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Unit 04 • 15 Lecture Hours (The Mega Numerical Unit)

Data Entry, Analysis and Interpretation

Hypothesis testing, parametric & non-parametric tests, and 100% verified step-by-step mathematical calculations.

TU Weightage: 20 - 25 Marks (Highest Weightage)✓ 5 Step-by-Step Solved Numericals
Core Memory Anchors: Statistical Hypothesis Testing & Numericals
20-25 Marks Total (Theory + Numerical Problems)

💡 Elaboration Strategy: परीक्षा हलमा यी मुख्य Anchor Points स्मरण राख्नुभयो भने प्रत्येक बुँदालाई प्राज्ञिक रूपमा विस्तार गरेर सजिलै २ देखि ३ पृष्ठको पूर्ण १०-मार्क्सको उत्तर तयार गर्न सकिन्छ।

1

Universal Decision Rule: Calculated vs Critical Value

कुनै पनि test मा यदि तपाईँले हिसाब गरेर निकालेको मान (|Calculated Value|) टेबलको मान (|Critical Table Value|) भन्दा ठूलो आयो भने सिधै H₀ लाई Reject गर्ने! यसको अर्थ हाम्रा दुई समूहबीच साच्चिकै महत्त्वपूर्ण (statistically significant) भिन्नता छ।

निष्कर्ष लेख्दा सधैं दुईवटा कुरा लेख्नुस्: पहिलो 'H₀ is Rejected at 5% significance level', र दोस्रो व्यावहारिक अर्थ 'नयाँ algorithm पुरानो भन्दा साँच्चिकै छिटो छ'।
2

Type I (झुटो आरोप / α) vs Type II (दोषी छुट्ने / β) Errors

Type I error (False Positive) भनेको सत्य कुरालाई अस्वीकार गर्नु हो (जस्तै निर्दोष मान्छेलाई जेल हाल्नु)। Type II error (False Negative) भनेको गलत कुरालाई पनि स्वीकार गर्नु हो (दोषी अपराधीलाई प्रमाण नपुगेर छाड्नु)।

Hypothesis testing को कुनै पनि १०-मार्क्स प्रश्नमा यो २x२ Decision Table बनाउनुस्: [Reality H₀ True/False vs Decision Reject/Accept]।
3

कुन बेला कुन Test चलाउने? (Test Selection Rule)

Categorical/Count डेटा छ भने Chi-Square (χ²); दुईवटा समूहको औषत (mean) दाँज्नु छ भने Student's t-test; तीन वा सोभन्दा बढी समूह छन् भने ANOVA (F-test); दुई समूहको variance को स्थिरता दाँज्नु छ भने F-test; र Ranking/Ordinal डेटा छ भने Spearman's Rank (ρ)।

Sample size n < 30 र जनसंख्याको variance थाहा नहुँदा सधैं t-test प्रयोग हुन्छ भनेर सम्झिनुस्।
4

हिसाब गर्दा पूरा गर्नुपर्ने ५ वटा पक्का Steps

Step 1: H₀ र H₁ प्रस्ट लेख्ने -> Step 2: Formula लेखेर value calculate गर्ने -> Step 3: Degrees of Freedom (df) निकाल्ने -> Step 4: Critical Value (टेबल मान) टिप्ने -> Step 5: Decision र व्यावहारिक Conclusion लेख्ने।

यो ५-step को ढाँचामा हिसाब गर्नुभयो भने परीक्षामा १ मार्क्स पनि काटिने ठाउँ रहँदैन।
Section 01

Master Statistical Test Decision Tree (“कुन Test कहिले लगाउने?”)

