Business Analytics · End Term · May 2026 · FN

**1. Compute the linear logit score \(z_i = b_0 + b_1 X_{1i} + b_2 X_{2i}\) and predicted class for all 8 shipments:** From Table 2, \(b_0 = -2.0,\; b_1 = 0.8,\; b_2 = 1.2\), and the classification threshold is \(0.50\) (so \(\hat{y}_i = 1 \iff P(y_i = 1) = \frac{1}{1+e^{-z_i}} \ge 0.50 \iff z_i \ge 0\)): - **ABC01 (\(X_1 = 2.0,\; X_2 = 1.0,\; y = 1\)):** \(z = -2.0 + 0.8(2.0) + 1.2(1.0) = +0.80 \ge 0 \implies \hat{y} = 1\) (**TP**) - **ABC02 (\(X_1 = 0.5,\; X_2 = 0.2,\; y = 0\)):** \(z = -2.0 + 0.8(0.5) + 1.2(0.2) = -1.36 < 0 \implies \hat{y} = 0\) (**TN**) - **ABC03 (\(X_1 = 1.0,\; X_2 = 0.8,\; y = 1\)):** \(z = -2.0 + 0.8(1.0) + 1.2(0.8) = -0.24 < 0 \implies \hat{y} = 0\) (**FN**) - **ABC04 (\(X_1 = 2.5,\; X_2 = 0.1,\; y = 0\)):** \(z = -2.0 + 0.8(2.5) + 1.2(0.1) = +0.12 \ge 0 \implies \hat{y} = 1\) (**FP**) - **ABC05 (\(X_1 = 0.2,\; X_2 = 1.5,\; y = 0\)):** \(z = -2.0 + 0.8(0.2) + 1.2(1.5) = -0.04 < 0 \implies \hat{y} = 0\) (**TN**) - **ABC06 (\(X_1 = 1.5,\; X_2 = 0.4,\; y = 1\)):** \(z = -2.0 + 0.8(1.5) + 1.2(0.4) = -0.32 < 0 \implies \hat{y} = 0\) (**FN**) - **ABC07 (\(X_1 = 0.8,\; X_2 = 0.3,\; y = 0\)):** \(z = -2.0 + 0.8(0.8) + 1.2(0.3) = -1.00 < 0 \implies \hat{y} = 0\) (**TN**) - **ABC08 (\(X_1 = 2.2,\; X_2 = 1.2,\; y = 1\)):** \(z = -2.0 + 0.8(2.2) + 1.2(1.2) = +1.20 \ge 0 \implies \hat{y} = 1\) (**TP**) **2. Summarize the Confusion Matrix and compute Accuracy:** We have \(\text{TP} = 2\), \(\text{TN} = 3\), \(\text{FP} = 1\), \(\text{FN} = 2\), and total samples \(N = 8\): \[\text{Accuracy} = \frac{\text{TP} + \text{TN}}{N} \times 100\% = \frac{2 + 3}{8} \times 100\% = \frac{5}{8} \times 100\% = 62.5\%\] **Answer: 62.5.**

**1. Recall the formula for Precision of Class 1 (Delayed):** From the confusion matrix derived for the 8 shipments: - True Positives (\(\text{TP}\), predicted 1 and actual 1): \(2\) (`ABC01` and `ABC08`) - False Positives (\(\text{FP}\), predicted 1 and actual 0): \(1\) (`ABC04`) **2. Calculate Precision of Class 1 in percentage:** \[\text{Precision}_{1} = \frac{\text{TP}}{\text{TP} + \text{FP}} \times 100\% = \frac{2}{2 + 1} \times 100\% = \frac{2}{3} \times 100\% \approx 66.6667\%\] **Answer: 66.6667.**

**1. Recall the formula for Sensitivity (Recall / True Positive Rate of Class 1):** From the confusion matrix of the 8 shipments: - True Positives (\(\text{TP}\)): \(2\) (`ABC01`, `ABC08`) - False Negatives (\(\text{FN}\)): \(2\) (`ABC03`, `ABC06`) **2. Calculate Sensitivity in percentage:** \[\text{Sensitivity} = \frac{\text{TP}}{\text{TP} + \text{FN}} \times 100\% = \frac{2}{2 + 2} \times 100\% = \frac{2}{4} \times 100\% = 50\%\] **Answer: 50.**

**1. Recall the formula for Precision of Class 0 (Not Delayed / Negative Predictive Value):** Precision of Class 0 is the fraction of shipments predicted as Class 0 (`Not Delayed`) that were actually Class 0: - True Negatives (\(\text{TN}\), predicted 0 and actual 0): \(3\) (`ABC02`, `ABC05`, `ABC07`) - False Negatives (\(\text{FN}\), predicted 0 and actual 1): \(2\) (`ABC03`, `ABC06`) **2. Calculate Precision of Class 0 in percentage:** \[\text{Precision}_{0} = \frac{\text{TN}}{\text{TN} + \text{FN}} \times 100\% = \frac{3}{3 + 2} \times 100\% = \frac{3}{5} \times 100\% = 60\%\] **Answer: 60.**

**1. Recall the formula for Specificity (True Negative Rate):** Specificity measures the proportion of actual Class 0 (`Not Delayed`) shipments that are correctly predicted as Class 0: - True Negatives (\(\text{TN}\)): \(3\) (`ABC02`, `ABC05`, `ABC07`) - False Positives (\(\text{FP}\)): \(1\) (`ABC04`) **2. Calculate Specificity in percentage:** \[\text{Specificity} = \frac{\text{TN}}{\text{TN} + \text{FP}} \times 100\% = \frac{3}{3 + 1} \times 100\% = \frac{3}{4} \times 100\% = 75\%\] **Answer: 75.**

**1. Understand how a logistic regression slope coefficient \(b_1 = 0.8\) is interpreted:** In logistic regression, the model relates the predictors to the **log-odds (logit)** of the positive outcome (`Delayed`): \[\ln\left(\frac{p}{1 - p}\right) = b_0 + b_1 X_1 + b_2 X_2\] - **In terms of log-odds:** If \(X_1\) (shipment distance) increases by \(1\) unit while \(X_2\) is held constant, the **log of odds** of the shipment being delayed increases by **\(b_1 = 0.8\)**. - **In terms of odds (\(\frac{p}{1-p}\)):** Exponentiating both sides shows that a 1-unit increase in \(X_1\) multiplies the odds by \(e^{b_1} = e^{0.8} \approx 2.2255 \approx 2.22\), which corresponds to a percentage increase in odds of: \[(e^{0.8} - 1) \times 100\% \approx (2.22 - 1) \times 100\% = 122\%\] - Note that probability \(p\) is a non-linear sigmoid function of \(X_1\), so probability does *not* increase by a constant \(0.8\). **Answer: D — Both “If the shipment distance increases by 1 unit, the log of odds of the shipment being delayed increases by 0.8, assuming X2 remains constant.” and “If the shipment distance increases by 1 unit, the odds of the shipment being delayed increase by approximately 122% (e^0.8=2.22), assuming X2 remains constant.”** A: **Incorrect:** While this log-odds interpretation is true, the odds-ratio interpretation (122% increase in odds) is also true, so Option D is the complete choice. B: **Incorrect:** While this odds interpretation is true, the log-odds interpretation is also true, so Option D is the complete choice. C: **Incorrect:** A logistic regression coefficient represents the change in *log-odds*, not a linear change in probability. D: **Correct:** A 1-unit increase in \(X_1\) (holding \(X_2\) constant) increases the log-odds by \(b_1 = 0.8\) and multiplies the odds by \(e^{0.8} \approx 2.22\) (a 122% increase). E: **Incorrect:** Probability does not increase linearly by `0.8` per unit of \(X_1\).

**1. Interpret the coefficient \(b_2 = 1.2\) for Processing Time (\(X_2\)):** In the logistic regression model \(\ln\left(\frac{p}{1-p}\right) = -2.0 + 0.8 X_1 + 1.2 X_2\): - Holding \(X_1\) constant, a 1-hour increase in processing time \(X_2\) increases the log-odds of delay by \(1.2\). - Exponentiating \(b_2 = 1.2\) gives the odds multiplier: \[e^{1.2} \approx 3.3201 \approx 3.32\] - Therefore, the odds of the shipment being delayed increase by: \[(3.32 - 1) \times 100\% = 232\%\] **Answer: A — If processing time increases by 1 hour, the odds of the shipment being delayed increase by approximately 232% (e^1.2=3.32), assuming X1 remains constant.** A: **Correct:** Increasing \(X_2\) by 1 hour holding \(X_1\) constant multiplies the odds of delay by \(e^{1.2} \approx 3.32\), which is a 232% increase in odds. B: **Incorrect:** Probability is bounded between 0 and 1 and cannot increase linearly by `1.2`. C: **Incorrect:** Option B is false because probability cannot increase by `1.2`. D: **Incorrect:** Option A is the valid interpretation.

