Business Analytics · Quiz 2 · Jan 2025

Let \( T \) be the total life (in kms) of a new car battery. We are given: - \( P(T > 10,000) = 0.8 \) - \( P(T > 20,000) = 0.4 \) - \( P(T > 30,000) = 0.1 \) **1. Formulate the Conditional Probability:** We want to find \( P(T > 20,000 \mid T > 10,000) \). By the definition of conditional probability: \[ P(T > 20,000 \mid T > 10,000) = \frac{P(T > 20,000 \cap T > 10,000)}{P(T > 10,000)} \] **2. Simplify the Numerator and Compute:** Since the event \( \{T > 20,000\} \) is a subset of \( \{T > 10,000\} \), their intersection is simply \( \{T > 20,000\} \): \[ P(T > 20,000 \mid T > 10,000) = \frac{P(T > 20,000)}{P(T > 10,000)} = \frac{0.4}{0.8} = 0.50 \] **Answer: 0.50.**

For a **non-linear (constant-elasticity / power) demand-response curve**, demand \( D \) and price \( P \) follow the relationship: \[ D(P) = C \cdot P^{-b} \] We are given two points on the curve: - At \( P_1 = 80 \), \( D_1 = 1000 \) - At \( P_2 = 40 \), \( D_2 = 2500 \) **1. Find the Price Elasticity Exponent \( b \):** Taking the ratio \( \frac{D_2}{D_1} \): \[ \frac{2500}{1000} = \left(\frac{40}{80}\right)^{-b} \implies 2.5 = 2^b \] Taking the natural logarithm on both sides: \[ b = \frac{\ln(2.5)}{\ln(2)} = \frac{0.9162907}{0.6931472} \approx 1.321928 \] **2. Find the Constant \( C \):** Using \( D_2 = 2500 \) at \( P_2 = 40 \): \[ C = 2500 \times 40^{1.321928} \approx 2500 \times 131.188 = 327,970.5 \] **3. Predict Demand at \( P = 60 \):** \[ D(60) = C \cdot 60^{-b} = 2500 \times \left(\frac{60}{40}\right)^{-1.321928} = 2500 \times (1.5)^{-1.321928} \approx 2500 \times 0.58548 = 1463.70 \] **Answer: 1463.70.**

In pricing and revenue analytics, **latent demand** (also called unrealized demand) on a demand-response curve refers to the potential customer demand that exists at prices **below the identified optimal selling price** \( p^* \) (down to zero price). **Explanation:** - When a firm sets its price at the revenue-maximizing optimal price \( p^* \), consumers who are willing to pay a positive price less than \( p^* \) do not purchase the product. - If the price is reduced below the identified optimal price \( p^* \), this previously untapped (latent) demand is revealed. **Answer: Price is reduced below the identified optimal price.** A: **Correct:** Latent demand corresponds to the region of the demand-response curve obtained when the price is reduced below the identified optimal price. B: **Incorrect:** Increasing price above the optimal price reduces quantity demanded toward the willingness-to-pay ceiling (consumer surplus region), not latent demand. C: **Incorrect:** The optimal price is always less than the maximum willingness-to-pay price. D: **Incorrect:** Reducing quantity below the optimal quantity corresponds to charging a price higher than optimal. E: **Incorrect:** Demand cannot exceed the maximum demand at zero price.