One of the most critical examination skills in Unit 4 is inspecting the problem statement and instantly determining which statistical formula to apply:

FLOWCHART: SELECTING THE CORRECT STATISTICAL TEST
What is the nature and level of your research data?
├── 1. Categorical / Frequencies (Counts in rows & columns):
└── Use Chi-Square (χ²) Test of Independence [df = (r - 1)(c - 1)]
├── 2. Ordinal / Ranks (Rankings given by judges/evaluators):
└── Use Spearman's Rank Order Correlation (ρ) [ρ = 1 - 6Σd² / n(n² - 1)]
└── 3. Continuous Numeric Data (Interval or Ratio Scale):
├── Want to compare Variability (Variance) of two systems?
└── Use F-Test (Variance Ratio) [F = S₁² / S₂²]
├── Want to compare Means of TWO groups (n < 30, unknown σ)?
├── Same subjects Before vs After? ──> Paired t-Test
└── Two distinct independent groups? ──> Two-Sample Independent t-Test
└── Want to compare Means of THREE or MORE groups?
└── Use One-Way ANOVA (F-Test) [F = MS_between / MS_within]
👨‍🏫शिक्षकको बोर्ड नोट (Teacher's Board Note)
यो फ्लोचार्ट दिमागमा राख्नुभयो भने परीक्षामा प्रश्न देख्ने बित्तिकै १० सेकेन्डभित्र कुन सूत्र लगाउने भन्ने स्पष्ट हुन्छ!
Section 02

Type I vs Type II Errors Decision Matrix

Whenever an empirical test of significance is performed, two errors can occur:

Researcher DecisionNull Hypothesis (H₀) is Actually TRUENull Hypothesis (H₀) is Actually FALSE
Reject H₀TYPE I ERROR (α)False Positive (False Alarm): Claiming an algorithm is better when it is not.CORRECT DECISIONPower of Test (1 - β): Successfully detecting a genuine breakthrough.
Fail to Reject H₀CORRECT DECISIONConfidence Level (1 - α): Correctly maintaining the status quo.TYPE II ERROR (β)False Negative (Missed Detection): Missing a truly effective algorithm.
🎯 Must-Draw Examination Diagram:TU को ५-मार्क्सको प्रश्नमा यो २×२ म्याट्रिक्स टेबल अनिवार्य कोर्नुहोस् र अल्फा (α) र बिटा (β) लाई स्पष्ट रेखाङ्कन गर्नुहोस्।
Numerical 01

Chi-Square (χ²) Test of Independence (Fully Solved Model)

Non-Parametric TestTU Model Numerical

Chi-Square (χ²) Test of Independence

🎯 कुन बेला यो Test प्रयोग गर्ने? (When to use):

दुईवटा Categorical variables (जस्तै: Developer Level र Code Review Result) बीच कुनै सम्बन्ध (Association) छ कि छैन भनी जाँच्न।

👨‍🏫 शिक्षकको कोर लजिक (Teacher's Mental Model):

परीक्षकले Observed Frequency (O) को टेबल दिन्छन्। हामीले पहिले Expected Frequency (E = Row Total × Col Total / Grand Total) निकाल्नुपर्छ। त्यसपछि (O - E)² / E को जोड निकालेर Critical Value सँग तुलना गर्ने हो।

📝 Question Statement (परीक्षाको प्रश्न):A software engineering lab tests whether developer experience (Junior vs Senior) influences Code Review Acceptance on the first attempt. Sample of 120 submissions showed: • Junior: 20 Accepted, 40 Major Rework • Senior: 50 Accepted, 10 Major Rework Test at α = 0.05 whether developer experience and code review acceptance are independent.
Given Data:Junior Developers: 20 Accepted, 40 Rework (Total = 60)Senior Developers: 50 Accepted, 10 Rework (Total = 60)Accepted Submissions Total: 70Major Rework Total: 50Grand Total (N): 120

Step-by-Step Verified Solution:

1State Hypotheses (H₀ and H₁)

Formulate standard null and alternative hypotheses stating independence vs dependence.

👨‍🏫 Step Tip: H₀ मा सधैं 'कुनै सम्बन्ध छैन (Independent)' लेख्नुपर्छ र H₁ मा 'सम्बन्ध छ (Dependent)' लेख्नुपर्छ।
• Null Hypothesis (H₀): Developer experience and Code Review outcome are independent (no association).
• Alternative Hypothesis (H₁): Developer experience and Code Review outcome are dependent (significant association exists).
2Calculate Expected Frequencies (E)
E = (Rᵢ × Cⱼ) / N

Apply the universal formula E = (Row Total × Column Total) / Grand Total for all 4 cells.

