Business Analytics · Quiz 2 · Jan 2026

**1. Interpret the regression standard error:** For simple regression with an intercept and one predictor, the residual degrees of freedom are n−2. The regression standard error satisfies \(s=\sqrt{SSE/(n-2)}\). **2. Substitute Model-1 data:** There are n=50 observations and s=9.4164. Hence \[SSE=s^2(50-2)=9.4164^2(48)=4256.09227008.\] **3. Round the requested total:** The total unexplained variability is the residual sum of squares, approximately 4256.09. It is not s itself or the residual mean square. **Answer: 4256.09.** **Convention note:** This calculation uses the displayed standard error; its last digits are subject to rounding in the source output.

**1. Use the simple-regression correlation:** Model-2 has one predictor, Design, with reported correlation r=0.20 with Profit. In simple regression with an intercept, \(R^2=r^2=0.04\). **2. Form the ANOVA F ratio:** With n=50, regression degrees of freedom 1 and residual degrees of freedom 48, \[F=\frac{R^2/1}{(1-R^2)/48}=\frac{0.04\cdot48}{0.96}=2.\] **3. State the requested entry:** P2 is the F statistic, approximately 2.00 using the reported correlation. The common total sum of squares cancels from this ratio. **Answer: 2.00.** **Convention note:** The correlation is rounded, so 2.00 is an approximation from the displayed r=0.20. No precise unreported total sum of squares is assumed.

**1. Identify the coefficient t statistic:** To test the default null hypothesis that the Delivery Time slope is zero, use \(t=(\hat\beta-0)/SE(\hat\beta)\). **2. Substitute the displayed slope and standard error:** \[P3=\frac{0.25}{0.032}=7.8125.\] **3. Round:** The t statistic is 7.81 to two decimal places. Its sign is positive because the estimated slope is positive. **Answer: 7.81.** **Convention note:** The coefficient and standard error are displayed in rounded form, so the calculation follows those printed values.

**1. Count Model-4 predictors:** The explanatory variables are Quality, Design and Delivery Time, giving k=3 predictors. **2. Apply ANOVA degrees of freedom:** The regression degrees of freedom equal the number of fitted slopes, excluding the intercept. Thus P4=3. The intercept contributes to the parameter count used for residual degrees of freedom, not to this regression entry. **Answer: 3.**

**1. Count fitted coefficients:** Model-4 uses three slopes and an intercept, so it estimates four coefficients from 50 observations. **2. Subtract fitted parameters:** \[df_{\rm residual}=n-k-1=50-3-1=46.\] As a cross-check, regression df 3 plus residual df 46 equals total df 49. **Answer: 46.**

**1. Define the production quantities:** Let x₁, x₂ and x₃ denote the daily numbers of Phones, TVs and Computers produced. These are the independent business choices. **2. Write the profit objective:** \(\max\ 300x_1+700x_2+1000x_3\), with nonnegative quantities. The operation times constrain those quantities but are not themselves additional production decisions. **3. Count the canonical primal variables:** There is one variable per product, giving 3 decision variables. Auxiliary slack variables used in a solution method are not counted here. **Answer: 3.**

**1. Identify the stated resource budget:** The passage supplies one total working-time budget of 10 hours, or 600 minutes, rather than four independently specified workstation capacities. **2. Aggregate the operation times per product:** A Phone uses 3+2+4+5=14 minutes, a TV uses 20+24+22+7=73, and a Computer uses 30+17+15+9=71. The primal resource constraint is \[14x_1+73x_2+71x_3\le600.\] **3. Apply primal–dual correspondence:** One primal resource inequality creates one dual decision variable u. For example, the dual minimizes 600u subject to 14u≥300, 73u≥700, 71u≥1000 and u≥0. Thus the count is 1. **Answer: 1.** **Convention note:** This follows the stated single aggregate time budget. If four separate machine-capacity constraints were supplied instead, the corresponding dual would have four resource variables; those independent capacities are not specified here.

**1. Calculate from the printed coefficients:** The modeled positive event is No Heart Attack=1. Its log-odds are \(z=-14+2\,Age+2\,bpm-0.5\,Cholesterol\). For P1–P5, substitution gives z=126.5, 143, 109.5, 125.5 and 105.5. Every value is far above \(\log(0.8/0.2)=\log4\approx1.3863\), so all five patients are predicted as class 1 at threshold 0.8. **2. Construct the confusion counts:** Actual class 1 consists of P2 and P3; P1, P4 and P5 are class 0. Predicting everyone as 1 gives TP=2, FP=3, TN=0 and FN=0. These counts use No Heart Attack as the positive class, as specified. **3. Calculate overall accuracy:** \[100\frac{TP+TN}{TP+TN+FP+FN}=100\frac{2+0}{5}=40.00.\] Two of the five patient classifications are correct. **Answer: 40.00.** **Convention note:** The calculation follows the displayed coefficients literally. No unreported feature scaling or decimal-point correction is introduced.