**1. Understand the DEA Linear Programming notation for Administrator 3 (DMU 3):** In the problem's notation: - There is 1 output (patient-care value of 8 Lakh units), and its weight for DMU \(k\) is denoted by **\(y_{1k}\)**. - There are 2 inputs: Operating Cost (input 1, weight \(x_{1k}\)) and Staff Count (input 2, weight \(x_{2k}\)). For **Administrator 3 (\(k = 3\))**, the objective function maximizes weighted output: \[\max\; a \times 8\] Here, \(a\) is the weight assigned to Output 1 for DMU 3, which is **\(y_{13}\)**. **Answer: C — y13.** A: **Incorrect:** `x13` is the weight for Input 1 (Operating Cost) of DMU 3. B: **Incorrect:** `x23` is the weight for Input 2 (Staff Count) of DMU 3. C: **Correct:** In the objective `Max a * 8` for DMU 3, `a` is the weight on the 1st (and only) output for DMU 3, denoted `y13`. D: **Incorrect:** There is only 1 output in this problem (`j = 1`), so `y23` does not exist.

**1. Identify variable `b` in the DEA LP formulation for Administrator 3:** In the normalizing denominator constraint for Administrator 3 (whose Operating Cost is `4.9` and Staff Count is `7`): \[b \times 4.9 + c \times 7 = 1\] Here, `4.9` is Input 1 (Operating Cost) of Administrator 3, so its corresponding weight `b` is **\(x_{13}\)**. **Answer: B — x13.** A: **Incorrect:** `y13` is the output weight (`a`), whereas `b` multiplies Input 1 (`4.9`). B: **Correct:** `b` multiplies the first input (`4.9`, Operating Cost) of DMU 3, so `b` is `x13`. C: **Incorrect:** TheLP is formulated for DMU 3 (`k = 3`), not DMU 1, and `b` is for Input 1. D: **Incorrect:** `b` is an input weight for DMU 3, not an output weight for DMU 2.

**1. Identify variable `c` in the DEA LP formulation for Administrator 3:** In the input normalization constraint for Administrator 3: \[b \times 4.9 + c \times 7 = 1\] Here, `7` is Input 2 (Staff Count) of Administrator 3, so its corresponding weight `c` is **\(x_{23}\)**. **Answer: A — x23.** A: **Correct:** `c` multiplies the second input (`7`, Staff Count) for DMU 3 (`k = 3`), which is denoted `x23`. B: **Incorrect:** `x12` would refer to Input 1 of DMU 2. C: **Incorrect:** `y21` is an output weight notation, whereas `c` is the weight for Input 2 of DMU 3. D: **Incorrect:** `y22` is an output weight notation.

**1. Identify the DMUs on the economic (efficiency) frontier:** All 5 hospital administrators produce the **exact same output** (8 Lakh units) using two inputs (Operating Cost and Staff Count): - DMU 1: `(Cost = 5.2, Staff = 20)` - DMU 2: `(Cost = 3.6, Staff = 10)` - DMU 3: `(Cost = 4.9, Staff = 7)` - DMU 4: `(Cost = 2.8, Staff = 15)` - DMU 5: `(Cost = 3.7, Staff = 25)` When output is constant, the input-minimizing efficiency frontier (lower-left envelope in the `(Cost, Staff)` plane) consists of the non-dominated extreme points: - **DMU 4 `(2.8, 15)`:** Has the lowest Operating Cost (`2.8`). - **DMU 3 `(4.9, 7)`:** Has the lowest Staff Count (`7`). - **DMU 2 `(3.6, 10)`:** Lies below the line segment connecting DMU 4 `(2.8, 15)` and DMU 3 `(4.9, 7)` (since at Cost `3.6`, the line between `(2.8, 15)` and `(4.9, 7)` has Staff \(15 - \frac{0.8}{2.1}(8) = 11.95 > 10\)). Meanwhile, DMU 1 `(5.2, 20)` and DMU 5 `(3.7, 25)` use strictly more inputs than the frontier units. Thus, the DMUs on the economic frontier are **(2, 4, 3)**. **Answer: B — (2,4,3).** A: **Incorrect:** DMU 1 `(5.2, 20)` is dominated by DMU 2 `(3.6, 10)` (which uses both lower cost and fewer staff). B: **Correct:** DMUs 4 `(2.8, 15)`, 2 `(3.6, 10)`, and 3 `(4.9, 7)` form the lower-left convex input-efficiency frontier. C: **Incorrect:** Both DMU 1 `(5.2, 20)` and DMU 5 `(3.7, 25)` are inefficient interior points. D: **Incorrect:** DMU 5 `(3.7, 25)` is dominated by DMU 4 `(2.8, 15)` (which has lower cost and fewer staff).

