Research Methodology - Dr. Krishan K. Pandey
Research Methodology - Dr. Krishan K. Pandey Research Methodology - Dr. Krishan K. Pandey
Testing of Hypotheses-II 291 The test statistic, utilising the McNemer test, can be worked out as under: Degree of freedom = 1. χ 2 = c h b g c h b g 2 2 A − D − 1 200 − 100 − 1 = A + D 200 + 100 99 × 99 = = 300 32. 67 From the Chi-square distribution table, the value of χ 2 for 1 degree of freedom at 5% level of significance is 3.84. The calculated value of χ 2 is 32.67 which is greater than the table value, indicating that we should reject the null hypothesis. As such we conclude that the change in people’s attitude before and after the experiment is significant. 4. Wilcoxon Matched-pairs Test (or Signed Rank Test) In various research situations in the context of two-related samples (i.e., case of matched paires such as a study where husband and wife are matched or when we compare the output of two similar machines or where some subjects are studied in context of before-after experiment) when we can determine both direction and magnitude of difference between matched values, we can use an important non-parametric test viz., Wilcoxon matched-paires test. While applying this test, we first find the differences (d i ) between each pair of values and assign rank to the differences from the smallest to the largest without regard to sign. The actual signs of each difference are then put to corresponding ranks and the test statistic T is calculated which happens to be the smaller of the two sums viz., the sum of the negative ranks and the sum of the positive ranks. While using this test, we may come across two types of tie situations. One situation arises when the two values of some matched pair(s) are equal i.e., the difference between values is zero in which case we drop out the pair(s) from our calculations. The other situation arises when two or more pairs have the same difference value in which case we assign ranks to such pairs by averaging their rank positions. For instance, if two pairs have rank score of 5, we assign the rank of 5.5 i.e., (5 + 6)/2 = 5.5 to each pair and rank the next largest difference as 7. When the given number of matched pairs after considering the number of dropped out pair(s), if any, as stated above is equal to or less than 25, we use the table of critical values of T (Table No. 7 given in appendix at the end of the book) for the purpose of accepting or rejecting the null hypothesis of no difference between the values of the given pairs of observations at a desired level of significance. For this test, the calculated value of T must be equal to or smaller than the table value in order to reject the null hypothesis. In case the number exceeds 25, the sampling distribution of T is taken as approximately normal with mean U T = n(n + 1)/4 and standard deviation b gb g/ , σT = nn+ 1 2n+ 1 24 where n = [(number of given matched pairs) – (number of dropped out pairs, if any)] and in such situation the test statistic z is worked out as under:
292 Research Methodology T − UT z = σ We may now explain the use of this test by an example. Illustration 4 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 obtained: Pair Brand A Brand B 1 73 51 2 43 41 3 47 43 4 53 41 5 58 47 6 47 32 7 52 24 8 58 58 9 38 43 10 61 53 11 56 52 12 56 57 13 34 44 14 55 57 15 65 40 16 75 68 Test the hypothesis, using Wilcoxon matched-pairs test, that there is no difference between the perceived quality of the two samples. Use 5% level of significance. Solution: Let us first write the null and alternative hypotheses as under: H 0 : There is no difference between the perceived quality of two samples. H a : There is difference between the perceived quality of the two samples. Using Wilcoxon matched-pairs test, we work out the value of the test statistic T as under: T Table 12.4 Pair Brand A Brand B Difference Rank of Rank with signs d i |d i | + – 1 73 51 22 13 13 … 2 43 41 2 2.5 2.5 … Contd. ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○
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Testing of Hypotheses-II 291<br />
The test statistic, utilising the McNemer test, can be worked out as under:<br />
Degree of freedom = 1.<br />
χ 2<br />
=<br />
c h<br />
b g<br />
c h<br />
b g<br />
2 2<br />
A − D − 1 200 − 100 − 1<br />
=<br />
A + D 200 + 100<br />
99 × 99<br />
= =<br />
300<br />
32. 67<br />
From the Chi-square distribution table, the value of χ 2 for 1 degree of freedom at 5% level of<br />
significance is 3.84. The calculated value of χ 2 is 32.67 which is greater than the table value,<br />
indicating that we should reject the null hypothesis. As such we conclude that the change in people’s<br />
attitude before and after the experiment is significant.<br />
4. Wilcoxon Matched-pairs Test (or Signed Rank Test)<br />
In various research situations in the context of two-related samples (i.e., case of matched paires<br />
such as a study where husband and wife are matched or when we compare the output of two similar<br />
machines or where some subjects are studied in context of before-after experiment) when we can<br />
determine both direction and magnitude of difference between matched values, we can use an<br />
important non-parametric test viz., Wilcoxon matched-paires test. While applying this test, we first<br />
find the differences (d i ) between each pair of values and assign rank to the differences from the<br />
smallest to the largest without regard to sign. The actual signs of each difference are then put to<br />
corresponding ranks and the test statistic T is calculated which happens to be the smaller of the two<br />
sums viz., the sum of the negative ranks and the sum of the positive ranks.<br />
While using this test, we may come across two types of tie situations. One situation arises when<br />
the two values of some matched pair(s) are equal i.e., the difference between values is zero in which<br />
case we drop out the pair(s) from our calculations. The other situation arises when two or more pairs<br />
have the same difference value in which case we assign ranks to such pairs by averaging their rank<br />
positions. For instance, if two pairs have rank score of 5, we assign the rank of 5.5 i.e., (5 + 6)/2 = 5.5<br />
to each pair and rank the next largest difference as 7.<br />
When the given number of matched pairs after considering the number of dropped out pair(s), if<br />
any, as stated above is equal to or less than 25, we use the table of critical values of T (Table No. 7<br />
given in appendix at the end of the book) for the purpose of accepting or rejecting the null hypothesis<br />
of no difference between the values of the given pairs of observations at a desired level of significance.<br />
For this test, the calculated value of T must be equal to or smaller than the table value in order to<br />
reject the null hypothesis. In case the number exceeds 25, the sampling distribution of T is taken as<br />
approximately normal with mean U T = n(n + 1)/4 and standard deviation<br />
b gb g/ ,<br />
σT = nn+ 1 2n+ 1 24<br />
where n = [(number of given matched pairs) – (number of dropped out pairs, if any)] and in such<br />
situation the test statistic z is worked out as under: