Unit V · Lecture 34 of 40 · B.Sc. Data Science, 1st Year, Semester II
A characteristic that can be measured numerically.
Examples:
• Height (in cm)
• Weight (in kg)
• Exam score (out of 100)
• Age (in years)
Tools: Mean, variance, correlation, regression (Units I–IV)
A characteristic that cannot be measured numerically but can only be classified into categories.
Examples:
• Gender (Male / Female)
• Blood group (A / B / AB / O)
• Pass / Fail
• Smoker / Non-smoker
Tools: Theory of Attributes (this unit!)
An attribute is a qualitative characteristic of a statistical unit that can be observed but not measured numerically. Each unit either possesses or does not possess the attribute. The Theory of Attributes deals with the classification and analysis of data based on such qualitative characteristics.
Dichotomous classification divides a population into two mutually exclusive and exhaustive groups based on the presence or absence of an attribute. Every member of the population belongs to exactly one of the two groups.
Think of each attribute as a yes/no question asked to every member of the population: "Do you smoke?" → Yes (smoker) or No (non-smoker). Every person must answer either yes or no — there is no third option, and no one can answer both. This is dichotomous classification.
For any attribute A:
(A) = the class that possesses attribute A
(α) = the class that does not possess attribute A (complement of A)
α is read as "not-A" or "alpha".
Key rule: (A) and (α) are mutually exclusive (no overlap) and exhaustive (cover the entire population).
N = (A) + (α) where N = total population size.
Attribute A = Literate
(A) = Literate people, (α) = Illiterate people
Attribute B = Employed
(B) = Employed people, (β) = Unemployed people
Attribute C = Vaccinated
(C) = Vaccinated, (γ) = Not vaccinated
Attribute D = Smoker
(D) = Smoker, (δ) = Non-smoker
When we classify a population simultaneously by two attributes A and B, we get four classes (called cross classes or dichotomies):
| (B) — Has B | (β) — Not B | Total | |
|---|---|---|---|
| (A) — Has A | (AB) | (Aβ) | (A) |
| (α) — Not A | (αB) | (αβ) | (α) |
| Total | (B) | (β) | N |
(AB) = has both A and B
(Aβ) = has A but not B
(αB) = has B but not A
(αβ) = has neither A nor B
Total frequencies:
(A) = (AB) + (Aβ)
(α) = (αB) + (αβ)
(B) = (AB) + (αB)
(β) = (Aβ) + (αβ)
N = (A) + (α) = (B) + (β)
A = Literate, B = Employed in a town of N = 1000:
(AB) = 400 — Literate AND Employed
(Aβ) = 200 — Literate AND Unemployed
(αB) = 150 — Illiterate AND Employed
(αβ) = 250 — Illiterate AND Unemployed
Check:
(A) = 400 + 200 = 600 ✓
(α) = 150 + 250 = 400 ✓
(B) = 400 + 150 = 550 ✓
(β) = 200 + 250 = 450 ✓
N = 600 + 400 = 1000 ✓
With three dichotomous attributes, we get 2³ = 8 cross classes. Each class is denoted by a combination of the attribute letters (present) and Greek letters (absent).
| Class | Has A? | Has B? | Has C? | Description |
|---|---|---|---|---|
| (ABC) | Yes | Yes | Yes | Has all three |
| (ABγ) | Yes | Yes | No | Has A and B, not C |
| (AβC) | Yes | No | Yes | Has A and C, not B |
| (Aβγ) | Yes | No | No | Has only A |
| (αBC) | No | Yes | Yes | Has B and C, not A |
| (αBγ) | No | Yes | No | Has only B |
| (αβC) | No | No | Yes | Has only C |
| (αβγ) | No | No | No | Has none |
In a survey of 200 students:
• 120 are male (M), 80 are female (F)
• 90 passed (P), 110 failed (Q)
• 70 are male AND passed
Construct the 2×2 contingency table.
Let A = Male, B = Pass. Then α = Female, β = Fail.
Given: (A) = 120, (α) = 80, (B) = 90, (β) = 110, (AB) = 70
Step 1: (Aβ) = (A) − (AB) = 120 − 70 = 50
Step 2: (αB) = (B) − (AB) = 90 − 70 = 20
Step 3: (αβ) = (α) − (αB) = 80 − 20 = 60
Check: (αβ) = (β) − (Aβ) = 110 − 50 = 60 ✓
| Pass (B) | Fail (β) | Total | |
|---|---|---|---|
| Male (A) | 70 | 50 | 120 |
| Female (α) | 20 | 60 | 80 |
| Total | 90 | 110 | 200 |
| Symbol | Meaning |
|---|---|
| A, B, C | Attributes (qualitative characteristics) |
| α, β, γ | Complementary classes (absence of A, B, C respectively) |
| (A) | Frequency (number of individuals) possessing attribute A |
| (α) | Frequency NOT possessing attribute A |
| (AB) | Frequency possessing BOTH A and B |
| (Aβ) | Frequency possessing A but NOT B |
| (αB) | Frequency possessing B but NOT A |
| (αβ) | Frequency possessing NEITHER A nor B |
| N | Total population size = (A) + (α) = (B) + (β) |
| (A) + (α) | Always equals N (exhaustive classification) |
| (AB) + (Aβ) | Always equals (A) (row total) |
(A) = (AB) + (Aβ) — Row total for A
(α) = (αB) + (αβ) — Row total for α
(B) = (AB) + (αB) — Column total for B
(β) = (Aβ) + (αβ) — Column total for β
N = (A) + (α) = (B) + (β) = (AB) + (Aβ) + (αB) + (αβ)
Q1: An attribute is:
Q2: If A is an attribute, (α) represents:
Q3: With two dichotomous attributes, how many cross classes exist?
Q4: (AB) + (Aβ) always equals:
Q5: In the notation (AβC), which attributes are present?
1. In a group of 500 people: 300 are vaccinated (V), 200 are not. 250 are above 40 years (O), 250 are not. 180 are vaccinated AND above 40. Construct the 2×2 table for attributes V and O.
2. Given (A) = 200, (B) = 150, (AB) = 80, N = 400. Find (Aβ), (αB), (αβ), (α).
3. List all 8 cross classes for three attributes A = Literate, B = Employed, C = Male. Describe what each class represents.
4. Explain why (A) + (α) = N always holds, but (AB) + (αβ) does NOT necessarily equal N.
5. In a class of 60 students, 35 study Maths (M), 28 study Physics (P), and 15 study both. Find (MP), (Mρ), (μP), (μρ) and verify N = 60.
Consistency of Data — Checking whether a given set of frequencies is logically consistent