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🔷 TOPIC (vii) — THEORY OF ATTRIBUTES

Syllabus topic (vii) of 10 · runs until Topic (viii)

Covers: Attributes & notation · number of classes · order of frequencies · algebraic expressions · contingency tables · consistency of data · independence & association — proportion method and (AB) = (A)(B)/N · Yule's coefficient of association 🎯 Asked in the papers: 2024 · Q77 — class of 100 students, match the boys/girls figures · 2022 · Q39 — ultimate class frequency for independent attributes

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Deals with the qualitative characteristics calculated using quantitative measurements. e.g. honesty, habit of smoking etc.

1. Attributes and Notation

Attributes "Qualitative characteristics of an individual."

Dicotomony / Dichotomous classification Attribute divides class into two at each level.

graph TD
    A[Population] --> B[Male]
    A --> C[Female]
    B --> D[Literate]
    B --> E[Illiterate]
    C --> F[Literate]
    C --> G[Illiterate]

Types of Attributes

graph TD
    A[Types of Attributes] --> B["Positive Attributes<br>(Presence)<br>Attribute"]
    A --> C["Negative Attributes<br>(Absence)<br>Attribute"]

Symbols / Notations / Mnemonics Presence of an attribute — Capital Letters; $A, B, C \text{ etc.}$ Absence of an attribute — Greek Letters; $\alpha, \beta, \gamma \text{ etc.}$

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e.g $A \rightarrow \text{Presence of honesty}$ $\alpha \rightarrow \text{Absence of honesty}$ $B \rightarrow \text{Literate}$ $\beta \rightarrow \text{Illiterate (Absence of Literacy)}$ $C \rightarrow \text{Employed}$ $\gamma \rightarrow \text{Unemployed (Absence of employment)}$

• class Homogeneous & Mutually exclusive group.
• class Frequency Number of items in each class.
  • denoted by brackets over class symbols. e.g: $\rightarrow (A), (AB), (BC), (Ac) \rightarrow \text{==+ve frequencies==}$ $\rightarrow (\alpha), (\beta), (\gamma), (\alpha\beta) \text{ etc } \rightarrow \text{==-ve frequencies==}$ $\rightarrow (AB\gamma), (\alpha B\gamma), (A\beta) \text{ etc } \rightarrow \text{==Contrarary frequencies==}$

2. Classes and Frequencies

2.1 Number of Classes

1 - Attribute — 3 classes — $A, \alpha, N$ 2 - Attribute — 9 classes $\rightarrow N, A, B, \alpha, \beta, AB, A\beta, \alpha B, \alpha\beta$. 3 - Attributes — 27 classes. n - Attributes = $3^{\text{No. of Attributes}}$

e.g No. of Attributes = 7. $\therefore \text{No. of classes} = 3^7 = \text{==2,187 classes==}$

2.2 Order of Frequencies

Order - 0 : $N$ Order - 1 : $(A) (\alpha) : (B) (\beta) : (C) (\gamma)$ Order - 2 : $(AB) (A\beta), (\alpha B), (\alpha\beta) : (AC), (A\gamma), (\alpha C) (\alpha\gamma)$ Order - 3 : $(ABC), (\alpha BC) / (ABC), (AB\gamma) \dots$

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$N = \text{==}(A), (\alpha)\text{==} : \text{==}(B), (\beta)\text{==} : \text{==}(C), (\gamma)\text{==}$

$A = (AB) + (A\beta)$ $B = (AB) + (\alpha B)$

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2.3 Some Algebraic Expressions

$(A) = (AB) + (A\beta)$ $(B) = (AB) + (\alpha B)$ $(\alpha) = (\alpha B) + (\alpha\beta)$ $(\beta) = (A\beta) + (\alpha\beta)$

$(ABC) + (AB\gamma) + (\alpha B)$ $= (AB) + (\alpha B) = (B)$

2.4 Formula for Any Order of Frequency Classes

No. of classes of $n^{\text{th}}$ order frequency $n! = n \times n-1 \times n-2 \times \dots \times 1$ $1! = 1$ $0! = 1$ $^nC_n = 1$ $^nC_r = \frac{n!}{(n-r)! r!}$

$^nC_r \times 2^r \text{ classes}$

• $n \rightarrow$ No. of attributes
• $r \rightarrow$ order

Q. Find the number of Order-2 classes if the number of attributes is 2. Sol Number of 2 order class = $^nC_r \times 2^r$ $= ^2C_2 \times 2^2$ $= 1 \times 2^2 = 4$

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Q. Calculate the 1st-order classes if the number of attributes = 2. Sol No. of Order-1 classes = $^nC_r \times 2^r$ $= ^2C_1 \times 2^1$ $= \frac{2!}{(2-1)!(1!)} \times 2^1$ $= \frac{2 \times 1}{(1!)(1!)} \times 2$ $= \frac{2}{1} \times 2 = \text{==4==}$

Q. calculate No. of order-2 classes if the number of attributes = 3 Sol No. of order-2 classes = $^nC_r \times 2^r$ $= ^3C_2 \times 2^2$ $= \frac{3!}{(3-2)! (2!)} \times 4$ $= \frac{3 \times 2 \times 1}{(1!) (2 \times 1)} \times 4$ $= \frac{6}{2} \times 4 = \text{==12==}$

