Read this article to learn about:- 1. Meaning of Non-Parametric Tests 2. Important Types of Non-Parametric Tests 3. Non-Parametric Vs. Distribution-Free Tests.
Meaning of Non-Parametric Tests:
Statistical tests that do not require the estimate of population variance or mean and do not state hypotheses about parameters are considered non-parametric tests. Non-parametric tests are also referred to as distribution-free tests.
These tests have the obvious advantage of not requiring the assumption of normality or the assumption of homogeneity of variance. They compare medians rather than means and, as a result, if the data have one or two outliers, their influence is negated.
When do you use non-parametric tests? Non-parametric tests are appropriately used when one or more of the assumptions underlying a particular parametric test has been violated.
Generally, however the t-test is fairly robust to all but the severest deviations from the assumptions. Non-parametric tests do not make the assumption of normality about the population distribution. The hypotheses of a non-parametric test are concerned with something other than the value of a population parameter.
The main advantage of non-parametric methods is that they do not require that the underlying population have a normal or any other shaped distribution. The main disadvantages of these tests are that they ignore a certain amount of information. For example to convert data to non-parametric data we can convert numerical data to non-parametric form by replacing numerical values such as 113.45, 189.42, 76.5, 101.79 by either ascending or descending order ranks.
Therefore we can replace them by 1, 2, 3, 4, and 5. However if we represent 189.42 by 5, we lose some information, which is contained in the value 189.42. 189.42 are the largest value and this represented by the rank 5. However the rank 5 could also represent 1189.42, as that would also be the largest value. Therefore use of ranked data leads to some loss of information.
The second disadvantage is that these tests are not as sharp or efficient as parametric tests. The estimate of an interval using a non-parametric test may be twice as large as for the parametric case.
What happens when we use the wrong test in the wrong situation? Generally, parametric tests are more powerful than non-parametric tests (e.g., the non-parametric method will have a greater probability of committing a type II error – accepting a false null hypothesis).
Important Types of Non-Parametric Tests:
Since the theory behind these tests is beyond the scope of our course we shall look at relevant applications and methodologies for carrying out some of the more important non parametric tests.
Non-parametric tests are frequently used to test hypotheses with dependent samples.
The dependent sample tests are:
(i) Mc Nemar test
(iii) Wilcox on test
Non-Parametric Tests for Independent Samples are given below:
(i) Chi square test
(ii) Kolmogorov -Smirnov one sample test
Table of equivalent parametric and non-parametric tests:
When do we use Non-Parametric Tests?
We use non-parametric tests in least one of the following five types of situations:
1. The data entering the analysis are enumerative; that is, counted data represent the number of observations in each category or cross-category.
2. The data are measured and or analyzed using a nominal scale of measurement.
3. The data are measured and or analyzed using an ordinal scale of measurement.
4. The inference does not concern a parameter in the population distribution; for example, the hypothesis that a time-ordered set of observations exhibits a random pattern.
5. The probability distribution of the statistic upon which the analysis is based is not dependent upon specific information or conditions (i.e., assumptions) about the population(s) from which the sample(s) are drawn, but only upon general assumptions, such as a continuous and/or symmetric population distribution. According to these criteria, the distinction of non-parametric is accorded either because of the level of measurement used or required for the analysis, as in types 1, 2, 3.
That is we use either counted or ordinal or nominal scale data. We do not make inferences about population parameters such as the mean. The generality of the assumptions made about the population distribution, as in type 5. That is we do not know or make assumptions about the specific form of the underlying population distribution.
Non-Parametric Vs. Distribution-Free Tests:
The test technique makes use of one or more values obtained from sample data to arrive at probability statement about the hypothesis. As we have seen non-parametric tests are those used when some specific conditions for ordinary tests are violated? Distribution-free tests are those for which the procedure is valid for all different shapes of the population distribution.
For example, the Chi-square test concerning the variance of a given population is parametric since this test requires that the population distribution be normal. On the other hand the Chi-square test of independence does not assume normality condition, or even that the data are numerical.
The Kolmogorov-Smirnov test is a distribution-free test, which is applicable to comparing two populations with any distribution of a continuous random variable:
1. Standard uses of Non-Parametric Tests.
To be used with two independent groups (analogous to the independent groups T-test) we may use the Mann-Whitney Rank Test as a non-parametric alternative to Students T-test when one does not have normally distributed data.
1. Wilcox on.
2. To be used with two related (i.e., matched or repeated) groups (analogous to the dependent samples t-test).
To be used with two or more independent groups (analogous to the single-factor between- subjects ANOVA).
To be used with two or more related groups (analogous to the single-factor within-subjects (ANOVA). We now look at a few of the non-parametric tests in more details including their applications.