👨‍🏫 Step Tip: Row Total र Column Total गुणन गरेर जम्मा Grand Total (120) ले भाग गर्नुहोस्। यहाँ सबै Row Total = 60 भएकाले हिसाब धेरै सफा हुन्छ।
• E(Junior, Accepted) = (60 × 70) / 120 = 4200 / 120 = 35.0
• E(Junior, Rework) = (60 × 50) / 120 = 3000 / 120 = 25.0
• E(Senior, Accepted) = (60 × 70) / 120 = 4200 / 120 = 35.0
• E(Senior, Rework) = (60 × 50) / 120 = 3000 / 120 = 25.0
3Compute Chi-Square Test Statistic (χ²)
χ² = Σ [ (O - E)² / E ]

Construct calculation table for observed (O) and expected (E) frequencies.

👨‍🏫 Step Tip: कापीमा यो ५-कोलम भएको टेबल अनिवार्य बनाउनुहोस्। परीक्षकले सिधै यो टेबल हेरेर Full Marks दिन्छन्।
CellObserved (O)Expected (E)(O - E)(O - E)²(O - E)² / E
Junior, Accepted2035.0-15.0225.06.429
Junior, Rework4025.0+15.0225.09.000
Senior, Accepted5035.0+15.0225.06.429
Senior, Rework1025.0-15.0225.09.000
Total (χ² calculated)120120.00.0-30.858
χ² calculated = 30.858
4Degrees of Freedom (df) & Critical Table Lookup
df = (r - 1)(c - 1)

Determine df and find critical value at α = 0.05.

👨‍🏫 Step Tip: 2×2 टेबलमा df सधैं (2 - 1)(2 - 1) = 1 हुन्छ। α = 0.05 मा Table Value सधैं 3.841 हुन्छ।
• Rows (r) = 2, Columns (c) = 2
• df = (2 - 1) × (2 - 1) = 1 × 1 = 1
• Critical χ² at α = 0.05 and df = 1 is 3.841
Final Step: Decision Rule & Academic ConclusionReject H₀
Calculated:30.858
Critical Table:3.841
Degrees of Freedom:1
Alpha (α):0.05
✍️ What to Write in the TU Exam Answer Sheet (Final Report):Since the calculated Chi-Square value (30.858) is far greater than the critical value (3.841) at α = 0.05, we reject the null hypothesis. There is a statistically significant association between developer experience and code review acceptance. Senior developers have significantly higher first-time acceptance rates.
Exam Hall Verification Checklist (विद्यार्थीले जाँच्नुपर्ने कुराहरू):
  • Always verify that Σ (O - E) = 0. If it is not zero, your calculation has an arithmetic error.
  • Always state both H₀ and H₁ explicitly before touching numbers.
  • Remember: For df = 1 and α = 0.05, the critical value is always 3.841.
Numerical 02

One-Sample Student's t-Test (Fully Solved Model)

Parametric TestTU Model Numerical

One-Sample Student's t-Test

🎯 कुन बेला यो Test प्रयोग गर्ने? (When to use):

जब Sample Size सानो (n < 30) हुन्छ, Population Standard Deviation (σ) थाहा हुँदैन, र Sample Mean (X̄) लाई कुनै तोकिएको Benchmark (μ₀) सँग दाँज्नुपर्ने हुन्छ।

👨‍🏫 शिक्षकको कोर लजिक (Teacher's Mental Model):

पहिले Sample Mean (X̄) निकाल्ने, त्यसपछि Sample Variance s² = Σ(X - X̄)² / (n - 1) निकाल्ने, त्यसपछि Standard Error SE = s / √n ले भाग गरेर t-value निकाल्ने।

📝 Question Statement (परीक्षाको प्रश्न):A software lab claims a new caching algorithm reduces database query latency to an average of μ₀ = 50 ms. A random test of n = 10 query executions recorded the following response latencies (in ms): 44, 46, 43, 48, 45, 47, 42, 49, 44, 42. Test at α = 0.05 whether the algorithm significantly outperforms the 50 ms benchmark.
Given Data:Sample size (n): 10Hypothesized Population Mean (μ₀): 50 msSignificance Level (α): 0.05

Step-by-Step Verified Solution:

1State Hypotheses (H₀ and H₁)

Formulate null and alternative claims.

