In: Statistics and Probability
The following data on x = score on a measure of test
anxiety and y = exam score for a sample of n = 9
students are consistent with summary quantities given in a paper.
Higher values for x indicate higher levels of anxiety.
Compute the value of the correlation coefficient. (Give the answer
to four decimal places.)
r = ________
x | 23 | 13 | 15 | 0 | 17 | 21 | 19 | 14 | 20 |
y | 44 | 59 | 49 | 76 | 49 | 51 | 46 | 52 | 51 |
Solution:
The correlation coefficient formula is given as below:
Correlation coefficient = r = [n∑xy - ∑x∑y]/sqrt[(n∑x^2 – (∑x)^2)*(n∑y^2 – (∑y)^2)]
The calculation table is given as below:
No. |
x |
y |
x^2 |
y^2 |
xy |
1 |
23 |
44 |
529 |
1936 |
1012 |
2 |
13 |
59 |
169 |
3481 |
767 |
3 |
15 |
49 |
225 |
2401 |
735 |
4 |
0 |
76 |
0 |
5776 |
0 |
5 |
17 |
49 |
289 |
2401 |
833 |
6 |
21 |
51 |
441 |
2601 |
1071 |
7 |
19 |
46 |
361 |
2116 |
874 |
8 |
14 |
52 |
196 |
2704 |
728 |
9 |
20 |
51 |
400 |
2601 |
1020 |
Total |
142 |
477 |
2610 |
26017 |
7040 |
From above table, we have
n = 9
∑x = 142
∑y = 477
∑x^2 = 2610
∑y^2 = 26017
∑xy = 7040
r = [n∑xy - ∑x∑y]/sqrt[(n∑x^2 – (∑x)^2)*(n∑y^2 – (∑y)^2)]
r = [9*7040 - 142*477]/sqrt[(9*2610– (142)^2)*(9*26017– (477)^2)]
r = [-4374]/sqrt[(23490 – 20164)*( 234153 – 227529)]
r = -4374 /sqrt(22031424)
r = -4374 /4693.764
r = -0.93187
r = -0.9319
There is a very strong negative linear relationship exists between the given two variables.