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- 1. Correlating Test Scores
- 2. In testing teachers are frequently called upon to describe the relationship between two sets of measures. These two measures might be scores by the same set of students on two separate tests. This chapter shall focus on two of the most commonly used measures of relationship in educational measurement and evaluation, namely: Pearson-Product-Moment Correlation, Spearman rho.
- 3. Correlation Correlation is the relationship between two or more paired factors or two or more sets (Best and Khan, 19980. The degree of relationship is usually measured and represented by a correlation coefficient.
- 4. A correlation coefficient is numerical measure of the linear relationship between two factors or sets of scores (Deauna, 1996). Coefficient can be identified by either the letter r or the Greek letter rho. Or other symbols, depending on the manner the coefficient has been computed. Obtained correlation coefficient can range from a – 1.00 or a + 1.00 toward zero. The sign of the coefficient indicates the directions of the relationship and the numerical value of Its strength.
- 5. Obtained correlation coefficient can be interpreted with the use of a scale like the ones presented below (Best & kahn, 1998). Correlation Coefficient Degree of Relationship .00 - .20 Negligible .21 - .40 Low .41 - .60 Moderate .61 - .80 Substantial .81 – 1.100 High to Very High
- 6. The correlation between two sets of scores can either be4 positive or negative (Gracia, 2003). Positive correlation means that high scores in one variable (X) are associated with high scores in other variable (Y). Conversely a negative correlation means that high scores on the variable are associated with the low scores in another variable vise-versa.
- 7. Pearson’s Product-Moment Correlation This measure of relationship is used when focus to be correlated are both metric data. By metric data are meant measurements, which can be subjected to the four fundamental operations. To compute the correlation coefficient using the aforementioned test statistics, follow these steps:
- 8. 1. Compute the sum of each set scores (SX, SY). 2. Square each score and sum the squares (SX², SY²). 3. Count the number of scores in each group (N) 4. Multiply each X score by its corresponding Y score. 5. Sum the cross products of X (SXY). 6. Calculate the correlation, following the formula.
- 9. r = [NSXY- (SX) (SY)] [(NSX² - (SX²) - (NSY² - (SY) ²)] Where: N = Number of paired observation SXY = sum of the cross products of C and Y SX = sum of the scores under Variable X SY = sum of the scores under variable Y (SX)² = Sum of x scores acquired (SY) = sum of y equated SX² = sum of squared X scores SY² = Sum of squared Y scores
- 10. Let us illustrate how Pearson’s is computed. Table 9.1 shows the computational procedures in determining the degree of relationship between test scores of 10 students in English (x) and mathematics (Y)
- 11. TABLE 10.1 Computation of correlation Coefficient Using Pearson’s r
- 12. X Y X² Y² XY 90 80 9100 6400 7200 85 72 7225 5184 6120 80 70 6400 4900 5600 75 65 3625 4255 4875 70 68 4900 4624 4760 65 55 4225 3025 3575 60 60 3600 3600 3600 55 50 3025 2500 2750 50 53 2500 2809 2650 45 44 2025 1936 1980 SX=675 SY=617 SX²=47625 SY²=39203 SXY=43110
- 13. N = 10 N SXY = 10 (43110) = 431,100 (SX)(SY)= (675)(617) = 416,475 {N SXY = (SX)(SY) = 431,100 – 416,475 = 14,625 N SX² = 10(47625) = 476,250 (SX)² = (675)(675) = 455,625 = 20.625 N SY² = 10 (39203) = 392,030 (SY)² = (617) (617) = 380,689 ( SY² = (SY)² = 392,030 = 380,689 = 11,341 {(N SX² - (SY)² (N SY² - (SY)²)} = (20,625)(11,341) = 233,906,125 = SQUARE ROOT OF 233,908,125 = 15294,05522 r = 14,625/15294.05522 r = 0.956 or 0.96
- 14. Results of computation of person’s r yielded a computed r of 0.96. this indicates that a very high degree of relationship exists between the test scores in English and mathematics. A students who scored high in English also obtained a high score in Mathematics.
- 15. Spearman Rho This measure relationship is used when test scores are ordinal or rank- ordered. In computing rho, the following steps have to be observed: 1. Rank the process for in distribution X, giving the highest score a rank of 1 2. Repeat the process for the scores in distribution of Y. 3. Obtain the difference between the two sets of ranks (D). 4. Square each of these differences and sum up squared differences (SD²). 5. Solve for rho, the following the formula: rho = 1 – 6SD² N³ - N Where : rho = rank-order correlation coefficient D = difference between paired ranks SD² = number of paired of ranks The computational procedures for the calculation of rho are reflected in Table 9.2
- 16. TABLE 10.2 Computation of Correlation Coefficient Using Spearman Rho
- 17. X Y Rank of X Rank Of Y D D 90 80 1 1 0 0 85 72 2 2 0 0 80 70 3 3 0 0 75 65 4 5 -1 1 70 68 5 4 1 1 65 55 6 7 -1 1 60 60 7 6 1 1 55 50 8 9 -1 1 50 53 9 8 1 1 45 44 10 10 0 0 SD² = 6
- 18. N = 10 6SD² 6(6) 36 Rho = 1– N³ - N = 1- 10³ - 10 = 1 - 1000 -10 = 1 – 36 996 = 1 – 0. 0364 = 0.964 or 0.96

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