Mc Nemer Test:
This test is one of the most important non parametric tests often used when the data happen to be nominal and relate to two related samples.
This test is used for analyzing research designs of the before and after format where the data are measured nominally. The samples, therefore, become dependent or related samples. The use of this test is limited to the case where a 2×2 contingency table is involved. The test is most popularly used to test response to a pre and post situation of a control group.
We can illustrate its use with an example:
A survey of 260 consumers was taken to test the effectiveness of mailed coupons and its effect on individuals who changed their purchase rate for the product. The researcher took a random sample of consumers before the release of the coupons to assess their purchase rate. On the basis of their responses they were divided into groups as to their purchase rate (low, high). After the campaign they were again asked to complete the forms and again classified on their purchase rate.
Table given below shows the results from our sample:
Cases that showed a change in the before and after the campaign in terms of their purchase response were placed in cells A and D. this was done as follows:
a. An individual is placed in cell A if he or she changed from a low purchase to a high purchase rate.
b. Similarly he’s placed in D if he goes from high to a low rate.
c. If no change is observed in his rate he is placed on cells B or C. The research wishes to determine if the mail order campaign was a success. We shall now briefly outline the various steps involved in applying the Mc Nemar test.
We state the null hypotheses. This essentially states that there is no perceptible or significant change in purchase behavior of individuals. Thus for individuals who change their purchase rate this means that the probability of those changing from high to low equals low to high. Ho: P (A) = P (D) Ha: P (A) = P (D). To test the null hypotheses we would examine the cases of change from cells A to D.
The level of significance is chosen, for example = .05.
We now have to decide on the appropriate test to be used. The McNemar test is appropriate because the study is a before and after study and the data are nominal. The study involves the study of two related variables.
The McNemar test involves calculating the chi square value as given by the formula below:
Chi-square = (|A-D | -1)2/A + D i.e., we calculate the absolute difference between the A and D cells.
The decision rule for a=. 05, the critical value of Chi-square is 3.84 for degree of freedom one. Therefore we will reject the null hypotheses if the calculated value of Chi-square. Exceeds the critical value from the tables.
We now actually calculate the test statistic.
The calculated value of Chi-square;
= (|A-D | -1) 2/A+D = (70-30 -1)2/100=15.21
Draw a Statistical Conclusion:
Since calculated value of Chi-square exceeds the critical value we reject the null hypotheses and we can infer that the mail coupon campaign was successful in increasing the purchase rate of the product under study.
When an analysis involves more than two variables we use the Cochran value of Chi-square test. For situations are involving to repeated observations, the dependent variable can take on only two values; either 0 or 1.
Wilcoxon-Matched Pairs Test (or Signed Rank Test):
In a various research situation in the context of two related samples (i.e., case of matched pairs such as a study where men and women are matched or when we compare the output of two similar machines or where some subjects are studied in context of before and after experiment) when we can evaluate both direction and magnitude of difference between matched values, we can use an important non- parametric test viz., Wilcoxon matched pairs test.
While applying this test, we first find the difference (di) between each pair of values and assign rank to differences from the smallest to the largest without regard sign. The actual signs of each difference are then put to corresponding ranks and the test statistic T is computed which happens to be the smaller of the two sums viz., the sum of the negative ranks and sum of positive ranks.
The standard deviation is given by:
Where n = [(no. of given matched pairs) – (no. of dropped out pairs, if any)] and in such situation the test statistics z is worked-out as under:
An experiment is conducted to judge the effect of brand name on quality perception. 16 subjects are recruited for the purpose and are asked to taste and compare two samples of product on a set of scale items judged to be ordinal.
The following data are given below:
Test of hypothesis, using Wilcoxon matched pairs test, that there is no difference between the perceived quality of the two samples. Use 5% level of significance.
H0: There is no difference between the perceived qualities of two samples.
H1: There is difference between the perceived qualities of the two samples.
Using Wilcoxon matched pairs test is:
We drop out pair 8 as d value for this is zero and as such as our n = (16 – 1) = 15 in the given problem.
Now from tabulated value of Tat 5% level of significance is 25 (n = 15), and the calculated value of J is 19.5, which is less than of tabulated value of T.
Since we observed that the reject of the null hypothesis and conclude that there is difference between the perceived quality of the two samples.
Tests for Ordinal Data:
So far the test we have discussed is applicable only to nominal data. We now look at a test, which is specifically designed for ordinal data. Kolmogorov-Smirnov one Sample Test
This test is similar to the chi square test of goodness of fit. This is because it looks at the degree of agreement between the distribution of observed values and some specified theoretical distribution (expected frequencies). The Kolmogorov Smirov test is used if we want to compare the distribution on an ordinal scale.
