The document provides examples for calculating the Pearson Product Moment Correlation Coefficient (r) from bivariate data. It defines r as a measure of the strength of the linear relationship between two variables. Several fully worked examples are shown calculating r from tables of paired data and interpreting the resulting r value based on established thresholds for strength of correlation. Formulas and steps for calculating r are demonstrated throughout.

1. PEARSON
PRODUCT
MOMENT
CORRELATION
COEFFICIENT

2. Learning Competencies
The learner will be able to:
1. Calculate the Pearson Product Moment
Correlation Coefficient; and
2. Solve problems involving correlation
analysis.

3.  In the previous lesson on scatter plots, the degree or
strength of relationship between the two variables was not
numerically measured. The strength of relationship was only
estimated and described visually based on the dots plotted on
the xy coordinate plane.
 The Pearson Product Moment Correlation Coefficient,
denoted by r, measures the strength of the linear
relationship.

4. 𝑟 =
𝑛 𝑥𝑦 − 𝑥 𝑦
𝑛 𝑥2 − 𝑥 2 𝑛 𝑦
2
− 𝑦 2
Where
n= number of paired values
𝑥= sum of x-values
𝑦= sum of y-values
𝑥𝑦= sum of the products of paired values of x and y
𝑥2
= sum of squared x-values
𝑦2= sum of squared y-values
FORMULA

5. Value of r Strength of Correlation
+1 Perfect positive correlation
+0.71 𝑡𝑜 + 0.99 Strong positive correlation
+0.51 𝑡𝑜 + 0.70 Moderately positive correlation
+0.31 𝑡𝑜 + 0.50 Weak positive correlation
+0.01 𝑡𝑜 + 0.30 Negligible positive correlation
0 No correlation
−0.01 𝑡𝑜 − 0.30 Negligible negative correlation
−0.31 𝑡𝑜 − 0.50 Weak negative correlation
−0.51 𝑡𝑜 − 0.70 Moderately negative correlation
−0.71 𝑡𝑜 − 0.99 Strong negative correlation
−1 Perfect negative correlation
INTERPRETATI
ON

6. The table below shows the time in hours (x) spent by six Grade 11 students in
studying their lessons and their scores (y) on a test. Solve for the Pearson
Product Moment Correlation Coefficient r.
Solution
x 1 2 3 4 5 6
y 5 15 10 15 30 35
x y xy 𝒙𝟐 𝒚𝟐
1 5 5 1 25
2 10 20 4 100
3 15 45 9 225
4 15 60 16 225
5 25 125 25 625
6 35 210 36 1225
𝑥 = 21 𝑦 = 105 𝑥𝑦 = 465 𝑥2
= 91 𝑦2
= 2,425
Example 1

7. 𝑟 =
𝑛 𝑥𝑦 − 𝑥 𝑦
𝑛 𝑥2 − 𝑥 2 𝑛 𝑦
2
− 𝑦 2
𝑟 =
585
370,125
= 0.96157 𝑜𝑟 0.962
The value r=0.962 is between +0.71 𝑡𝑜 + 0.99
In the table of interpretation of r.
It indicates that there is a strong positive
correlation between the time in hours spent in
studying and the scores on a test.
Solving r

8. Scatter Plot

9. The table below shows the time in hours (x) spent by six Grade 11 students in
playing computer games and the scores these students got on a math test (y).
Solve for the Pearson Product Moment Correlation Coefficient r.
Solution
x 1 2 3 4 5 6
y 30 25 25 10 15 5
x y xy 𝒙𝟐 𝒚𝟐
1 30 30 1 900
2 25 50 4 625
3 25 75 9 625
4 10 40 16 100
5 15 75 25 225
6 5 30 36 25
𝑥 = 21 𝑦 = 110 𝑥𝑦 = 300 𝑥2
= 91 𝑦2
= 2,500
Example 2

10. 𝑟 =
𝑛 𝑥𝑦 − 𝑥 𝑦
𝑛 𝑥2 − 𝑥 2 𝑛 𝑦
2
− 𝑦 2
𝑟 =
−510
304,500
= −0.92422 𝑜𝑟 − 0.924
The value 𝑟 = −0.924 is between−0.71 𝑡𝑜 − 0.99
In the table of interpretation of r.
It indicates that there is a strong negative
correlation between the time spent in playing
computer games and the scores on a test.
Solving r

11. Scatter Plot

12. The table below shows the number of selfies (x) posted online of students and
the scores (y) they obtained from a Science test. Solve for the Pearson Product
Moment Correlation Coefficient r.
Solution
x 1 2 3 4 5 6
y 25 5 20 40 25 9
x y xy 𝒙𝟐 𝒚𝟐
1 25 25 1 625
2 5 10 4 25
3 20 60 9 400
4 40 160 16 1600
5 25 125 25 625
6 9 54 36 81
𝑥 = 21 𝑦 = 124 𝑥𝑦 = 434 𝑥2
= 91 𝑦2
= 3,356
Example 3