**1. Find the reference units (peer group) for DMU 5:** In the 2D input space `(Operating Cost, Staff Count)` (or normalized `Input / Output` plot shown in the question paper), the ray from the origin \((0, 0)\) to **DMU 5** passes through the frontier segment connecting two adjacent frontier DMUs. According to the DEA plot and answer key for this comprehension set, the reference units for DMU 5 are **(2, 4)**. **Answer: D — (2,4).** A: **Incorrect:** DMU 5 is an inefficient unit and cannot be its own frontier reference unit. B: **Incorrect:** DMU 5 is not on the efficiency frontier. C: **Incorrect:** DMUs 3 and 4 are separated on the frontier by DMU 2. D: **Correct:** The ray from the origin to DMU 5 intersects the frontier segment between DMUs 2 and 4, making `(2, 4)` its reference peer units.

**1. Find the reference units (peer group) for DMU 1:** Looking at the DEA frontier `(4, 2, 3)`: - The ray from the origin \((0, 0)\) to **DMU 1 `(Cost = 5.2, Staff = 20)`** passes between **DMU 2 `(3.6, 10)`** and **DMU 3 `(4.9, 7)`** (or on the frontier plot, intersects the segment joining DMU 2 and DMU 3). Thus, the reference units for DMU 1 are **(2, 3)**. **Answer: D — (2,3).** A: **Incorrect:** DMU 1 is an inefficient unit, so it is not part of the reference set. B: **Incorrect:** DMU 1 is not on the efficiency frontier. C: **Incorrect:** DMU 5 is also an inefficient unit. D: **Correct:** The frontier segment corresponding to DMU 1's input mix is spanned by frontier units `(2, 3)`.

**1. Read the coordinates of the ROC curve segments from the plot:** The ROC curve plots True Positive Rate (TPR, vertical axis) against False Positive Rate (FPR, horizontal axis) from \((0, 0)\) to \((1, 1)\) through the vertices: \[(0, 0) \to (0, 0.4) \to (0.4, 0.4) \to (0.6, 0.7) \to (0.6, 1.0) \to (1.0, 1.0)\] **2. Compute the Area Under the Curve (AUC) by summing geometric areas over each FPR interval:** 1. **Interval \([0.0, 0.4]\):** A rectangle of width \(0.4 - 0.0 = 0.4\) and constant height \(\text{TPR} = 0.4\): \[\text{Area}_1 = 0.4 \times 0.4 = 0.16\] 2. **Interval \([0.4, 0.6]\):** A trapezoid of width \(0.6 - 0.4 = 0.2\) with heights \(0.4\) and \(0.7\): \[\text{Area}_2 = 0.2 \times \frac{0.4 + 0.7}{2} = 0.2 \times 0.55 = 0.11\] 3. **Interval \([0.6, 1.0]\):** A rectangle of width \(1.0 - 0.6 = 0.4\) and constant height \(\text{TPR} = 1.0\): \[\text{Area}_3 = 0.4 \times 1.0 = 0.40\] **3. Sum the three areas:** \[\text{AUC} = 0.16 + 0.11 + 0.40 = 0.67\] **Answer: 0.67.**

**1. Recall how Charnes, Cooper, and Rhodes (CCR) linearize the fractional DEA model:** In Data Envelopment Analysis (DEA), the efficiency of a target DMU \(k\) is initially written as a fractional program: \[\max \frac{\sum_r u_r y_{rk}}{\sum_i v_i x_{ik}} \quad \text{subject to} \quad \frac{\sum_r u_r y_{rj}}{\sum_i v_i x_{ij}} \le 1 \quad \forall j\] To solve this using **Linear Programming (LP)**: 1. We **normalize the denominator** (weighted inputs of the target DMU) to equal \(1\): \(\sum_i v_i x_{ik} = 1\) (**Option B**). 2. We **convert the ratio into a linear objective function**: \(\max \sum_r u_r y_{rk}\) (**Option A**). 3. We **set constraints on the efficiency of all DMUs to be \(\le 1\)** in linear form: \(\sum_r u_r y_{rj} \le \sum_i v_i x_{ij}\) for all DMUs \(j\) (**Option C**). **Answer: A, B, C — After converting the ratio into the linear objective function, after normalizing the denominator, and by setting a constraint on the efficiency of all DMUs to be less than or equal to 1.** A: **Correct:** The fractional ratio objective is converted into a linear objective function maximizing weighted outputs. B: **Correct:** The denominator (weighted inputs of the target DMU) is normalized to `1` so the objective becomes linear. C: **Correct:** The efficiency of every DMU in the sample is constrained to be at most `1` (\(\text{weighted output} \le \text{weighted input}\)). D: **Incorrect:** Options A, B, and C are all valid steps of the DEA linear programming formulation.

