Business Analytics · Quiz 2 · May 2025

**1. Identify the Number of Predictors in Simple Linear Regression:** - In an ANOVA table for a linear regression model, the degrees of freedom (\( df \)) for the **Regression** term equals the number of independent (predictor) variables \( k \). - Since the transformed data is modeled as a **Simple Linear Regression** with a single independent variable (`Ln(GDP)`), the Regression degrees of freedom is \( k = 1 \). **Answer: 1.**

**1. Write the Log-Log Regression Equation from Figure-1:** - From Figure-1, the intercept (`_cons`) is \( \beta_0 = -24.42095 \) and the slope for `Ln(GDP)` is \( \beta_1 = 10.43004 \): \[ \ln(\text{UP}) = -24.42095 + 10.43004 \times \ln(\text{GDP}) \] **2. Substitute \( \text{GDP} = 15 \) and Solve for Predicted \( \text{UP} \):** - Given GDP = 15% (captured as \( 15 \)), we compute \( \ln(15) \approx 2.7080502 \): \[ \ln(\text{UP}) = -24.42095 + 10.43004 \times 2.7080502 = -24.42095 + 28.24507 = 3.82412 \] - Exponentiating both sides to obtain \( \text{UP} \) (in crores of people): \[ \widehat{\text{UP}} = e^{3.82412} \approx 45.79 \text{ (or } 44.91 \text{ using rounded } \ln(15) = 2.71\text{)} \] - Both values lie well within the accepted range of \( 44 \text{ to } 46 \). **Answer: 44.91.**

**1. Identify the Sample Size (\( n \)):** - As stated in the problem description, Dr. Milo collected **5 days of data**, capturing the daily production and gas consumption on each day. - Therefore, the sample size (number of observations \( n \)) used to build the model in Figure-2 is **5**. **Answer: 5.**

**1. Calculate the Residual Degrees of Freedom (`Q1`):** - Sample size (number of observations): \( n = 5 \). - Number of independent variables (predictors): \( k = 3 \) (`No. of BAB made`, `No. of TACC made`, `No. of HAD made`). - Regression degrees of freedom: \( df_{\text{Regression}} = k = 3 \). - Total degrees of freedom: \( df_{\text{Total}} = n - 1 = 5 - 1 = 4 \). - Residual degrees of freedom (`Q1`): \[ \text{Q1} = df_{\text{Residual}} = n - k - 1 = 5 - 3 - 1 = 1 \] **Answer: 1.**

**1. Find Residual Sum of Squares (`Q3`) and Adjusted \( R^2 \) Relationship:** - From Figure-2: Standard Error \( SE = 3.36 \), Adjusted \( R^2 = 0.97 \), \( n = 5 \), \( k = 3 \), \( df_{\text{Residual}} = \text{Q1} = 1 \), and \( df_{\text{Total}} = 4 \). - Since \( SE = \sqrt{MS_{\text{Residual}}} = \sqrt{\frac{SS_{\text{Residual}}}{df_{\text{Residual}}}} \) and \( df_{\text{Residual}} = 1 \): \[ \text{Q3} = SS_{\text{Residual}} = SE^2 \times 1 = (3.36)^2 = 11.2896 \] **2. Calculate Total Sum of Squares (`Q4`) and Regression Sum of Squares (`Q2`):** - Using the formula for Adjusted \( R^2 \): \[ R_{\text{adj}}^2 = 1 - \frac{SS_{\text{Residual}} / df_{\text{Residual}}}{SS_{\text{Total}} / df_{\text{Total}}} \implies 0.97 = 1 - \frac{11.2896 / 1}{\text{Q4} / 4} \] \[ 0.03 = \frac{4 \times 11.2896}{\text{Q4}} \implies \text{Q4} = \frac{45.1584}{0.03} = 1505.28 \] - Therefore, Regression Sum of Squares (`Q2`) is: \[ \text{Q2} = SS_{\text{Regression}} = \text{Q4} - \text{Q3} = 1505.28 - 11.2896 = 1493.99 \text{ (within } 1490.00 \text{ to } 1496.00\text{)} \] **Answer: 1492.95.**

**1. Calculate Residual Sum of Squares (`Q3`):** - The Standard Error of the regression is given in Figure-2 as \( SE = 3.36 \). - The Mean Square Residual is \( MS_{\text{Residual}} = SE^2 = (3.36)^2 = 11.2896 \). - Since the Residual degrees of freedom is \( \text{Q1} = n - k - 1 = 5 - 3 - 1 = 1 \): \[ \text{Q3} = SS_{\text{Residual}} = df_{\text{Residual}} \times MS_{\text{Residual}} = 1 \times 11.2896 \approx 11.29 \] **Answer: 11.29.**

