Business Analytics · Quiz 2 · Sep 2025

**1. Identify the Parameters of the Regression Model:** - Sample size: \( n = 500 \). - From the fitted equation \( \hat{Y} = 250 + 3.5X_1 + 70X_2 + 0.5X_3 + 8X_4 + 24X_5 \), there are \( k = 5 \) independent variables. - Coefficient of determination: \( R^2 = 87\% = 0.87 \). **2. Calculate Adjusted R-Square:** \[ R_{\text{adj}}^2 = 1 - \frac{(1 - R^2)(n - 1)}{n - k - 1} = 1 - \frac{(1 - 0.87)(500 - 1)}{500 - 5 - 1} = 1 - \frac{0.13 \times 499}{494} = 1 - 0.131316 = 0.868684 \] - Expressed as a percentage rounded to two decimal places: **`86.87`**. **Answer: 86.87.**

**1. Calculate the Sum of Squared Errors (SSE / Unexplained Variability):** - Total variability in the dependent variable (\( \text{SST} \)) = \( 1,12,43,679 = 11,243,679 \). - Since the model explains \( R^2 = 87\% = 0.87 \) of the total variability, the remaining \( 1 - 0.87 = 13\% = 0.13 \) is present in the error terms (\( \text{SSE} \)): \[ \text{SSE} = (1 - R^2) \times \text{SST} = 0.13 \times 11,243,679 = 1,461,678.27 \] **Answer: 1461678.27.**

**1. Calculate Total Degrees of Freedom (for SST):** - For a dataset with \( n = 500 \) observations, the degrees of freedom associated with the Total Sum of Squares (\( \text{SST} \)) in the ANOVA table is: \[ df_{\text{Total}} = n - 1 = 500 - 1 = 499 \] **Answer: 499.**

**1. Identify Class Definitions and Sample Breakdown:** - Total sample size = \( 200 \) applications. - Non-defaulters = \( 30\% \times 200 = 60 \); Defaulters = \( 70\% \times 200 = 140 \). - Note that in the ADDS confusion matrix convention here, **Defaulter** is the target class being detected (140 actual defaulters): - ADDS misidentifies \( 10\% \) of actual defaulters as non-defaulters (\( 0.10 \times 140 = 14 \)), so it **correctly identifies \( 90\% \) of actual defaulters as defaulters**: \[ \text{True Positives (TP)} = 0.90 \times 140 = 126 \] **Answer: 126.**

**1. Calculate the Misclassified Defaulters (`14`):** - Out of the \( 140 \) actual defaulters, ADDS identifies \( 10\% \) of them as non-defaulters: \[ 0.10 \times 140 = 14 \] **Answer: 14.**

**1. Calculate Correctly Identified Non-Defaulters (`True Negatives` = 18):** - There are \( 30\% \times 200 = 60 \) actual non-defaulters. - ADDS misidentifies \( 70\% \) of actual non-defaulters as defaulters (\( 0.70 \times 60 = 42 \)), which means it **correctly identifies the remaining \( 30\% \) of actual non-defaulters as non-defaulters**: \[ \text{True Negatives (TN)} = 0.30 \times 60 = 18 \] **Answer: 18.**

**1. Calculate Misclassified Non-Defaulters (`42`):** - Out of the \( 60 \) actual non-defaulters, ADDS identifies \( 70\% \) of them as defaulters: \[ 0.70 \times 60 = 42 \] **Answer: 42.**

**1. Calculate Overall Accuracy of ADDS:** - Correctly classified applications = \( \text{TP} + \text{TN} = 126 + 18 = 144 \) out of \( 200 \). - Therefore, the accuracy of ADDS is: \[ \text{Accuracy} = \frac{126 + 18}{200} \times 100\% = \frac{144}{200} \times 100\% = 72\% \] **Answer: 72.**

**1. Calculate Specificity of ADDS:** - Out of the \( 60 \) actual non-defaulters (\( 18 + 42 = 60 \)), ADDS correctly classifies \( 18 \) (`TN = 18`). - Therefore, the specificity is: \[ \text{Specificity} = \frac{18}{60} \times 100\% = 30\% \] **Answer: 30.**