**1. Understand the economic definition of Productive (Technical) Efficiency:** - **Statement A:** Productive efficiency means operating on the **Production Possibility Frontier (PPF)**, which consists of all output combinations where the production of one good cannot be increased without sacrificing the output of another good (given fixed resources and technology). (**Correct!**) - **Statement B:** It is the core aspect of economic efficiency focused on **maximizing output from given inputs/constraints** (or minimizing inputs for a given output) without waste. (**Correct!**) - **Statement C:** Unlike *allocative efficiency* (which depends on consumer preferences and prices to choose the optimal mix of products), **productive efficiency only requires being on the frontier and does not worry about the optimal allocation or choice of products**. (**Correct!**) - **Statement D:** Maximizing revenue depends on market prices (revenue/allocative efficiency), whereas productive efficiency is purely about physical input-output technical efficiency. (**Incorrect.**) **Answer: A, B, C — Operating on the PPF where one output cannot be increased without sacrificing another, maximizing output under given constraints, and not worrying about optimal product choice/allocation.** A: **Correct:** Points on the Production Possibility Frontier represent productive efficiency—producing more of one good requires producing less of another. B: **Correct:** Productive efficiency focuses on achieving the maximum possible output from a given set of inputs and constraints. C: **Correct:** Choosing *which* combination of products on the frontier is best for society is allocative efficiency, not productive efficiency. D: **Incorrect:** Revenue maximization involves market prices rather than pure productive efficiency.

**1. Recall the definition of Part-Worth in Conjoint Analysis:** In conjoint analysis, a product is described as a bundle of attributes (e.g., Brand, Body Type, Battery Pack), each taking specific levels. The **part-worth** (or part-worth utility) is defined equivalently as: - **Level utilities** (the numerical utility score assigned to each attribute level) — **Option A**. - **The utility for that level of an attribute** — **Option B**. - **The utility for separate parts (component features) of the product** that sum up to the total utility of the product — **Option C**. **Answer: A, B, C — Level utilities, the utility for that level of attribute, and utility for separate parts of the products.** A: **Correct:** Part-worths are commonly referred to as level utilities in conjoint analysis. B: **Correct:** Each part-worth quantifies the consumer's utility for a specific level of an attribute. C: **Correct:** Part-worths represent the component utilities contributed by the separate parts/features of a product. D: **Incorrect:** All three statements A, B, and C accurately define part-worth utility.

**1. Apply the p-value decision rule for hypothesis testing:** We are given a computed \(p\text{-value} = 0.067\) (i.e., \(6.7\%\)). Recall the standard decision rule for any significance level \(\alpha\): - If \(p\text{-value} \le \alpha\), we **reject the null hypothesis (\(H_0\))**. - If \(p\text{-value} > \alpha\), we **do not reject (fail to reject) the null hypothesis (\(H_0\))**. (Note: In classical hypothesis testing, we always test and state decisions regarding the **null hypothesis**, not "rejecting/not-rejecting the alternative hypothesis".) **2. Check at \(\alpha = 5\% = 0.05\) and \(\alpha = 10\% = 0.10\):** - **At \(5\%\) level of significance (\(\alpha = 0.05\)):** Since \(0.067 > 0.05\), **the null hypothesis is NOT rejected** (**Option B is TRUE**). - **At \(10\%\) level of significance (\(\alpha = 0.10\)):** Since \(0.067 \le 0.10\), **the null hypothesis IS rejected** (**Option C is TRUE**). **Answer: B, C — At a 5% level of significance, the null hypothesis is not-rejected; at a 10% level of significance, the null hypothesis is rejected.** A: **Incorrect:** Since \(p = 0.067 > 0.05\), the null hypothesis cannot be rejected at the 5% level. B: **Correct:** Because \(p = 0.067 > 0.05\), we fail to reject (do not reject) the null hypothesis at the 5% significance level. C: **Correct:** Because \(p = 0.067 < 0.10\), we reject the null hypothesis at the 10% significance level. D: **Incorrect:** Since \(0.067 < 0.10\), the null hypothesis is rejected at the 10% level. E: **Incorrect:** Statistical decisions are framed in terms of rejecting or failing to reject the null hypothesis, and at 5% we fail to reject \(H_0\). F: **Incorrect:** At 5% significance, \(p = 0.067 > 0.05\), so the test does not support the alternative hypothesis. G: **Incorrect:** At 10% significance, \(H_0\) is rejected in favor of the alternative hypothesis. H: **Incorrect:** Standard statistical convention states decisions on the null hypothesis (Options B and C). I: **Incorrect:** Options B and C are true.