Let us analyze the primal-dual relationship and **Complementary Slackness** for the given Linear Program: **1. Number of Dual Decision Variables (Option C):** - In Linear Programming duality, each **constraint** in the primal (excluding non-negativity restrictions) corresponds to **one decision variable** in the dual. - Since the primal has **3 decision variables** and **2 constraints**, the dual formulation has **2 decision variables** and **3 constraints**. - Thus, **Option C is correct** (and Option A is incorrect). **2. Complementary Slackness for the Binding Second Constraint (Option D):** - By the **Complementary Slackness Theorem**, if a primal constraint is **not binding** (strict inequality \( \text{LHS} < \text{RHS} \), positive slack), its corresponding dual variable must be **zero**. - Conversely, when the second primal constraint is **binding** (\( \text{LHS} = \text{RHS} \), zero slack) at the non-degenerate optimal solution, the dual variable (shadow price) corresponding to the second constraint is **non-zero**. - Thus, **Option D is correct** (and Option B is incorrect). **Answer: There are two decision variables in the dual formulation, and The dual variable corresponding to the second constraint is non-zero (Options C and D).** A: **Incorrect:** The number of dual decision variables equals the number of primal constraints (2), not the number of primal variables (3). B: **Incorrect:** The first constraint is non-binding (so its dual variable is zero), whereas the second constraint is binding (so its dual variable is non-zero). C: **Correct:** Because the primal has 2 functional constraints, the dual has 2 decision variables. D: **Correct:** Since the second primal constraint is binding (`LHS = RHS`), its corresponding dual variable (shadow price) is non-zero.

In the standard (canonical) form of the primal Linear Program for **FUNKY**: **Identify the Decision Variables:** The company manufactures three products to maximize profit: - \( x_1 = \) number of **Purses** produced per day - \( x_2 = \) number of **Shoes** produced per day - \( x_3 = \) number of **Backpacks** produced per day Therefore, there are **3** decision variables in the standard form of the primal. **Answer: 3.**

Excluding the non-negativity constraints (\( x_1, x_2, x_3 \ge 0 \)), each limited resource in Table-2 gives one availability constraint in the primal LP: **1. Leather Constraint:** \[ 2x_1 + 1x_2 + 3x_3 \le 42 \] **2. Sewing Hours Constraint:** \[ 2x_1 + 1x_2 + 2x_3 \le 40 \] **3. Polishing Hours Constraint:** \[ 1x_1 + 2x_2 + 1x_3 \le 42 \] Thus, there are **3** functional constraints in the standard form of the primal. **Answer: 3.**

Let us check which primal resource constraints are **binding** vs **non-binding** when the company produces \( x_1 = 10 \) Purses, \( x_2 = 15 \) Shoes, and \( x_3 = 2 \) Backpacks: **1. Evaluate the LHS of Each Resource Constraint:** - **Leather (Available = 42):** \[ \text{LHS}_1 = 2(10) + 1(15) + 3(2) = 20 + 15 + 6 = 41 < 42 \quad (\text{Non-binding, slack} = 1) \] - **Sewing (Available = 40):** \[ \text{LHS}_2 = 2(10) + 1(15) + 2(2) = 20 + 15 + 4 = 39 < 40 \quad (\text{Non-binding, slack} = 1) \] - **Polishing (Available = 42):** \[ \text{LHS}_3 = 1(10) + 2(15) + 1(2) = 10 + 30 + 2 = 42 = 42 \quad (\text{Binding, slack} = 0) \] **2. Apply Complementary Slackness:** By complementary slackness, any primal constraint with strictly positive slack (\( \text{LHS} < \text{RHS} \)) has a corresponding dual variable equal to **0**. Since **2** constraints (Leather and Sewing) are non-binding, **2** decision variables in the dual will have a value of `0`. **Answer: 2.**

The primal objective function maximizes total daily profit. From Table-2, the unit profits are: - Purse: Rs. \( 24 \)/unit - Shoes: Rs. \( 22 \)/unit - Backpacks: Rs. \( 45 \)/unit **Compute Total Profit for \( x_1 = 10, \; x_2 = 15, \; x_3 = 2 \):** \[ Z = 24 x_1 + 22 x_2 + 45 x_3 \] \[ Z = 24(10) + 22(15) + 45(2) = 240 + 330 + 90 = 660 \] **Answer: 660.**

