. Guilford, J.P and Fruchter, B. (1965), Fundamental Statistics in Psychology and Education. McGraw Hill Book Company: Singapore, p. 64.
3. Koul, L. (1992), Methodology of Education Research. Vikas Publishing House: New Delhi, p. 48.
4. Lindquist, E.F. (1960), Statistical Analysis in Education Research. Oxford & I.B.H. Publishing Co.: New Delhi, p. 60.
5. Mangal, S.K. (1987), Statistics in Psychology and Education. Tata McGraw Hill Publishing Co. Ltd: New Delhi, p. 53.
Additional Readings
1. Bloom, M.F. (1968), Educational Statistics. American Book, Co.: New York, Cincinnati, Chicago, Boston.
2. Ferguson, G.A. (1980), Statistical Analysis in Psychology and Education. McGraw Hill Book Co.: New York.
3. Fisher, R.A. (1936), Statistical Method for Research Workers. Oliver and Boyd: Edinburgh.
4. Villars, D.S. (1951), Statistical Design and Analysis of Experiments for Development Research. Brown: Iowa.
5. Winer, B.J. (1971), Statistical Principle in Experimental Design. McGraw Hill Hogakusha Ltd: London.