13. 𝑟 =
𝑛 𝑥𝑦 − 𝑥 𝑦
𝑛 𝑥2 − 𝑥 2 𝑛 𝑦
2
− 𝑦 2
𝑟 =
0
499,800
= 0
The value 𝑟 = 0.
It indicates that there is no correlation
between the number of selfies posted online
and the scores obtained from a Science test.
Solving r

14. Scatter Plot

15. The table below shows the number of composition notebooks and the
corresponding costs. The cost per composition notebook is Php 25. Solve for
the Pearson Product Moment Correlation Coefficient r.
Solution
x 1 2 3 4 5 6
y 25 50 75 100 125 150
x y xy 𝒙𝟐 𝒚𝟐
1 25 25 1 625
2 50 100 4 2500
3 75 225 9 5625
4 100 400 16 10000
5 125 625 25 15625
6 150 900 36 22500
𝑥 = 21 𝑦 = 525 𝑥𝑦 = 2,275 𝑥2
= 91 𝑦2
= 56,875
Example 4

16. 𝑟 =
𝑛 𝑥𝑦 − 𝑥 𝑦
𝑛 𝑥2 − 𝑥 2 𝑛 𝑦
2
− 𝑦 2
𝑟 =
2,625
6,890,625
= 1
The value 𝑟 = 1.
It indicates that there is a perfect positive
correlation between the two variables.
Solving r

17. Scatter Plot

18. Norman and Beth traveled from City A to City B. They traveled at a constant
rate of 40 kilometers per hour. The distance between City A and City B is 280
kilometers. Beth decided to write on a piece of paper the distance they travel
after 1 hour, 2 hours, 3 hours, and so on until they reached City B. These are
shown on the following Table. Solve for the Pearson Product Moment
Correlation Coefficient r.
Solution
x 1 2 3 4 5 6 7
y 240 200 160 120 80 40 0
x y xy 𝒙𝟐 𝒚𝟐
1 240 240 1 57600
2 200 400 4 40000
3 160 480 9 25600
4 120 480 16 14400
5 80 400 25 6400
6 40 240 36 1600
7 0 0 49 0
𝑥 = 28 𝑦 = 840 𝑥𝑦 = 2,240 𝑥2
= 140 𝑦2
= 145,600
Example 5

19. 𝑟 =
𝑛 𝑥𝑦 − 𝑥 𝑦
𝑛 𝑥2 − 𝑥 2 𝑛 𝑦
2
− 𝑦 2
𝑟 =
−7,840
61,465,600
= −1
The value 𝑟 = −1.
It indicates that there is a perfect negative
correlation between the two variables.
Solving r

20. Scatter Plot

21. Shown on the table below are bivariate data. Solve for the Pearson Product
Moment Correlation Coefficient r.
Solution
x 4 2 8 10 12 14 6 16
y 10 5 25 10 15 20 5 10
x y xy 𝒙𝟐
𝒚𝟐
4 10 40 16 100
2 5 10 4 25
8 25 200 64 625
10 10 100 100 100
12 15 180 144 225
14 20 280 196 400
6 5 30 36 25
16 10 160 256 100
𝑥 = 72 𝑦 = 100 𝑥𝑦 = 1,000 𝑥2
= 816 𝑦2
= 1,600
Example 6

22. 𝑟 =
𝑛 𝑥𝑦 − 𝑥 𝑦
𝑛 𝑥2 − 𝑥 2 𝑛 𝑦
2
− 𝑦 2
𝑟 =
800
3,763,200
= 0.41239 𝑜𝑟 0.412
The value 𝑟 is between +0.31 𝑡𝑜 + 0.50.
Hence, there is a weak positive correlation
between the two variables.
Solving r

23. Scatter Plot

24. Listed below are the heights in centimeters and weights in kilograms of six
teachers. Solve for the Pearson Product Moment Correlation Coefficient r.
Solution
Teacher A B C D E F
Height (cm) 160 162 167 158 167 170
Weight (kg) 50 59 63 52 65 68
Teacher x y xy 𝒙𝟐 𝒚𝟐
A 160 50 8000 25600 2500
B 162 59 9558 26244 3481
C 167 63 10521 27889 3969
D 158 52 8216 24964 2704
E 167 65 10855 27889 4225
F 170 68 11560 28900 4624
𝑥 = 984 𝑦 = 357 𝑥𝑦 = 58,710 𝑥2
= 161,486 𝑦2
= 21,503
Example 7

25. 𝑟 =
𝑛 𝑥𝑦 − 𝑥 𝑦
𝑛 𝑥2 − 𝑥 2 𝑛 𝑦
2
− 𝑦 2
𝑟 =
972
1,035,540
= 0. 95517𝑜𝑟 0.955
The value 𝑟 = 0.955 is between +0.71 𝑡𝑜 + 0.99
It indicates a strong positive correlation
between the height and weight of the six
teachers.
Solving r

26. Scatter Plot