We are given the linear demand function: \[ D(p) = 2570 - 4p \] **1. Formulate the Revenue Function \( R(p) \):** \[ R(p) = p \cdot D(p) = p(2570 - 4p) = 2570p - 4p^2 \] **2. Maximize \( R(p) \) by Setting the First Derivative to Zero:** \[ R'(p) = 2570 - 8p = 0 \] \[ p^* = \frac{2570}{8} = 321.25 \] (Since \( R''(p) = -8 < 0 \), \( p^* = 321.25 \) maximizes revenue.) **Answer: 321.25.**

We are given the demand function \( D(p) = 2570 - 4p \) and a unit production cost of \( c = 250 \). **1. Formulate the Profit Function \( \Pi(p) \):** Profit equals unit margin \( (p - 250) \) multiplied by demand \( D(p) \): \[ \Pi(p) = (p - 250)(2570 - 4p) = 2570p - 4p^2 - 642500 + 1000p = -4p^2 + 3570p - 642500 \] **2. Maximize \( \Pi(p) \) by Setting the First Derivative to Zero:** \[ \Pi'(p) = -8p + 3570 = 0 \] \[ p^* = \frac{3570}{8} = 446.25 \] (Alternatively, for a linear demand curve, the profit-maximizing price is \( p^*_{\text{profit}} = p^*_{\text{rev}} + \frac{c}{2} = 321.25 + \frac{250}{2} = 321.25 + 125 = 446.25 \).) **Answer: 446.25.**

In Table-1 (`Confusion Matrix`), rows represent the **Actual Class** and columns represent the **Predicted Class**. **1. Formula for Precision of "Hearts":** Precision for a class is the number of True Positives for that class divided by the **total number of samples predicted as that class** (the column sum for `Hearts`): - True Positives for `Hearts` (Actual = Hearts, Predicted = Hearts) \( = 100 \). - Total Predicted as `Hearts` (Column 1 sum) \( = 100 + 15 + 20 + 5 = 140 \). **2. Calculate Percentage Precision:** \[ \text{Precision}_{\text{Hearts}} = \frac{100}{140} \times 100\% = \frac{5}{7} \times 100\% \approx 71.4286\% \approx 71.43\% \] **Answer: 71.43.**

**1. Formula for Recall of "Spades":** Recall (or sensitivity) for a class is the number of True Positives for that class divided by the **total number of actual samples in that class** (the row sum for `Spades`): - True Positives for `Spades` (Actual = Spades, Predicted = Spades) \( = 40 \). - Total Actual `Spades` (Row 2 sum) \( = 15 + 40 + 10 + 35 = 100 \). **2. Calculate Percentage Recall:** \[ \text{Recall}_{\text{Spades}} = \frac{40}{100} \times 100\% = 40\% \] **Answer: 40.**

**1. Formula for Sensitivity of "Diamonds":** Sensitivity is another name for **Recall**: the proportion of actual `Diamonds` cards that are correctly predicted as `Diamonds`: - True Positives for `Diamonds` (Actual = Diamonds, Predicted = Diamonds) \( = 130 \). - Total Actual `Diamonds` (Row 3 sum) \( = 20 + 5 + 130 + 5 = 160 \). **2. Calculate Percentage Sensitivity:** \[ \text{Sensitivity}_{\text{Diamonds}} = \frac{130}{160} \times 100\% = 81.25\% \] **Answer: 81.25.**