3. Contingency Tables

Note: Ultimate frequencies $\rightarrow$ Highest order Frequencies $2^n$

• Contingency matrix shows the relation between two variables / attributes.
• Used by Karl Pearson for first time — $\rightarrow$ theory of contingency $\rightarrow$ Relation to Association & Normal correlation

Contingency table / Cross-Tabulation / Cross-Tab.

graph TD
    A[Contingency table] --> B["==2-Attributes=="]
    A --> C[3-Attributes]
    
    B -.-> B1[* 9 square table]
    B -.-> B2[* 2x2 table]

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3.1 Practice Questions

Q1. If (A) = 40, (AB) = 50, B = 80 & N = 160. Find out the remaining values. Sol

Q2. If the values (AB) = 60, (A$\beta$) = 40, ($\alpha$B) = 30, ($\alpha\beta$) = 20. Find out the rest of values. Sol

H/W Q3:$\rightarrow$ If (A)=130, (B)=100, ($\beta$)=120 and ($\alpha$)=110 find the other values.

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4. Applications of the Theory of Attributes

4.1 Consistency of Data

Rule

1. No class frequency can be negative. Frequency of every class $\ge 0$ (OR)
2. No class frequency can be greater than N. (Each class frequency $\le N$)

Q. Determine consistency in the given data: (AB) = 70, ($\alpha$B) = 40, ($\alpha\beta$) = 60, B = 100.

Conclusion: No frequency is negative, nor any frequency is greater than N. $\therefore$ Data is consistent

Q. Find out if the data is consistent or not. values given are N = 300, (A) = 200, B = 180, (AB) = 170.

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• Data is consistent

Q. N = 400, (A) = 300, (B) = 280, (AB) = 170. Determine consistency. Sol

• Data Inconsistent $\because \text{==} (\alpha\beta) = -10 \text{==}$

Homework Q. (A) = 200, (B) = 100, ($\alpha$B) = 80, N = 300. Determine consistency.

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==Trick == To check consistency of data. $\downarrow$ Check for the ultimate frequencies, If any ultimate frequency is "negative", the data is inconsistent otherwise not.

Q. If (AB)=100, ($\alpha$B)=40, (A$\beta$)=30 & ($\alpha\beta$)=60. Find if Data is consistent or not. Sol No ultimate frequency is negative. $\therefore$ Data is consistent.

Homework Q. Determine the consistency of Data. (A$\beta$)=220, ($\alpha$B)=130, ($\alpha\beta$)=110, (A)=180. Find consistency of Data.

4.2 Independence and Association

• Attributes are said to be Independent if there does not exist any relation between them. e.g. Gender and success, Beauty and Intelligence.

• Two attributes are said to be associated if they are related in one way or other. e.g.

• Positive Association: Present or Absent together. e.g.: unemployment & poverty.

• Negative Association: one is present & another absent. e.g.: Education & Ignorance.

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Methods for checking association and independence

(1) ==Proportion Method ==

(i) $\frac{(AB)}{(B)} = \frac{(A\beta)}{(\beta)}$ $\rightarrow$ Independent (ii) $\frac{(AB)}{(B)} > \frac{(A\beta)}{(\beta)}$ $\rightarrow$ Positive Association (iii) \frac{(AB)}{(B)} < \frac{(A\beta)}{(\beta)} $\rightarrow$ Negative Association

Note: ($\alpha$,$\beta$), (A,$\beta$), ($\alpha$,B) are also Independent.

Q. (AB) = 100, (B) = 10, (A$\beta$) = 150 & ($\beta$) = 15. Find out how A & B are Associated? Sol $\frac{(AB)}{(B)} = \frac{100}{10} = 10$ $\frac{(A\beta)}{(\beta)} = \frac{150}{15} = 10$ $\therefore \frac{(AB)}{(B)} = \frac{(A\beta)}{(\beta)} \rightarrow$ Independent.

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Q. (AB) = 200, (B)= 10, (A$\beta$) = 250 and ($\beta$) = 25 Sol $\frac{(AB)}{(B)} = \frac{200}{10} = 20$ $\frac{(A\beta)}{(\beta)} = \frac{250}{25} = 10$ $\therefore \frac{(AB)}{(B)} > \frac{(A\beta)}{(\beta)} \rightarrow$ Positively Associated.

H/W Q// (AB) = 210, (B) = 10, (A$\beta$) = 310, ($\beta$) = 10. Find association?

(2) ==Comparison Method ==

(i) $(AB) = \frac{(A) \times (B)}{N} \rightarrow$ Independent (ii) $(AB) > \frac{(A) \times (B)}{N} \rightarrow$ Positively Associated (iii) (AB) < \frac{(A) \times (B)}{N} \rightarrow Negatively Associated

Q. Type of Association in the given data N = 106, (A) = 70, (B) = 36, (AB) = 20. Sol $(AB) = 20$ $\frac{(A) \times (B)}{N} = \frac{70 \times 36}{106} = 23.77$

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\therefore (AB) < \frac{(A) \times (B)}{N} \rightarrow Negatively associated.