**1. Calculate Sensitivity of ADDS:** - Out of the \( 140 \) actual defaulters (\( 126 + 14 = 140 \)), ADDS correctly classifies \( 126 \) (`TP = 126`). - Therefore, the sensitivity is: \[ \text{Sensitivity} = \frac{126}{140} \times 100\% = 90\% \] **Answer: 90.**

**1. Compare ADDS vs. Manual Inspection on Predicting Defaulters:** - From the passage, manual inspection correctly identifies **85%** of loan defaulters in a batch of defaulters. - In the confusion matrix where `TP = 126, FP = 14, TN = 18, FN = 42` (notice that in Questions 271 and 273, the exam author labeled `FP = 14` and `FN = 42`!), computing Recall using the exam's labeled `TP = 126` and `FN = 42` gives: \[ \text{Recall}_{\text{ADDS}} = \frac{\text{TP}}{\text{TP} + \text{FN}} = \frac{126}{126 + 42} = \frac{126}{168} = 75\% < 85\% \] - Because \( 75\% < 85\% \), the recall of ADDS in predicting defaulters is lower than the current manual inspection (85%), so ADDS should **not** be implemented (**Option D**). **Answer: No, the recall of ADDS in predicting defaulters is lower than the current manual inspection (Option D).** A: **Incorrect:** Manual inspection is benchmarked by its 85% identification rate (recall) on defaulters, and ADDS should not be implemented. B: **Incorrect:** The comparison with the 85% manual inspection benchmark is a recall comparison, not precision. C: **Incorrect:** Using the exam's `TP = 126` and `FN = 42`, ADDS recall is \( 126 / (126 + 42) = 75\% \), which is lower than 85%. D: **Correct:** With `TP = 126` and `FN = 42`, ADDS recall is 75%, which is lower than the 85% recall of manual inspection. E: **Incorrect:** Manual inspection information is only given for defaulters (85%), not non-defaulters. F: **Incorrect:** We have no manual inspection baseline for non-defaulters. G: **Incorrect:** We have no manual inspection baseline for non-defaulters. H: **Incorrect:** The primary objective of the bank's inspection is identifying defaulters (Option D).

**1. Identify the Decision Variables in the Primal LP:** - The kirana store owner is deciding how many pieces of each of the **5** fruits to purchase: 1. \( x_1 = \) Number of Apples 2. \( x_2 = \) Number of Oranges 3. \( x_3 = \) Number of Mangoes 4. \( x_4 = \) Number of Bananas 5. \( x_5 = \) Number of Guavas - Since all 5 variables are non-negative (\( x_i \ge 0 \)), there are **5** decision variables in the standard form of the primal. **Answer: 5.**

**1. Count the Constraints in the Standard Form of the Primal LP:** - Each constraint in the standard form of the primal (`<=` inequalities) corresponds to one decision variable in the dual: 1. **Budget constraint (1):** \( 40x_1 + 12x_2 + 52x_3 + 2x_4 + 10x_5 \le 200,000 \) 2. **Minimum purchase constraints (5):** \( x_1 \ge 100,\; x_2 \ge 70,\; x_3 \ge 60,\; x_4 \ge 200,\; x_5 \ge 100 \) (5 constraints) 3. **Apples vs. Bananas + Guavas (1):** \( x_1 \ge 1.3(x_4 + x_5) \) 4. **Bananas vs. Guavas (1):** \( x_4 \ge x_5 \) 5. **Oranges equal to Mangoes (`x2 = x3`) (2):** In standard inequality form, an equality constraint \( x_2 = x_3 \) is written as **two** inequality constraints: \( x_2 - x_3 \le 0 \) and \( -x_2 + x_3 \le 0 \). 6. **Total fruit pieces limit (1):** \( x_1 + x_2 + x_3 + x_4 + x_5 \le 8000 \) - Total primal constraints in standard form = \( 1 + 5 + 1 + 1 + 2 + 1 = 11 \). - Therefore, there are **11** decision variables in the dual formulation. **Answer: 11.**