👨‍🏫 Step Tip: H₀: μ = 50 ms (कुनै सुधार छैन), H₁: μ ≠ 50 ms (दुई-तर्फी / Two-tailed परीक्षण)।
• Null Hypothesis (H₀): μ = 50 ms (Mean query latency equals 50 ms).
• Alternative Hypothesis (H₁): μ ≠ 50 ms (Mean query latency is significantly different from 50 ms).
2Calculate Sample Mean (X̄) and Standard Deviation (s)
X̄ = ΣX / n , s = √[ Σ(X - X̄)² / (n - 1) ]

Compute mean and unbiased sample standard deviation.

👨‍🏫 Step Tip: n - 1 = 9 ले भाग गर्न नबिर्सिनुहोस्। Population मा n ले भाग हुन्छ तर Sample मा सधैं n - 1 हुन्छ।
Sample (X)Deviation (X - X̄)(X - X̄)²
4444 - 45 = -11
4646 - 45 = +11
4343 - 45 = -24
4848 - 45 = +39
4545 - 45 = 00
4747 - 45 = +24
4242 - 45 = -39
4949 - 45 = +416
4444 - 45 = -11
4242 - 45 = -39
ΣX = 450Σ(X - X̄) = 0Σ(X - X̄)² = 54
• Sample Mean: X̄ = 450 / 10 = 45.0 ms
• Sample Variance: s² = 54 / (10 - 1) = 54 / 9 = 6.0
• Sample Standard Deviation: s = √6.0 ≈ 2.449 ms
• Standard Error of Mean: SE = s / √n = 2.449 / √10 = 2.449 / 3.1623 ≈ 0.7746 ms
3Calculate t-Statistic
t = (X̄ - μ₀) / (s / √n)

Substitute values into the t-test formula.

👨‍🏫 Step Tip: ऋणात्मक चिन्ह आए पनि Two-tailed test मा हामी Absolute value |t| लिन्छौं।
• t = (45.0 - 50.0) / 0.7746 = -5.0 / 0.7746 = -6.455
• Absolute value: |t calculated| = 6.455
|t calculated| = 6.455
4Degrees of Freedom & Critical Value

Determine df = n - 1 and lookup critical t.

👨‍🏫 Step Tip: Two-tailed test मा α = 0.05 र df = 9 हुँदा Table Value सधैं 2.262 हुन्छ।
• df = n - 1 = 10 - 1 = 9
• For two-tailed test at α = 0.05 and df = 9, critical t = 2.262
Final Step: Decision Rule & Academic ConclusionReject H₀
Calculated:6.455
Critical Table:2.262
Degrees of Freedom:9
Alpha (α):0.05
✍️ What to Write in the TU Exam Answer Sheet (Final Report):Because the absolute calculated t-statistic (6.455) exceeds the critical table value (2.262) at α = 0.05 with 9 degrees of freedom, we reject H₀. The new caching algorithm achieves a mean latency of 45 ms, which is statistically significantly faster than the 50 ms benchmark.
Exam Hall Verification Checklist (विद्यार्थीले जाँच्नुपर्ने कुराहरू):
  • Divide by n - 1 = 9 when calculating sample variance s², not 10.
  • Check that the sum of deviations Σ(X - X̄) equals exactly 0.
  • State whether you are using a two-tailed or one-tailed test.
Numerical 03