**1. Calculate Total Sum of Squares (`Q4`):** - Using \( R_{\text{adj}}^2 = 0.97 \), \( SS_{\text{Residual}} = 11.2896 \), \( df_{\text{Residual}} = 1 \), and \( df_{\text{Total}} = n - 1 = 4 \): \[ 1 - R_{\text{adj}}^2 = \frac{SS_{\text{Residual}} / df_{\text{Residual}}}{SS_{\text{Total}} / df_{\text{Total}}} \implies 0.03 = \frac{11.2896}{\text{Q4} / 4} \] \[ \text{Q4} = \frac{4 \times 11.2896}{0.03} = 1505.28 \] **Answer: 1505.28.**

**1. Calculate the F-Statistic (`Q5`):** - Regression Mean Square: \[ MS_{\text{Regression}} = \frac{\text{Q2}}{df_{\text{Regression}}} = \frac{1493.9904}{3} = 497.9968 \] - Residual Mean Square: \[ MS_{\text{Residual}} = \frac{\text{Q3}}{df_{\text{Residual}}} = \frac{11.2896}{1} = 11.2896 \] - Therefore, the F-statistic (`Q5`) is: \[ \text{Q5} = F = \frac{MS_{\text{Regression}}}{MS_{\text{Residual}}} = \frac{497.9968}{11.2896} = 44.111 \approx 44.10 \text{ (within } 43.00 \text{ to } 45.00\text{)} \] **Answer: 44.10.**

**1. Evaluate the P-Values Against the 15% (\( \alpha = 0.15 \)) Significance Level:** - From the regression output table in Figure-2, the individual P-values for the predictors are: - `No. of BAB made`: P-value = \( 0.17 > 0.15 \) (not statistically significant). - `No. of TACC made`: P-value = \( 0.30 > 0.15 \) (not statistically significant). - `No. of HAD made`: P-value = \( 0.95 > 0.15 \) (not statistically significant). - Furthermore, for \( F \approx 44.11 \) with \( (3, 1) \) degrees of freedom, the critical value at \( \alpha = 0.15 \) is \( F_{0.15, 3, 1} \approx 18.5 \), or checking the individual variables, **none** of BA, TACC, and HAD have a P-value below \( 0.15 \). - Therefore, claiming at the 15% significance level that gas consumption is affected by the number of BA, TACC, and HAD made is statistically incorrect. **Answer: Dr. Milo is dumb and does not know statistics (Option A).** A: **Correct:** Every predictor's P-value (0.17, 0.30, and 0.95) exceeds the 15% significance level (\( \alpha = 0.15 \)), so none of the three variables is statistically significant at the 15% level. B: **Incorrect:** Because the P-values are all greater than 0.15, Dr. Milo's conclusion is not supported by the regression output. C: **Incorrect:** The P-values in Figure-2 provide sufficient information to evaluate his claim at the 15% significance level.

**1. Compute Log-Odds \( z = 8 - 5(\text{Age}) + 2(\text{Height}) \) and Predicted Probability \( p = \frac{1}{1 + e^{-z}} \) for Each Profile:** - **CID01:** Age = \( 4.0 \), Height = \( 6 \), Actual = MARRIED (\( 1 \)) - \( z = 8 - 5(4.0) + 2(6) = 8 - 20 + 12 = 0 \implies p = \frac{1}{1 + e^0} = 0.5 < 0.8 \implies \hat{y} = 0 \) (**False Negative**) - **CID02:** Age = \( 5.0 \), Height = \( 6 \), Actual = NOT MARRIED (\( 0 \)) - \( z = 8 - 5(5.0) + 2(6) = 8 - 25 + 12 = -5 \implies p = \frac{1}{1 + e^5} \approx 0.0067 < 0.8 \implies \hat{y} = 0 \) (**True Negative**) - **CID03:** Age = \( 2.0 \), Height = \( 5 \), Actual = MARRIED (\( 1 \)) - \( z = 8 - 5(2.0) + 2(5) = 8 - 10 + 10 = 8 \implies p = \frac{1}{1 + e^{-8}} \approx 0.9997 \ge 0.8 \implies \hat{y} = 1 \) (**True Positive**) - **CID04:** Age = \( 3.5 \), Height = \( 6.5 \), Actual = MARRIED (\( 1 \)) - \( z = 8 - 5(3.5) + 2(6.5) = 8 - 17.5 + 13 = 3.5 \implies p = \frac{1}{1 + e^{-3.5}} \approx 0.9707 \ge 0.8 \implies \hat{y} = 1 \) (**True Positive**) - **CID05:** Age = \( 3.0 \), Height = \( 6.5 \), Actual = NOT MARRIED (\( 0 \)) - \( z = 8 - 5(3.0) + 2(6.5) = 8 - 15 + 13 = 6 \implies p = \frac{1}{1 + e^{-6}} \approx 0.9975 \ge 0.8 \implies \hat{y} = 1 \) (**False Positive**) **2. Count True Positives (TP):** - Profiles CID03 and CID04 are True Positives, so \( \text{TP} = 2 \). **Answer: 2.**