We can look at an example to see how this test works; a paint manufacturer is interested in testing for four different shades of a colour- very light, very light, bright and dark. Each respondent is shown four prints of the shades and asked to indicate his preference. If colour shade is unimportant, the photos of each shade should be chosen equally except for random differences. If colour shades are important the respondents should prefer one of the extreme shades.
Since shade represents a natural ordering, the Kolmogorov test is applied to test the preference hypothesis. The test involves specifying the cumulative frequency distribution. We then refer to the sampling distribution indicates where there is a divergence between the two distributions is likely to occur- due to chance or if the observed difference is due to a result of preference.
Suppose for example that a sample of 200 homeowners and we got the following shade preference distribution:
The manufacturer asks whether these results indicate a preference.
The data is shown in table given below:
We would carry out the test as follows:
1. Specification of null hypotheses; H0 is that there would be no difference among the shades. Ha – there is a difference in the shades of the new colour.
2. The level of significance: The test would be conducted at the 55 level.
3. Decision regarding statistical test. Here the Kolmogorov t-Smirnov test is appropriate because the data measured are ordinal and we are interested in comparing the above frequency distribution with a theoretical distribution.
4. This test focuses on the largest value of the deviations among observed and theoretical proportions. D = max |Fo (X)-Sn (X)|. Where Fo (X) is the specified cumulative frequency distribution under H0 for any value of x and is the proportion of cases expected to have scores equal to or less than x. So(X): observed cumulative frequency distribution of a random sample of N observations where X is any possible score.
5. The Decision Rule:
If the researcher chooses a=. 05, the critical value of D for large samples is given by the formula 1.36. On where n is the sample size. For example the critical value is .96. The decision rule, therefore, states that null hypotheses will be rejected if computed D >.96.
6. Calculation of Test Statistic:
The theoretical probability distribution is calculated by taking each class frequency as if it were under the null hypotheses, i.e., in our case they would be equal. It is then expressed as a fraction of the total sample. Thus row 2 would be calculated as 50/200 = .25, 100/200 = .5, 150/200 = .75, 200/200 = 1.
The observed cumulative frequency of row 3 is found by 80/ 200 = .40, 140/200 = .70, 180/ 200 = .90 the calculated D value is the point of greatest divergence between cumulative observed and theoretical distributions. In our example this is .20.
Mann-Whitney or U Test:
This is a very popular test amongst the rank sum tests. This test in used to evaluate whether two independent samples have been drawn from the same population.
For the statistic U = n1. n2 + (n) (n1 +1)/ 2 – R where n1 and n2 are the sample size and R1 in the sum of ranks assigned to the values of first sample.
This is used for data, which is ordinal and can be ranked. This test makes use of the actual ranks of the observations as a means of testing the hypotheses about the identity of two population distributions.
The null hypothesis Ho is that there is no difference in attitudes of the two groups of accounts towards bank services. Ha is that there is a significant difference in attitudes of the two groups. The level of significance for this test is. 05.
Kruskal-Wallis Test (or H Test):
Kruskal Wallis test is conducted is a way similar to the G’-test described above. This test is used to test the null hypothesis that k independent random samples come from identical universes against the alternative hypothesis that the means of these universes are not equal.
In this test like the U-test the data are ranked jointly from the low to high or high to low as if they constituted a single sample.
Where n = n1+ n2 + n3 + … + nk and Ri being the sum of rank assigned to ni observations in the ith sample.
The F-test uses the variation ratio to test the significant of difference between two sample variance. This test based on f-distribution is known as so in honour of a great statistician even agriculturist Prof. R.A. Fisher.
Suppose x1, x2, … , xn1 and y2, …y2, …, yn1 independent random samples drawn from the normal population with the some variance, i.e., H0: σx2 = σy2 = σ2
Alternative Hypothesis (H1): σx ≠ σy or σ2
or σx > σy and σy > σx
For the test of statistics
Where x̅ and y̅ are mean of samples are
For the F-statistics follow, F-distribution with (n1 -1), (n2 -1) degree of freedom (v). Since the tabulated value of F-test is taken from for table at (n1 -1), (n2 -1) degree of freedom at given level of significance (generally 1%, 5%, and 10%).
Hence, if Fcat > Ftab, then the null hypothesis is rejected at given level of significance. Otherwise the null hypothesis is accepted at given level of significance.
Calculate the variances of the two samples
Compute the degree of freedom
V1= n1 – 1 and V2 – n2 – 1
Choose the level of significance.
Find the value of Fcal at the desired level of significance and the degree of freedom V1 and V2 from the F-distribution table.
Compare Fobs and Fcal. If Fobs < Fcal, then the difference is not significant.