Independent Two-Sample t-Test with Pooled Variance

Parametric TestTU Model Numerical

Two-Sample Independent t-Test (Pooled Variance)

🎯 कुन बेला यो Test प्रयोग गर्ने? (When to use):

दुई फरक स्वतन्त्र समूहहरू (जस्तै: Traditional UI प्रयोगकर्ता vs New AI UI प्रयोगकर्ता) को औसत समय वा प्रदर्शनमा भिन्नता छ कि छैन जाँच्न।

👨‍🏫 शिक्षकको कोर लजिक (Teacher's Mental Model):

पहिले दुवै समूहको Mean (X̄₁, X̄₂) निकाल्ने, त्यसपछि Pooled Variance Sₚ² = [Σ(X₁ - X̄₁)² + Σ(X₂ - X̄₂)²] / (n₁ + n₂ - 2) निकाल्ने, अनि t = (X̄₁ - X̄₂) / √[Sₚ²(1/n₁ + 1/n₂)] गर्ने।

📝 Question Statement (परीक्षाको प्रश्न):An HCI experiment compares user task completion times (seconds) between Traditional UI (Group 1, n₁ = 8) and AI-Assisted UI (Group 2, n₂ = 8). • Group 1 times: 32, 34, 30, 36, 31, 35, 33, 29 • Group 2 times: 26, 28, 25, 29, 27, 24, 28, 25 Assuming equal variances, test at α = 0.05 whether the AI UI significantly reduces task completion time.
Given Data:Group 1 (Traditional): n₁ = 8, ΣX₁ = 260, X̄₁ = 32.5 sGroup 2 (AI-Assisted): n₂ = 8, ΣX₂ = 212, X̄₂ = 26.5 sSignificance Level: α = 0.05

Step-by-Step Verified Solution:

1State Hypotheses

Formulate H₀ and H₁.

👨‍🏫 Step Tip: H₀: μ₁ = μ₂ (दुवै UI मा समय एउटै लाग्छ), H₁: μ₁ ≠ μ₂ (समयमा महत्त्वपूर्ण फरक छ)।
• Null Hypothesis (H₀): μ₁ = μ₂ (No significant difference in completion times).
• Alternative Hypothesis (H₁): μ₁ ≠ μ₂ (Significant difference exists).
2Compute Sum of Squared Deviations

Calculate Σ(X₁ - X̄₁)² and Σ(X₂ - X̄₂)².

👨‍🏫 Step Tip: दुवै समूहको deviation निकाल्ने: Group 1 को Σ(X₁ - 32.5)² = 42.0, Group 2 को Σ(X₂ - 26.5)² = 22.0।
• Group 1 squared deviations: Σ(X₁ - X̄₁)² = (-0.5)² + (1.5)² + (-2.5)² + (3.5)² + (-1.5)² + (2.5)² + (0.5)² + (-3.5)² = 0.25 + 2.25 + 6.25 + 12.25 + 2.25 + 6.25 + 0.25 + 12.25 = 42.0
• Group 2 squared deviations: Σ(X₂ - X̄₂)² = (-0.5)² + (1.5)² + (-1.5)² + (2.5)² + (0.5)² + (-2.5)² + (1.5)² + (-1.5)² = 0.25 + 2.25 + 2.25 + 6.25 + 0.25 + 6.25 + 2.25 + 2.25 = 22.0
3Calculate Pooled Variance (Sₚ²) & Standard Error
Sₚ² = [ Σ(X₁ - X̄₁)² + Σ(X₂ - X̄₂)² ] / (n₁ + n₂ - 2)

Combine sample variances with pooled degrees of freedom.

👨‍🏫 Step Tip: Pooled Variance Sₚ² = (42 + 22) / 14 = 64 / 14 = 4.5714।
• Sₚ² = (42.0 + 22.0) / (8 + 8 - 2) = 64.0 / 14 = 4.5714
• SE = √[ Sₚ² × (1/n₁ + 1/n₂) ] = √[ 4.5714 × (1/8 + 1/8) ] = √[ 4.5714 × 0.25 ] = √1.14285 ≈ 1.069 s
4Compute Test Statistic (t)
t = (X̄₁ - X̄₂) / SE

Calculate t-ratio.

