THEORY EXAMINATION (SEM–IV) 2016-17 STATISTICAL TECHNIQUES

Course: B.Tech (All Branches)
Subject Code: EAS402
Subject Title: Statistical Techniques
Exam Type: Theory
Duration: 3 Hours
Maximum Marks: 100

SECTION – A (10 × 2 = 20 Marks)

Short conceptual questions focusing on definitions and fundamentals

SECTION – B (5 × 10 = 50 Marks)

Descriptive and computational questions testing applied statistical analysis

(a) Histogram Construction

Data table on page 1 shows class intervals (0–10, 10–20, … 90–100) with student frequencies.
→ Plot rectangular bars along X-axis (marks) and Y-axis (number of students).
→ Highlight the modal class (highest frequency = 60 for 50–60).

(b) Fit Poisson Distribution

Given data:

Steps:

Calculate mean (λ).

Apply P(x)=e−λλxx!P(x) = e^{-λ} \frac{λ^x}{x!}P(x)=e−λx!λx​.

Compute expected frequencies = N×P(x)N × P(x)N×P(x).

Compare observed vs theoretical frequencies.

(c) Correlation Coefficient r12r_{12}r12​

Given paired data for x₁–x₂ and x₃–x₄ (Page 1).
Formula:

r=n(Σxy)−(Σx)(Σy)[nΣx2−(Σx)2][nΣy2−(Σy)2]r = \frac{n(\Sigma xy) - (\Sigma x)(\Sigma y)}{\sqrt{[n\Sigma x^2 - (\Sigma x)^2][n\Sigma y^2 - (\Sigma y)^2]}}r=[nΣx2−(Σx)2][nΣy2−(Σy)2]​n(Σxy)−(Σx)(Σy)​

Interpretation: r=+1r = +1r=+1 (perfect positive), r=−1r = -1r=−1 (perfect negative), r=0r = 0r=0 (no relation).

(d) Chi-Square (χ²) Test on Immunization Data

Steps:

Compute expected frequencies under null hypothesis (no effect).

Apply:

1. χ2=∑(O−E)2Eχ² = \sum \frac{(O - E)^2}{E}χ2=∑E(O−E)2​

Compare with 5% critical value = 3.84 (1 degree of freedom).

Interpretation: χ² > 3.84 ⇒ reject H₀ ⇒ inoculation has a significant effect.

(e) Latin Square Design (Page 2 Table)

4×4 arrangement with treatments A, B, C, D.

Calculate Sum of Squares (Rows, Columns, Treatments, Error).

Perform ANOVA (Analysis of Variance) to test treatment effect significance.

SECTION – C (2 × 15 = 30 Marks)

Comprehensive theory and numerical application problems

Q3. Descriptive Statistics

(a) Primary vs Secondary Data

Primary: Collected firsthand (survey, experiment).

Secondary: Already published (journals, records).

(b) Mean Calculation (Direct & Shortcut)
Use:

Xˉ=ΣfXΣfandXˉ=A+Σfd′Σf×c\bar{X} = \frac{\Sigma fX}{\Sigma f} \quad \text{and} \quad \bar{X} = A + \frac{\Sigma fd'}{\Sigma f} \times cXˉ=ΣfΣfX​andXˉ=A+ΣfΣfd′​×c

where AAA = assumed mean, ccc = class width.

(c) Central Moments and Shape
Given μ₁=0, μ₂=2.5, μ₃=0.7, μ₄=18.75.

Skewness: β1=(μ3)2/(μ2)3=0.0784β₁ = (μ₃)^2 / (μ₂)^3 = 0.0784β1​=(μ3​)2/(μ2​)3=0.0784 → nearly symmetric.

Kurtosis: β2=μ4/(μ2)2=3.0β₂ = μ₄ / (μ₂)^2 = 3.0β2​=μ4​/(μ2​)2=3.0 → mesokurtic (normal-like).

Q4. Probability & Theorems

(a) Classical & Empirical Probability:

P(E)=Favorable outcomesTotal outcomesP(E) = \frac{\text{Favorable outcomes}}{\text{Total outcomes}}P(E)=Total outcomesFavorable outcomes​

(b) Total Probability Theorem:

P(B)=∑P(Ai)P(B∣Ai)P(B) = \sum P(A_i)P(B|A_i)P(B)=∑P(Ai​)P(B∣Ai​)

(c) Poisson Problem: Finding probability of drawing ace of spades at least once in 104 trials using P=1−e−λP = 1 - e^{-λ}P=1−e−λ, λ=npλ = npλ=np.

Q5. Correlation & Regression

(a) Proof: r=bxy×byxr = \sqrt{b_{xy} × b_{yx}}r=bxy​×byx​​ (geometric mean of regression coefficients).
(b) Karl Pearson’s r: Compute from given frequency distribution (10–15).
(c) Angle Between Regression Lines:

tan⁡θ=(1−r2)∣r∣bxby\tanθ = \frac{(1 - r^2)}{|r|} \sqrt{b_x b_y}tanθ=∣r∣(1−r2)​bx​by​​

Significance: indicates strength of correlation — smaller θ = stronger correlation.

Q6. Hypothesis Testing

(a) Types of Errors:

Type I (α): Reject true H₀.

Type II (β): Accept false H₀.
(b) Example: Mortality of men (36 out of 1000 vs expected 3%) — test using binomial or normal approximation.
(c) χ² Test: Define χ² and degree of freedom (df = (r−1)(c−1)).

Q7. Design of Experiments (DOE)

(a) Latin Square Design Advantages:
Controls variability in two directions; reduces error variance.
Disadvantages: Complex setup, limited treatments.
(b) Completely Randomized Design:
Given crop yield data for A, B, C; perform ANOVA and interpret if yield difference is significant (F = 2.286 for n₁=2, n₂=9).
(c) Replication Formula:

n=(tσD)2n = \left( \frac{tσ}{D} \right)^2n=(Dtσ​)2

Given t=1.96, σ=12%, D=10% of mean → compute n (replications required for 5% significance).

Summary Table