**1. Formula for Specificity of "Clubs" (One-vs-Rest):** Specificity for `Clubs` is the proportion of **actual non-Clubs cards** that are **not predicted as Clubs**: \[ \text{Specificity}_{\text{Clubs}} = \frac{TN_{\text{Clubs}}}{TN_{\text{Clubs}} + FP_{\text{Clubs}}} \] **2. Compute \( TN + FP \) (Total Actual Non-Clubs) and \( FP \) (Non-Clubs Predicted as Clubs):** - Total cards \( = 600 \). - Total Actual `Clubs` (Row 4 sum) \( = 5 + 25 + 15 + 110 = 150 \). - Therefore, Total Actual Non-Clubs (\( TN + FP \)) \( = 600 - 150 = 450 \). - False Positives for `Clubs` (Actual \( \neq \) Clubs, Predicted = Clubs, in Column 4) \( = 25 + 35 + 5 = 65 \). - True Negatives for `Clubs` (\( TN_{\text{Clubs}} \)) \( = 450 - 65 = 385 \). **3. Calculate Percentage Specificity:** \[ \text{Specificity}_{\text{Clubs}} = \frac{385}{450} \times 100\% \approx 85.5556\% \approx 85.56\% \] **Answer: 85.56.**

**1. Formula for Overall Model Accuracy:** Overall accuracy is the sum of the main diagonal entries (all correctly classified cards across all 4 suits) divided by the total sample size (\( N = 600 \)): - Correctly predicted `Hearts` \( = 100 \) - Correctly predicted `Spades` \( = 40 \) - Correctly predicted `Diamonds` \( = 130 \) - Correctly predicted `Clubs` \( = 110 \) - Total Correct \( = 100 + 40 + 130 + 110 = 380 \). **2. Calculate Percentage Accuracy:** \[ \text{Accuracy} = \frac{380}{600} \times 100\% = 63.3333\% \approx 63.33\% \] **Answer: 63.33.**

In Picture-1 (ANOVA table for **Model-1**), **`A1`** is the **Regression Sum of Squares (\( SS_{\text{Regression}} \))**. **1. Method 1 (From Total SS and Residual SS):** \[ SS_{\text{Total}} = SS_{\text{Regression}} + SS_{\text{Residual}} \] \[ A_1 = 178836.1 - 112.4796 = 178723.6204 \approx 178723.62 \] **2. Method 2 (From \( df \) and \( MS \)):** \[ A_1 = df_{\text{Regression}} \times MS_{\text{Regression}} = 3 \times 59574.54 = 178723.62 \] **Answer: 178723.62.**

In Picture-2 (**Model-2**), **`A2`** is the **t-Statistic (`t Stat`)** for `X Variable 1`. **1. Formula for the t-Statistic of a Regression Coefficient:** \[ t = \frac{\text{Coefficient}}{\text{Standard Error}} \] **2. Substitute the Values for `X Variable 1`:** - `Coefficients` \( = -3.952958885 \) - `Standard Error` \( = 0.03846985 \) \[ A_2 = \frac{-3.952958885}{0.03846985} \approx -102.7547 \approx -102.75 \] **Answer: -102.75.**

In Picture-3 (**Model-3**), **`A3`** is the **Regression Sum of Squares (\( SS_{\text{Regression}} \))**. **1. Use the Total Sum of Squares (\( SS_{\text{Total}} = A_5 \)):** Since all three models are built on the exact same dependent variable and dataset (\( N = 10 \) observations), the Total Sum of Squares \( SS_{\text{Total}} \) is identical across all three models: From Model-1 (or Model-2: \( 178723.4403 + 112.6596987 = 178836.1 \)), we have: \[ A_5 = SS_{\text{Total}} = 178836.1 \] **2. Compute \( A_3 = SS_{\text{Regression}} \) Using \( R^2 \):** From the Regression Statistics table of Model-3, \( R^2 = 0.997442694 \). \[ A_3 = R^2 \times SS_{\text{Total}} = 0.997442694 \times 178836.1 \approx 178378.82 \] **Answer: 178378.82.**