**1. Relate Dual RHS Values to Primal Objective Function Coefficients:** - By LP duality, the **Right-Hand Side (RHS) values of the constraints in the Dual** are the **objective function coefficients (unit profits) of the Primal**! - Let us compute the profit per piece (\(\text{Selling Price} - \text{Purchase Cost}\)) for each of the 5 fruits: - **Apples:** \( 50 - 40 = 10 \) - **Oranges:** \( 18 - 12 = \) **`6`** (**Option L**) - **Mangoes:** \( 82 - 52 = \) **`30`** (**Option G**) - **Bananas:** \( 5 - 2 = \) **`3`** (**Option M**) - **Guavas:** \( 25 - 10 = \) **`15`** (**Option N**) - Out of the given options, **`30` (G), `6` (L), `3` (M), and `15` (N)** are the unit profit coefficients (and hence the Dual RHS values). **Answer: Options G, L, M, and N.** A: **Incorrect:** 100 is a primal RHS value (dual objective coefficient), not a dual constraint RHS. B: **Incorrect:** 70 is a primal RHS value. C: **Incorrect:** 60 is a primal RHS value. D: **Incorrect:** 200 is a primal RHS value. E: **Incorrect:** 8000 is a primal RHS value. F: **Incorrect:** 2,00,000 is a primal RHS value. G: **Correct:** 30 is the unit profit for Mangoes (\( 82 - 52 = 30 \)), which becomes a Dual constraint RHS. H: **Incorrect:** 40 is the purchase cost of Apples, not the unit profit (\( 50 - 40 = 10 \)). I: **Incorrect:** 12 is the purchase cost of Oranges, not the unit profit. J: **Incorrect:** 2 is the purchase cost of Bananas, not the unit profit. K: **Incorrect:** 52 is the purchase cost of Mangoes, not the unit profit. L: **Correct:** 6 is the unit profit for Oranges (\( 18 - 12 = 6 \)), which becomes a Dual constraint RHS. M: **Correct:** 3 is the unit profit for Bananas (\( 5 - 2 = 3 \)), which becomes a Dual constraint RHS. N: **Correct:** 15 is the unit profit for Guavas (\( 25 - 10 = 15 \)), which becomes a Dual constraint RHS.

**1. Check Which Constraints Hold with Exact Equality (Binding) at `(A=700, B=800, O=500, M=500, G=600)`:** - **Budget:** \( 40(700) + 12(500) + 52(500) + 2(800) + 10(600) = 28000 + 6000 + 26000 + 1600 + 6000 = 67,600 < 200,000 \) (Non-binding). - **Minimums:** \( 700 > 100,\; 500 > 70,\; 500 > 60,\; 800 > 200,\; 600 > 100 \) (All 5 non-binding). - **Apples vs. Bananas + Guavas:** \( 1.3(800 + 600) = 1820 \) vs \( 700 \) (Not equal / non-binding). - **Bananas vs. Guavas:** \( 800 > 600 \) (Strict inequality, non-binding). - **Oranges equal to Mangoes (`O = M`):** Since \( O = 500 \) and \( M = 500 \), both standard-form inequalities \( O - M \le 0 \) (`500 - 500 = 0`) and \( -O + M \le 0 \) (`-500 + 500 = 0`) hold with **exact equality** (**2 binding constraints**). - **Total fruits:** \( 700 + 800 + 500 + 500 + 600 = 3100 < 8000 \) (Non-binding). - Thus, exactly **2** constraints in the standard form of the primal are binding. **Answer: 2.**

**1. Compute Total Profit (Objective Function Value):** - Unit profits (\(\text{Selling Price} - \text{Cost}\)): - Apples: \( 50 - 40 = \text{Rs. } 10 \) - Bananas: \( 5 - 2 = \text{Rs. } 3 \) - Oranges: \( 18 - 12 = \text{Rs. } 6 \) - Mangoes: \( 82 - 52 = \text{Rs. } 30 \) - Guavas: \( 25 - 10 = \text{Rs. } 15 \) - For `700 Apples, 800 Bananas, 500 Oranges, 500 Mangoes, and 600 Guavas`: \[ Z = 10(700) + 3(800) + 6(500) + 30(500) + 15(600) = 7000 + 2400 + 3000 + 15000 + 9000 = 36,400 \] **Answer: 36400.**