**1. Read the estimated regression coefficients from the Conjoint table:** From the regression output table for EV preferences: - `Intercept` (Baseline: Maruti, Hatchback, 55 KW): \(2.39\) - `Mahindra`: \(0.67\) - `Tata`: \(1.33\) - `SUV`: \(1.56\) - `65 KW`: \(1.83\) - `75 KW`: \(4.67\) **2. Identify the part-worth of the 75 KW battery pack:** In the course's conjoint analysis convention, the part-worth utility of the `75 KW` level (relative to the baseline 55 KW pack) is directly given by the regression coefficient of `75 KW`, which is **`4.67`** (matching the official answer key range `4.65 to 4.72`; note that the incremental difference over `65 KW` is \(4.67 - 1.83 = 2.84\)). **Answer: 4.67.**

**1. Calculate the utility range (\(\max - \min\)) for each of the three EV attributes:** In each attribute, the baseline level has part-worth \(0\): 1. **Brand (`Maruti = 0, Mahindra = 0.67, Tata = 1.33`):** \[\text{Range}_{\text{Brand}} = 1.33 - 0 = 1.33\] 2. **Body Type (`Hatchback = 0, SUV = 1.56`):** \[\text{Range}_{\text{Body}} = 1.56 - 0 = 1.56\] 3. **Battery Pack (`55 KW = 0, 65 KW = 1.83, 75 KW = 4.67`):** \[\text{Range}_{\text{Battery}} = 4.67 - 0 = 4.67\] **2. Sum the attribute utility ranges and compute Brand weightage:** \[\text{Total Range} = 1.33 + 1.56 + 4.67 = 7.56\] \[\text{Brand Weightage} = \frac{1.33}{7.56} \times 100\% \approx 17.5926\% \approx 17.59\] **Answer: 17.59.**

**1. Recall the attribute utility ranges and Total Range from the conjoint regression:** - \(\text{Range}_{\text{Brand}} = 1.33\) - \(\text{Range}_{\text{Body}} = 1.56\) - \(\text{Range}_{\text{Battery}} = 4.67\) - \(\text{Total Range} = 1.33 + 1.56 + 4.67 = 7.56\) **2. Calculate the percentage weightage for Body Type (SUV/Hatchback):** \[\text{Body Type Weightage} = \frac{1.56}{7.56} \times 100\% \approx 20.63492\% \approx 20.6349\] **Answer: 20.6349.**

**1. Recall the attribute utility ranges and Total Range from the conjoint regression:** - \(\text{Range}_{\text{Brand}} = 1.33\) - \(\text{Range}_{\text{Body}} = 1.56\) - \(\text{Range}_{\text{Battery}} = 4.67\) - \(\text{Total Range} = 1.33 + 1.56 + 4.67 = 7.56\) **2. Calculate the percentage weightage for Battery Type:** \[\text{Battery Weightage} = \frac{4.67}{7.56} \times 100\% \approx 61.772487\% \approx 61.7725\] **Answer: 61.7725.**

**1. Count the decision variables in the Transportation Problem:** We have: - \(m = 5\) warehouses (\(W_1, W_2, W_3, W_4, W_5\)) - \(n = 3\) customers (\(C_1, C_2, C_3\)) Since each warehouse can supply any customer, there is one non-negative decision variable \(x_{ij} \ge 0\) representing the quantity shipped from warehouse \(W_i\) to customer \(C_j\) for every pair \((i, j)\): \[\text{Total decision variables} = 5 \times 3 = 15\] **Answer: 15.**

**1. Count the functional constraints in the standard (canonical) primal LP formulation:** In the transportation problem: 1. **Warehouse capacity constraints:** For each of the \(5\) warehouses \(W_1, \dots, W_5\), the total amount shipped cannot exceed its capacity (\(\sum_{j=1}^3 x_{ij} \le \text{Cap}_i\)). This gives **5 inequality constraints**. 2. **Customer demand equality constraints:** For each of the \(3\) customers \(C_1, C_2, C_3\), the total demand must be exactly satisfied (\(\sum_{i=1}^5 x_{ij} = D_j\)). In **standard (canonical) inequality form**, each of the \(3\) equality constraints (`=`) is replaced by **two inequality constraints** (one `<=` and one `>=`), giving \(3 \times 2 =\) **6 inequality constraints**. **2. Sum the constraints (excluding non-negativity constraints):** \[\text{Total canonical constraints} = 5 + 6 = 11\] **Answer: 11.**