4.3 Yule's Coefficient of Association

$Q_{AB} = \frac{(AB)(\alpha\beta) - (A\beta)(\alpha B)}{(AB)(\alpha\beta) + (A\beta)(\alpha B)}$

Representation of values of Q. (i) If Q lies between 0 to 1 $\rightarrow$ Positively Associated (ii) Q lies between -1 to 0 $\rightarrow$ Negatively Associated (iii) Q = 0 $\rightarrow$ Independent (iv) Q = 1 $\rightarrow$ completely (Perfectly) Associated (v) Q = -1 $\rightarrow$ completely (Perfectly) Disassociated.

  Completely
Disassociated      Negatively Associated         Positively Associated         Completely
      |----------------------|-----------------------------|----------------------| Associated
     -1                   -0.5             0              0.5                     1
      ^                      ^             ^               ^                      ^
   strong                  weak       Independent         weak                  strong

(AB) or (αβ)                                                                 (Aβ) or (αB)
    = 0                                                                           = 0

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Q. If (AB) = 300, (B) = 15, (A$\beta$) = 350 and $\beta$ = 25. Find the Association between A & B. Sol Since the values given are (AB), (B), (A$\beta$) & ($\beta$), we can use proportion method. $\frac{(AB)}{(B)} = \frac{300}{15} = 20$ $\frac{(A\beta)}{(\beta)} = \frac{350}{25} = 14$ $\therefore \frac{(AB)}{(B)} > \frac{(A\beta)}{(\beta)} \rightarrow$ Positive Association b/w A & B.

Q. If N = 100, (A) = 50, (B) = 30 & (AB) = 40. Find Association between A & B. Sol Since the values given are (AB), (A), (B) and N $\therefore$ Comparison method will be time saving. $(AB) = 40$ $\frac{(A) \times (B)}{N} = \frac{50 \times 30}{100} = \frac{1500}{100} = 15$ Here $(AB) > \frac{(A) \times (B)}{N} \rightarrow$ Positive Association between A and B.

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Question asked in PAA (JKSSB) 2020

Q:: If N = 200, which of the options match the ultimate class frequencies, given that there are two independent attributes A = 100, B = 140? (A) 60 (B) 70 (C) 80 (D) 90 Sol Given values are N, (A), (B) It is also given in question that Attributes A & B are independent. $\therefore$ As per the given values, comparison method is to be used. $(AB)$? $(AB) = \frac{(A) \times (B)}{N} \rightarrow$ Condition for independent Attributes in Comparison Method. $(AB) = \frac{100 \times 140}{200} = \frac{140}{2} = \text{==70== (b)}$

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4.4 Yule's Coefficient — Worked Examples

Q. (AB) = 60, ($\alpha\beta$) = 20, A$\beta$ = 80, ($\alpha$B) = 30. Find the Association. Sol Since the given values are (AB), ($\alpha\beta$), (A$\beta$) and ($\alpha$B) $\therefore$ Yule's coefficient method will be used. $Q = \frac{(AB)(\alpha\beta) - (A\beta)(\alpha B)}{(AB)(\alpha\beta) + (A\beta)(\alpha B)}$ $= \frac{60 \times 20 - 80 \times 30}{60 \times 20 + 80 \times 30}$ $= \frac{1200 - 2400}{1200 + 2400} = \frac{-1200}{3600} = \frac{-12}{36}$ $Q = \text{==-0.33==} \rightarrow \text{weakly negatively associated.}$

Q2. (AB) = 70, ($\alpha\beta$) = 30, (A$\beta$) = 0, ($\alpha$B) = 60. Find Association. Sol $Q = \frac{(AB)(\alpha\beta) - (A\beta)(\alpha B)}{(AB)(\alpha\beta) + (A\beta)(\alpha B)}$ $= \frac{70 \times 30 - 0 \times 60}{70 \times 30 + 0 \times 60}$ $= \frac{2100}{2100} = \text{==1==} \rightarrow \text{Perfectly / completely Associated.}$

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Q. Find the Association between Literate Husband and Literate wife. Literate husband with Literate wife = 80 Literate husband with illiterate wife = 30 Illiterate husband with Literate wife = 100 Illiterate husband with illiterate wife = 60.

Sol A : Literate Husband B : Literate Wife $\alpha$ : Illiterate Husband $\beta$ : Illiterate wife

(AB) = 80, (A$\beta$) = 30, ($\alpha$B) = 100, ($\alpha\beta$) = 60 $Q = \frac{(AB)(\alpha\beta) - (A\beta)(\alpha B)}{(AB)(\alpha\beta) + (A\beta)(\alpha B)}$ $= \frac{80 \times 60 - 30 \times 100}{80 \times 60 + 30 \times 100} = \frac{4800 - 3000}{4800 + 3000}$ $= \frac{1800}{7800} = \frac{18}{78} = \text{==0.23==} \rightarrow \text{weak positively Associated}$

H/W Q!- (AB) = 90, ($\alpha\beta$) = 40, $\alpha$B = 60 and A$\beta$ = 35. Find association?

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