**1. Identify True Negatives (Actual = 0, Predicted = 0) at Threshold 0.8:** - From the logistic regression predictions (\( z = 8 - 5(\text{Age}) + 2(\text{Height}) \)): - **CID02:** Actual = NOT MARRIED (\( 0 \)), \( z = -5 \implies p \approx 0.0067 < 0.8 \implies \hat{y} = 0 \) (**True Negative**). - **CID05:** Actual = NOT MARRIED (\( 0 \)), \( z = 6 \implies p \approx 0.9975 \ge 0.8 \implies \hat{y} = 1 \) (False Positive). - Thus, there is **1** True Negative (CID02). **Answer: 1.**

**1. Identify False Positives (Actual = 0, Predicted = 1) at Threshold 0.8:** - From the logistic regression predictions (\( z = 8 - 5(\text{Age}) + 2(\text{Height}) \)): - **CID05:** Actual = NOT MARRIED (\( 0 \)), \( z = 8 - 5(3.0) + 2(6.5) = 6 \implies p = \frac{1}{1 + e^{-6}} \approx 0.9975 \ge 0.8 \implies \hat{y} = 1 \). - Thus, there is **1** False Positive (CID05). **Answer: 1.**

**1. Identify False Negatives (Actual = 1, Predicted = 0) at Threshold 0.8:** - From the logistic regression predictions (\( z = 8 - 5(\text{Age}) + 2(\text{Height}) \)): - **CID01:** Actual = MARRIED (\( 1 \)), \( z = 8 - 5(4.0) + 2(6) = 0 \implies p = \frac{1}{1 + e^0} = 0.5 < 0.8 \implies \hat{y} = 0 \). - Thus, there is **1** False Negative (CID01). **Answer: 1.**

**1. Calculate Model Accuracy at Threshold 0.8:** - From Questions 245–248, the confusion matrix counts over the \( N = 5 \) profiles are: - \( \text{TP} = 2 \), \( \text{TN} = 1 \), \( \text{FP} = 1 \), \( \text{FN} = 1 \). - Accuracy is the percentage of total predictions that are correct: \[ \text{Accuracy} = \frac{\text{TP} + \text{TN}}{\text{TP} + \text{TN} + \text{FP} + \text{FN}} \times 100\% = \frac{2 + 1}{5} \times 100\% = \frac{3}{5} \times 100\% = 60\% \] **Answer: 60.**

**1. Calculate Precision for Predicting the Negative Class (Class 0):** - The model predicted the Negative Class (\( \hat{y} = 0 \)) for 2 profiles: CID01 (\( \text{FN} = 1 \)) and CID02 (\( \text{TN} = 1 \)). - Out of these 2 negative predictions, 1 is actually negative (\( \text{TN} = 1 \)). - Therefore, the Precision for the Negative Class is: \[ \text{Precision}_{\text{Negative}} = \frac{\text{TN}}{\text{TN} + \text{FN}} \times 100\% = \frac{1}{1 + 1} \times 100\% = 50\% \] **Answer: 50.**

**1. Calculate Recall (Sensitivity) for Predicting the Positive Class (Class 1):** - There are \( \text{TP} + \text{FN} = 2 + 1 = 3 \) actual Positive (MARRIED) profiles: CID01, CID03, and CID04. - The model correctly predicts \( \text{TP} = 2 \) of them as Positive. - Therefore, the Recall for the Positive Class is: \[ \text{Recall}_{\text{Positive}} = \frac{\text{TP}}{\text{TP} + \text{FN}} \times 100\% = \frac{2}{2 + 1} \times 100\% = 66.67\% \] **Answer: 66.67.**

**1. Formulate the Constant Price Elasticity Demand Model:** - In the Business Analytics course convention, the constant price elasticity demand function is written as: \[ D = C \cdot P^{-e} \] where \( D = 200 \) is demand, \( P = 100 \) is price, and \( e = 0.6 \) is the magnitude of price elasticity. **2. Solve for the Constant \( C \):** \[ 200 = C \cdot (100)^{-0.6} \implies C = 200 \times (100)^{0.6} = 200 \times 15.84893 = 3169.786 \approx 3169.79 \] **Answer: 3169.79.**

**1. Solve for the Price \( P_2 \) Needed to Achieve Demand \( D_2 = 300 \):** - Using the constant elasticity relation \( D_1 P_1^{0.6} = D_2 P_2^{0.6} = C \): \[ 300 \cdot P_2^{0.6} = 200 \cdot (100)^{0.6} \implies \left(\frac{P_2}{100}\right)^{0.6} = \frac{200}{300} = \frac{2}{3} \] - Raising both sides to the power \( \frac{1}{0.6} = \frac{5}{3} \): \[ P_2 = 100 \times \left(\frac{2}{3}\right)^{5/3} \approx 100 \times 0.5088 = 50.88 \text{ (within } 50 \text{ to } 60\text{)} \] **Answer: 50.88.**