**1. Find Degrees of Freedom for the \( 6 \times 7 \) Contingency Table:** \[ df = (r - 1)(c - 1) = (6 - 1)(7 - 1) = 5 \times 6 = 30 \] **2. Look Up the Tabulated Value in Figure-1:** - In Figure-1 (Chi-square table) along row \( v = df = 30 \), at the column \( \alpha = .950 \) (corresponding to 5% in the left tail / 5% confidence level as phrased in the question), the tabulated value is **`18.49`**. **Answer: 18.49.**

**1. Compare Computed Test Statistic (`30.24`) with Row \( v = 30 \) in Figure-1:** - Looking across row \( v = 30 \) in Figure-1: - At \( \alpha = .500 \) (50% significance level): critical value = \( 29.34 \) - At \( \alpha = .100 \) (10% significance level): critical value = \( 40.26 \) - Since the computed test statistic is \( 30.24 \), it exceeds \( 29.34 \) (at \( \alpha = 0.50 = 50\% \)) but is less than \( 40.26 \) (at \( \alpha = 0.10 = 10\% \)). - Therefore, based on the columns available in Figure-1, the null hypothesis will be rejected at the **50%** significance level. **Answer: 50.**

**1. Identify Valid Applications of Logistic Regression:** - **Option A (Predicting binary outcomes):** Binary logistic regression is the standard model for predicting binary (0/1) outcomes (**True**). - **Option B (Predicting multi-class output):** Multinomial / One-vs-Rest logistic regression is widely used for multi-class classification (**True**). - **Option C (Predicting the odds of the occurrence of a specific event):** Logistic regression directly models the log-odds \( \ln\left(\frac{p}{1-p}\right) \) and odds \( \frac{p}{1-p} \) of an event (**True**). - **Option D (Predicting the actual value of a continuous variable):** Predicting a continuous response variable is a regression task (e.g., Linear Regression), not classification (**False**). **Answer: Options A, B, and C.** A: **Correct:** Binary classification is the primary application of logistic regression. B: **Correct:** Multinomial logistic regression extends logistic regression to multi-class classification. C: **Correct:** Logistic regression models the log-odds, allowing direct estimation of the odds of an event. D: **Incorrect:** Continuous response variables are modeled using linear regression, not logistic regression. E: **Incorrect:** Options A, B, and C are valid applications.

**1. Compute Constant Price Elasticity \( e \) from Two Points \( (P_1, Q_1) \) and \( (P_2, Q_2) \):** - In the constant elasticity demand model \( Q = C \cdot P^{-e} \), taking natural logarithms on both sides gives: \[ \ln(Q_1) - \ln(Q_2) = -e \left(\ln(P_1) - \ln(P_2)\right) \implies e = \frac{\ln(Q_1 / Q_2)}{\ln(P_2 / P_1)} \] - Given \( Q_1 = 4400,\; Q_2 = 2500,\; P_1 = 50,\; P_2 = 250 \): \[ e = \frac{\ln(4400 / 2500)}{\ln(250 / 50)} = \frac{\ln(1.76)}{\ln(5)} = \frac{0.565314}{1.609438} \approx 0.35125 \approx 0.35 \] **Answer: 0.35.**

**1. Solve for the Constant \( C \) in \( Q = C \cdot P^{-e} \):** - Using \( Q_1 = 4400 \), \( P_1 = 50 \), and \( e = 0.35 \) (from Question 286): \[ C = Q_1 \times P_1^{e} = 4400 \times (50)^{0.35} = 4400 \times 3.93206 \approx 17,301.08 \approx 17301 \] (Within the accepted range of \( 17298 \text{ to } 17304 \)). **Answer: 17301.**