In Picture-3 (**Model-3**), **`A4`** is the **Residual Sum of Squares (\( SS_{\text{Residual}} \))**. **1. Method 1 (Using Standard Error and Residual \( df = 8 \)):** In Excel's Regression Summary, `Standard Error` \( s_e = 7.560908499 \) is the square root of \( MS_{\text{Residual}} \): \[ MS_{\text{Residual}} = s_e^2 = (7.560908499)^2 \approx 57.167337 \] Since \( df_{\text{Residual}} = 8 \): \[ A_4 = SS_{\text{Residual}} = df_{\text{Residual}} \times MS_{\text{Residual}} = 8 \times 57.167337 \approx 457.34 \] **2. Method 2 (Using \( SS_{\text{Total}} - SS_{\text{Regression}} \)):** \[ A_4 = 178836.1 - 178378.82 = 457.28 \] (Both \( 457.28 \) and \( 457.34 \) lie well inside the accepted range `455.00 to 459.00`.) **Answer: 457.28.**

In Picture-3 (**Model-3**), **`A5`** is the **Total Sum of Squares (\( SS_{\text{Total}} \))**. **Explanation:** Because all three models (Model-1, Model-2, and Model-3) are built on the exact same dependent variable \( Y \) with the same 10 observations, the Total Sum of Squares \( SS_{\text{Total}} = \sum_{i=1}^{10} (y_i - \bar{y})^2 \) is identical across all three ANOVA tables. From Picture-1 (Model-1), `Total SS` is directly given as **178836.1** (or `178836.10`). **Answer: 178836.10.**

To compare linear regression models with **different numbers of independent variables** built on the same dependent variable and dataset, we compare their **Adjusted R-Square (\( R^2_{\text{adj}} \))**, Mean Squared Error (\( MS_{\text{Residual}} \)), and overall F-statistic significance: **1. Compare Residual SS and Adjusted \( R^2 \) Across the Three Models:** - **Model-1 (\( k = 3 \) predictors, \( df_{\text{res}} = 6 \))**: - \( SS_{\text{Residual}} = 112.4796 \), \( MS_{\text{Residual}} = 18.7466 \). - \( R^2_{\text{adj}} = 1 - \frac{MS_{\text{Residual}}}{SS_{\text{Total}} / df_{\text{Total}}} = 1 - \frac{18.7466}{178836.1 / 9} = 1 - \frac{18.7466}{19870.68} = 0.999056 \). - **Model-2 (\( k = 2 \) predictors, \( df_{\text{res}} = 7 \))**: - \( SS_{\text{Residual}} = 112.6597 \), so \( MS_{\text{Residual}} = \frac{112.6597}{7} = 16.0942 \) (lowest \( MS_{\text{Residual}} \)!). - \( R^2_{\text{adj}} = 1 - \frac{16.0942}{19870.68} = 0.999190 \) (highest Adjusted \( R^2 \), highest \( F = 5552.40 \), and lowest p-value \( 6.27 \times 10^{-12} \)!). - **Model-3 (\( k = 1 \) predictor, \( df_{\text{res}} = 8 \))**: - \( SS_{\text{Residual}} = 457.34 \), \( MS_{\text{Residual}} = 57.17 \), \( R^2_{\text{adj}} = 0.997123 \). **2. Conclusion:** **Model-2** achieves the highest Adjusted \( R^2 \), lowest \( MS_{\text{Residual}} \), and highest F-statistic (adding a 3rd variable in Model-1 barely reduces \( SS_{\text{Residual}} \) from `112.66` to `112.48` and actually worsens Adjusted \( R^2 \)). Therefore, **Model-2** is the best model. **Answer: Model-2.** A: **Incorrect:** Adding the 3rd predictor in Model-1 increases `MS_Residual` from `16.09` to `18.75` and lowers the Adjusted R-Square compared to Model-2. B: **Correct:** Model-2 has the highest Adjusted R-Square (`0.99919`), lowest `MS_Residual` (`16.09`), and strongest F-statistic significance (`6.27E-12`). C: **Incorrect:** Model-3 has a much higher residual sum of squares (`457.34`) and lower Adjusted R-Square (`0.99712`) than Model-2. D: **Incorrect:** As noted in the option text itself, this is a humorous distractor option.