5. correlation between interval and any nominal variables (1) (1)
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SPEARMAN RANK CORRELATION COEFFICIENT
(rs)
SPEARMAN RANK CORRELATION COEFFICIENT (rs)
When the variables of interest are measured in an ordinal scale, the
spearman rank correlation coefficient (rs) maybe used instead of the Pearson r. To obtain the Spearman rs , apply the following steps summarized as follows:
SPEARMAN RANK CORRELATION COEFFICIENT (rs)
STEP 1. Rank the scores in distribution X giving the lowest a rank of 1 and the highest a rank of n. Repeat the process for the scores in distribution Y.
SPEARMAN RANK CORRELATION COEFFICIENT (rs)
STEP 2. Obtain the difference (di) between the two sets of ranks.STEP 3. Square each difference and then take the sum of the squared di
SPEARMAN RANK CORRELATION COEFFICIENT (rs)
STEP 4. Complete the formula
SPEARMAN RANK CORRELATION COEFFICIENT (rs)
STEP 5. If the proportion of tries in either X or the Y observations is large, use the formula.
*optional
SPEARMAN RANK CORRELATION COEFFICIENT (rs)
where:
tx = number of observations in X tied at a given rankty = number of observations in Y tied at a given rank
SPEARMAN RANK CORRELATION COEFFICIENT (rs)
STEP 6. To test whether the observed rs value indicates an association between variables , the following maybe applied:For n from 4 to 30, critical values of rs for .05 and .01 level of significance are shown in the table.For n>30, significance of the observed rs under the null hypothesis can be determined using the t-test using the formula:
SPEARMAN RANK CORRELATION COEFFICIENT (rs)
The sampling distribution of the test is the student t distribution with n-2 degrees of freedom.
SPEARMAN RANK CORRELATION COEFFICIENT (rs)
Math Rank X Stat Rank
Y di di2
1 18 7 24 4 3 92 17 6 28 6 0 03 14 5 30 7 -2 44 13 4 26 5 -1 15 12 3 22 3 0 06 10 2 18 2 0 07 8 1 15 1 0 0
COMPUTATION FORMAT:
SPEARMAN RANK CORRELATION COEFFICIENT (rs)
SPEARMAN RANK CORRELATION COEFFICIENT (rs)
This test of significance of the null hypothesis using the computed rs in the example above is:H0: r=0(There is no relationship between math scores and statistics scores of students)Ha: r(There is significant relationship between math scores and statistics scores of students)
SPEARMAN RANK CORRELATION COEFFICIENT (rs)
At α = .05 the tabular t is tα/2.(n-2)=t.025.5=2.571
Decision: The null hypothesis is accepted because the computed t-value is less than the tabular value.Conclusion: The math scores is not significantly correlated to the statistics scores obtained by students.
Correlation Between Interval and Any Nominal Variables
The use of Correlation Rho Formula
E=∑Niyi
2– Ny2
∑y2 – Ny2 Where: Ni = number of sample
per categoryyi = average obtained
per categoryN = total no. of samplesy = over-all averagey = individual item
y
2
Let us measure the degree of relationship between the civil status and the anual salary of the given samples.
Single 65 83 81 69 73 89 76 60
Married 70 67 90 84 78
Widowed 89 64 78
Solution:N1 = 8 y1 = 596/ 8 = 74.5
N2 = 5 y2 = 389/5= 77.8
N3 = 3 y3 = 231/3= 77
N = 16 y = 1216/16 =76.0
y2 = (65)2 + (83)2 + (81)2 + ... + (89)2 + (64)2 + (78)2 = 93,792
[8(74.5) 2 + 5(77.8) 2 + 3(77)2] – 16(76) 2E2
= 93,792 – 16(76) 2 E2 =
0.03
INTERPRETATION: There is a very small positive relationship between the civil status and the annual salary of the given samples
Let us measure the degree of relationship between the subjects and the scores of the given samples.
Science 15 20 9 3 12 16
Math 5 5 14 6
English 23 13 12
Solution:N1 = 6 y1 = 75/ 6 = 12.5
N2 = 4 y2 = 30/4= 7.5
N3 = 3 y3 = 48/3= 16
N = 13 y = 153/13 =11.77
y2 = (15)2 + (20)2 + (9)2 + ... + (23)2
+ (13)2 + (12)2 = 2239 [6(12.5) 2 + 4(7.5) 2 + 3(16) 2] – 13(11.77) 2E2
= 2239 – 13(11.77) 2 E2 =0.3
INTERPRETATION: There is a very small positive relationship between the subjects and the scores
Let us measure the degree of relationship between the performance rank obtained by the trainees during the first and second evaluation period.
Student Trainee Rank During 1st Evaluation
Rank During 2nd Evaluation
A 8 7B 2 5C 7 10D 1 4E 4 2F 9 6G 3 1H 6 9I 10 8J 5 3
SOLUTION:Student Trainee
Rank During 1st Evaluation
Rank During 2nd
Evaluation D D2
A 8 7 1 1B 2 5 -3 9C 7 10 -3 9D 1 4 -3 9E 4 2 2 4F 9 6 3 9G 3 1 2 4H 6 9 -3 9I 10 8 2 4J 5 3 2 4
∑D2= 62
ρ = 1- 6(62)10(102
-ρ = 1- 37210(102-1)
ρ = 0.62
1)
Interpretation:
There is a high positive correlation between the
student trainees’ performance rank during the first and
second evaluation period.
THANK YOU
CREDITS TO: Ian EstradaReported by: Christen Diana Gallinera
Jocelle Mella