👨‍🏫 Step Tip:
• t = (32.5 - 26.5) / 1.069 = 6.0 / 1.069 ≈ 5.613
t calculated = 5.613
5Degrees of Freedom & Critical Value

Find critical t for df = n₁ + n₂ - 2.

👨‍🏫 Step Tip: Pooled df = n₁ + n₂ - 2 = 8 + 8 - 2 = 14 हो। Table मा 14 df र α = 0.05 मा 2.145 हेर्ने।
• df = 8 + 8 - 2 = 14
• For two-tailed test at α = 0.05 and df = 14, critical t = 2.145
Final Step: Decision Rule & Academic ConclusionReject H₀
Calculated:5.613
Critical Table:2.145
Degrees of Freedom:14
Alpha (α):0.05
✍️ What to Write in the TU Exam Answer Sheet (Final Report):The calculated t-value (5.613) exceeds the critical value (2.145) at α = 0.05. We reject H₀ and conclude that the AI-assisted interface significantly reduces task completion time compared to the traditional interface.
Exam Hall Verification Checklist (विद्यार्थीले जाँच्नुपर्ने कुराहरू):
  • Always use pooled df = n₁ + n₂ - 2 = 14.
  • Verify Sₚ² calculation carefully before taking square roots.
Numerical 04

F-Test for Equality of Two Variances (Variance Ratio Test)

Parametric TestTU Model Numerical

F-Test for Equality of Two Variances

🎯 कुन बेला यो Test प्रयोग गर्ने? (When to use):

दुईवटा Independent Populations को Variance (एकरूपता / Consistency) बराबर छ कि छैन जाँच्न।

👨‍🏫 शिक्षकको कोर लजिक (Teacher's Mental Model):

F-Test मा सधैं ठूलो Variance भएको समूहलाई माथि (Numerator) राख्ने (S₁² > S₂²)। यसले गर्दा F को मान सधैं १ भन्दा ठूलो वा बराबर आउँछ।

📝 Question Statement (परीक्षाको प्रश्न):A cloud engineer evaluates latency consistency between Cloud Server Cluster A (n₁ = 10) and Cluster B (n₂ = 12). Sample variance of latency is S₁² = 28.5 (ms²) for Cluster A, and S₂² = 9.5 (ms²) for Cluster B. At α = 0.05, test whether the two server clusters have equal latency variability.
Given Data:Server Cluster A: n₁ = 10, S₁² = 28.5 ms²Server Cluster B: n₂ = 12, S₂² = 9.5 ms²Significance Level: α = 0.05

Step-by-Step Verified Solution:

1State Hypotheses

Formulate variance equality hypotheses.

👨‍🏫 Step Tip: H₀: σ₁² = σ₂² (दुवैको Variance बराबर छ), H₁: σ₁² ≠ σ₂² (Variance फरक छ)।
• Null Hypothesis (H₀): σ₁² = σ₂² (The two server clusters have equal variances).
• Alternative Hypothesis (H₁): σ₁² ≠ σ₂² (The two server clusters have unequal variances).
2Calculate F-Statistic
F = S₁² / S₂² (where S₁² > S₂²)

Divide the larger sample variance by the smaller sample variance.

👨‍🏫 Step Tip: ठूलो Variance (28.5) लाई माथि र सानो Variance (9.5) लाई तल राख्नुहोस्: F = 28.5 / 9.5 = 3.00।
• F = 28.5 / 9.5 = 3.00
F calculated = 3.00
3Determine Degrees of Freedom

Assign degrees of freedom for numerator and denominator.

