Chi – square
A chi-square (χ2) statistic is a test that measures how a model compares to actual observed data. The data
used in calculating a chi-square statistic must be random, raw, mutually exclusive, drawn from independent
variables, and drawn from a large enough sample.
Chi-square tests are often used to test hypotheses. The chi-square statistic compares the size of any
discrepancies between the expected results and the actual results, given the size of the sample and the
number of variables in the relationship.
For these tests, degrees of freedom are used to determine if a certain null hypothesis can be rejected based
on the total number of variables and samples within the experiment. As with any statistic, the larger the
sample size, the more reliable the results.
Decision:
If p-value ≤ 𝜒 2 , accept the null hypothesis
If p-value > 𝜒 2 , reject the null and accept the alternative hypothesis
Sample Problem:
100 learners are asked to their choice of learning modality: online, modular, blended learning, face – to –
face. The responses of the learners are indicated in the table below.
Category
Observed
Expected
(O-E)
(O-E)2
(𝟎 − 𝑬)𝟐
𝑬
Online
Modular
Blended
F2F
25
25
25 – 25 = 0
0
0/25 = 0
5
25
5 – 25 = -20
400
400/25 = 16
20
25
20 – 25 = -5
25
25/25 = 1
50
25
50 – 25 = 25
625
625/25 = 25
100
100
42
Ho (null hypothesis): The modality of learning has equal distribution
Ha (alternative hypothesis): The modality of learning has unequal distribution
𝜒 2 = 42 (chi -square)
∝ = 0.05 (critical value)
df = 4 – 1 = 3 (degree of freedom)  number of categories subtracted by 1
p-value = 7.81
Conclusion: Reject the null hypothesis and accept the alternative hypothesis. The modality of learning has
unequal distribution
Activities:
a. The enrollment of Kinder to Grade 3 learners are given below. Every year, they expect 2% increase in their
enrollment. Will there be a significant relationship between the enrollment this school year compared to the
previous?
Category
Observed
Expected
Kinder
Grade 1
Grade 2
Grade 3
Total
512
624
671
641
N=
523
644
695
612
N=
Ho:
Ha:
𝜒2 =
df =
Conclusion:
(O-E)
(O-E)2
(𝟎 − 𝑬)𝟐
𝑬
𝜒2 =
∝=
p-value =
b. In the Licensure Examination for Professional Teachers (LET), the applicants think that in a 100 – item test
quesitons, there is an equal distribution of answers per letter.
Category
Observed
Expected
A
B
C
D
E
Total
18
15
24
17
26
N=
20
20
20
20
20
N=
Ho:
Ha:
𝜒2 =
df =
Conclusion:
(O-E)
(O-E)2
(𝟎 − 𝑬)𝟐
𝑬
𝜒2 =
∝=
p-value =