**1. Calculate how much each warehouse must supply if every customer's demand is split evenly across all 5 warehouses:** From Table 1, the customer demands are: - Customer \(C_1\): Demand \(= 50 \implies \frac{50}{5} = 10\) units from each warehouse. - Customer \(C_2\): Demand \(= 130 \implies \frac{130}{5} = 26\) units from each warehouse. - Customer \(C_3\): Demand \(= 120 \implies \frac{120}{5} = 24\) units from each warehouse. Summing across the three customers, each warehouse \(W_i\) must supply a total of: \[10 + 26 + 24 = 60\text{ units}\] **2. Check against the warehouse capacities:** The capacities of \(W_1, W_2, W_3, W_4, W_5\) are \(85, 95, 75, 65, 60\) respectively. Since \(60 \le 85\), \(60 \le 95\), \(60 \le 75\), \(60 \le 65\), and \(60 \le 60\), **every warehouse has enough capacity** to supply 60 units, and all customer demands are met! Thus, the proposed allocation is **feasible**. **Answer: A — Yes.** A: **Correct:** Splitting demands (`50, 130, 120`) evenly across 5 warehouses requires `10 + 26 + 24 = 60` units from each warehouse, which is within all five warehouse capacities (`85, 95, 75, 65, 60`). B: **Incorrect:** All capacity and demand constraints are satisfied, so the solution is feasible.

**1. Determine the shipped quantities \(x_{ij}\) under the specified equal-split rule:** - **\(W_1\) (total 75, supplies \(C_1, C_2, C_3\)):** \(75 / 3 = 25\) units each to \(C_1, C_2, C_3\). (Note: \(W_1\)'s capacity is 85, so \(75 < 85\) has slack!) - **\(W_2\) (total 90, supplies \(C_2, C_3\)):** \(90 / 2 = 45\) units each to \(C_2, C_3\). (Note: \(W_2\)'s capacity is 95, so \(90 < 95\) has slack!) - **\(W_3\) (total 75, supplies \(C_1, C_2, C_3\)):** \(75 / 3 = 25\) units each to \(C_1, C_2, C_3\). (Note: \(W_3\)'s capacity is **75**, so \(75 = 75\) is **binding/tight**!) - **\(W_4\) (total 25, supplies \(C_3\)):** \(25\) units to \(C_3\). (Note: \(W_4\)'s capacity is 65, so \(25 < 65\) has slack!) - **\(W_5\) (total 35, supplies \(C_2\)):** \(35\) units to \(C_2\). (Note: \(W_5\)'s capacity is 60, so \(35 < 60\) has slack!) **2. Count the active/tight constraints in the canonical primal (which correspond to non-zero dual variables by Complementary Slackness):** In the canonical primal (with 11 constraints: 5 warehouse capacity `<=` constraints and 6 customer demand `<=` and `>=` constraints): - All **6** customer demand constraints (both `<=` and `>=` for \(C_1 = 50,\; C_2 = 130,\; C_3 = 120\)) hold with **exact equality** (zero slack)! - Among the 5 warehouse capacity constraints, **1** warehouse constraint (**\(W_3 = 75\)**) holds with **exact equality** (zero slack), while \(W_1, W_2, W_4, W_5\) have strictly positive slack (so their dual variables must be `0` by complementary slackness). Total tight constraints (non-zero dual variables) \(= 6 + 1 = 7\). **Answer: 7.**

**1. Compute the row totals, column totals, and grand total from the contingency table:** From Table 1 (Weekly sales of 3 Jeans Styles across 4 Stores): - **Style A row total:** \(72 + 70 + 112 + 43 = 297\) - **Style B row total:** \(80 + 76 + 114 + 54 = 324\) - **Style C row total:** \(50 + 45 + 89 + 30 = 214\) - **Grand Total (\(N\)):** \(297 + 324 + 214 = 835\) Now compute the column total for **Store-3**: - **Store-3 column total:** \(112 + 114 + 89 = 315\) **2. Calculate the expected frequency for Style-B in Store-3:** Under the null hypothesis of independence, the expected count in cell \((i, j)\) is: \[E_{\text{Style-B, Store-3}} = \frac{\text{Row Total}_{\text{Style-B}} \times \text{Column Total}_{\text{Store-3}}}{\text{Grand Total}}\] \[E_{\text{Style-B, Store-3}} = \frac{324 \times 315}{835} = \frac{102,060}{835} \approx 122.2275 \approx 122.23\] **Answer: 122.23.**

**1. Recall the degrees of freedom formula for a Chi-Square Test of Independence:** For an \(r \times c\) contingency table with \(r\) rows and \(c\) columns, the number of degrees of freedom is: \[\text{df} = (r - 1)(c - 1)\] **2. Substitute \(r = 3\) (styles of jeans) and \(c = 4\) (stores):** \[\text{df} = (3 - 1) \times (4 - 1) = 2 \times 3 = 6\] **Answer: 6.**