👨‍🏫 Step Tip: Numerator df₁ = n₁ - 1 = 9; Denominator df₂ = n₂ - 1 = 11।
• Numerator df (v₁) = n₁ - 1 = 10 - 1 = 9
• Denominator df (v₂) = n₂ - 1 = 12 - 1 = 11
• Critical F from F-distribution table at α = 0.05 with (9, 11) df is 2.90
Final Step: Decision Rule & Academic ConclusionReject H₀
Calculated:3.00
Critical Table:2.90
Degrees of Freedom:v₁ = 9, v₂ = 11
Alpha (α):0.05
✍️ What to Write in the TU Exam Answer Sheet (Final Report):Since the calculated F-statistic (3.00) is greater than the critical value (2.90) at α = 0.05, we reject the null hypothesis. There is a statistically significant difference in latency variance between the two clusters; Cluster B operates with significantly greater consistency.
Exam Hall Verification Checklist (विद्यार्थीले जाँच्नुपर्ने कुराहरू):
  • Rule of thumb: Always place the larger variance in the numerator so F ≥ 1.
  • Keep numerator df and denominator df in correct order when checking the F-table.
Numerical 05

Spearman's Rank Order Correlation Coefficient (ρ)

Non-Parametric TestTU Model Numerical

Spearman's Rank Order Correlation (ρ)

🎯 कुन बेला यो Test प्रयोग गर्ने? (When to use):

जब Data Ordinal (Rank गरिएको) हुन्छ, वा Normal distribution नभएको Continuous data लाई क्रमबद्ध गरेर दुई Variable बीचको सम्बन्ध नाप्नुपर्ने हुन्छ।

👨‍🏫 शिक्षकको कोर लजिक (Teacher's Mental Model):

दुवै निर्णायक वा Variable को Rank बीचको फरक d = R₁ - R₂ निकाल्ने, त्यसको वर्ग d² निकाल्ने, र सूत्र ρ = 1 - [6Σd² / n(n² - 1)] मा हाल्ने।

📝 Question Statement (परीक्षाको प्रश्न):Two expert HCI usability judges independently ranked n = 8 software prototypes (A through H) on aesthetic appeal as follows: Prototypes: [A, B, C, D, E, F, G, H] Judge 1 Ranks (R₁): [1, 2, 3, 4, 5, 6, 7, 8] Judge 2 Ranks (R₂): [2, 1, 4, 3, 6, 5, 8, 7] Calculate Spearman's Rank Correlation Coefficient and interpret their level of agreement.
Given Data:Number of Prototypes (n): 8Judge 1 Ranks (R₁): 1, 2, 3, 4, 5, 6, 7, 8Judge 2 Ranks (R₂): 2, 1, 4, 3, 6, 5, 8, 7

Step-by-Step Verified Solution:

1Construct Difference of Ranks Table
d = R₁ - R₂ , d² = (R₁ - R₂)²

Calculate rank difference d = R₁ - R₂ and squared differences d².

👨‍🏫 Step Tip: ध्यान दिनुहोस्: Σd को जोड सधैं 0 हुनुपर्छ। यहाँ ८ वटा आइटम भएकाले n = 8, n² - 1 = 63 हुन्छ।
PrototypeJudge 1 (R₁)Judge 2 (R₂)d = R₁ - R₂d²
A12-11
B21+11
C34-11
D43+11
E56-11
F65+11
G78-11
H87+11
Total--Σd = 0Σd² = 8
2Apply Spearman's Formula
ρ = 1 - [ (6 × Σd²) / (n × (n² - 1)) ]

Substitute Σd² = 8 and n = 8 into the formula.