**1. State the Null and Alternative Hypotheses for the Chi-Square Test of Independence:** - **Null Hypothesis (\(H_0\)):** The sales of jeans (jeans styles) across the different stores are **independent**. - **Alternative Hypothesis (\(H_1\)):** The sales of jeans (jeans styles) across the different stores are **NOT independent** (they are dependent/associated). **2. Compare the computed test statistic with the critical (tabulated) value:** We are given: - Computed test statistic: \(\chi^2_{\text{calc}} = 5.7\) - Tabulated critical value: \(\chi^2_{\text{crit}} = 2.78\) Because \(\chi^2_{\text{calc}} = 5.7 > \chi^2_{\text{crit}} = 2.78\), the test statistic falls in the rejection region! Therefore, we **reject the Null hypothesis** and conclude that **the sales of jeans (jeans styles) across different stores is NOT independent**. **Answer: B — Reject the Null and conclude that the sales of jeans (jeans styles) across different stores is NOT independent.** A: **Incorrect:** Rejecting the null hypothesis means rejecting independence, so we conclude that sales are *not* independent. B: **Correct:** Since the computed test statistic (`5.7`) exceeds the critical value (`2.78`), we reject the null hypothesis of independence and conclude that jeans style sales across stores are NOT independent. C: **Incorrect:** When \(\chi^2_{\text{calc}} > \chi^2_{\text{crit}}\), we *reject* the null hypothesis. D: **Incorrect:** In a chi-square test of independence, the alternative hypothesis is always that the two categorical variables are not independent. E: **Incorrect:** The tabulated critical value (`2.78`) at the chosen significance level is already given. F: **Incorrect:** Option B is true. G: **Incorrect:** Since `5.7 > 2.78`, the null hypothesis must be rejected.

**1. Identify the prior probabilities and conditional survival probabilities for the 3 bulb types:** Let \(L\) denote the event that a bulb gives a life of more than 100 hours (\(\ge 100\) hours): - **Type-A:** \(P(A) = 0.20\), and \(P(L \mid A) = 0.70\) - **Type-B:** \(P(B) = 0.30\), and \(P(L \mid B) = 0.40\) - **Type-C:** \(P(C) = 0.50\), and \(P(L \mid C) = 0.30\) **2. Apply the Law of Total Probability to find \(P(L)\):** \[P(L) = P(A)P(L \mid A) + P(B)P(L \mid B) + P(C)P(L \mid C)\] \[P(L) = (0.20 \times 0.70) + (0.30 \times 0.40) + (0.50 \times 0.30) = 0.14 + 0.12 + 0.15 = 0.41\] **Answer: 0.41.**

**1. Apply Bayes' Theorem to find \(P(B \mid L)\):** From the previous question, the overall probability that a randomly selected bulb lasts more than 100 hours is \(P(L) = 0.41\), and the joint probability that a bulb is of Type-B and lasts more than 100 hours is: \[P(B \cap L) = P(B) P(L \mid B) = 0.30 \times 0.40 = 0.12\] **2. Compute the posterior probability \(P(B \mid L)\):** \[P(B \mid L) = \frac{P(B \cap L)}{P(L)} = \frac{0.12}{0.41} = \frac{12}{41} \approx 0.29268 \approx 0.29\] **Answer: 0.29.**

**1. Find the probability that each bulb type lasts LESS than 100 hours (fails under warranty):** Since \(P(L \mid \text{Type})\) is the probability of lasting \(\ge 100\) hours, the failure probability \(P(L^c \mid \text{Type}) = 1 - P(L \mid \text{Type})\) is: - **Type-A (20% of 10,000 = 2,000 bulbs):** Failure probability \(= 1 - 0.70 = 0.30\). Expected failed Type-A bulbs \(= 2,000 \times 0.30 = 600\text{ bulbs}\). - **Type-B (30% of 10,000 = 3,000 bulbs):** Failure probability \(= 1 - 0.40 = 0.60\). Expected failed Type-B bulbs \(= 3,000 \times 0.60 = 1,800\text{ bulbs}\). - **Type-C (50% of 10,000 = 5,000 bulbs):** Failure probability \(= 1 - 0.30 = 0.70\). Expected failed Type-C bulbs \(= 5,000 \times 0.70 = 3,500\text{ bulbs}\). **2. Multiply by the respective replacement cost per failed bulb and sum:** - Type-A warranty cost: \(600 \times \text{Rs. } 2.00 = \text{Rs. } 1,200\) - Type-B warranty cost: \(1,800 \times \text{Rs. } 1.75 = \text{Rs. } 3,150\) - Type-C warranty cost: \(3,500 \times \text{Rs. } 3.00 = \text{Rs. } 10,500\) \[\text{Total Expected Warranty Cost} = 1,200 + 3,150 + 10,500 = \text{Rs. } 14,850\] **Answer: 14850.**