👨‍🏫 Step Tip: 6 × 8 = 48 लाई 8 × (64 - 1) = 504 ले भाग गर्नुहोस्: 48 / 504 = 0.0952। अनि 1 बाट घटाउनुहोस्।
• Numerator = 6 × Σd² = 6 × 8 = 48
• Denominator = n × (n² - 1) = 8 × (8² - 1) = 8 × (64 - 1) = 8 × 63 = 504
• Fraction = 48 / 504 ≈ 0.0952
• ρ = 1 - 0.0952 = +0.9048 ≈ +0.905
ρ = +0.905
Final Step: Decision Rule & Academic ConclusionVery Strong Positive Correlation
Calculated:+0.905
Critical Table:0.714 (at α = 0.05, n = 8)
Degrees of Freedom:n = 8
Alpha (α):0.05
✍️ What to Write in the TU Exam Answer Sheet (Final Report):The calculated Spearman Rank Correlation coefficient of +0.905 indicates a very high, positive, and statistically significant degree of agreement between the two HCI usability judges.
Exam Hall Verification Checklist (विद्यार्थीले जाँच्नुपर्ने कुराहरू):
  • Always verify that Σd = 0 in your exam calculation table.
  • Check that n² - 1 is calculated accurately before multiplying by n.
  • Interpret ρ within the range of -1 (perfect negative) to +1 (perfect positive).
Section 08

One-Way ANOVA (Analysis of Variance) & The Summary Table

ANOVA compares the means of three or more groups simultaneously by decomposing the total sum of squares into Between-Group (treatment effect) and Within-Group (error) variance:

Source of VariationSum of Squares (SS)Degrees of Freedom (df)Mean Square (MS)F-Ratio
Between Groups (Treatment)SSBk - 1MSB = SSB / (k - 1)F = MSB / MSW
Within Groups (Error)SSWN - kMSW = SSW / (N - k)
TotalSST = SSB + SSWN - 1--
🎯 TU Exam Tip:कहिलेकाहीँ TU परीक्षामा ANOVA को हिसाब नसोधेर केवल “ANOVA Summary Table को ढाँचा कोर्नुहोस्” भनेर ५ नम्बरमा सोधिन्छ। माथिको तालिका जस्ताको तस्तै कोर्नुहोस्।
Section 09

Machine Learning Approaches to Decision Making & Prediction

The syllabus highlights modern algorithmic approaches that transcend classical hypothesis testing when dealing with massive high-dimensional datasets:

Supervised Predictive Modeling

Trains models on historical labeled datasets to predict continuous numerical targets (Regression) or discrete categories (Classification). Evaluated via K-Fold Cross Validation, Precision, Recall, and ROC-AUC.

Unsupervised Data Mining & Clustering

Discovers hidden user segments, clusters of anomalous server intrusions, and latent thematic groupings without predefined target labels. Algorithms: K-Means, DBSCAN, Hierarchical Clustering.

High-Probability Exam Blueprint

TU Exam Model Q&A Bank — Unit 4

5 Past & Model Questions

💡 यी प्रश्नहरू विगतका TU परीक्षा तथा सिलेबसका मुख्य उद्देष्यहरूबाट संकलन गरिएका हुन्। उत्तरपुस्तिकामा Full Marks पाउन चाहिने मोडेल ढाँचा यहाँ प्रस्तुत गरिएको छ।

शिक्षकको छोटो सारांश (One-Minute Exam Recall):
अदालतको उदाहरण सम्झिनुहोस्: Type I Error (α) भनेको निर्दोषलाई जेल हाल्नु (सत्य H₀ लाई Reject गर्नु) हो। Type II Error (β) भनेको दोषीलाई प्रमाण नपुगेर छोडिदिनु (गलत H₀ लाई Accept गर्नु) हो।

Standard Model Answer Structure:

1. Type I Error (α - False Positive): Rejecting a True Null Hypothesis. Occurs when researcher concludes a new algorithm is superior when in reality the observed difference occurred purely by chance.

2. Type II Error (β - False Negative): Failing to reject a False Null Hypothesis. Occurs when researcher misses a genuinely effective algorithmic breakthrough because sample size was too small.

3. Courtroom Analogy: Type I is convicting an innocent citizen (False Alarm); Type II is acquitting a guilty criminal due to lack of evidence (Missed Detection).

4. Power of the Test (1 - β): The probability of correctly rejecting a false null hypothesis.

✍️ Essential Answer Keywords / Must-Draw Elements:
✓ Type I Error (α / False Positive)✓ Type II Error (β / False Negative)✓ Power of Test (1 - β)✓ 2x2 Decision Matrix Table
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