Business Analytics · Quiz 1 · Sep 2024

**Answer: C** **Topic:** Machine Learning Fundamentals — *Purpose of n_iter_no_change in SGDRegressor* ### Step-by-Step Solution **Step 1:** In `scikit-learn`'s `SGDRegressor`, early stopping monitors the validation score (or training loss) across epochs and halts training when the model stops improving. **Step 2:** Specifically: - `max_iter` controls the maximum total number of iterations (Option A). - `tol` specifies the tolerance threshold for considering an improvement significant (Option B). - `n_iter_no_change` **defines the number of consecutive iterations with no improvement (by at least `tol`) before stopping training** (Option C). - `learning_rate` sets the learning rate schedule (Option D). **Step 3:** Therefore, **C** is correct. **Key Takeaway / Tip:** Do not confuse `tol` (the minimum improvement threshold) with `n_iter_no_change` (the patience count of epochs without improvement). A: **Incorrect:** Option A (Controls the total number of iterations.) does not satisfy the conditions derived in the step-by-step analysis for Purpose of n_iter_no_change in SGDRegressor (correct choice: C). B: **Incorrect:** Option B (Specifies the tolerance level for early stopping.) does not satisfy the conditions derived in the step-by-step analysis for Purpose of n_iter_no_change in SGDRegressor (correct choice: C). C: **Correct:** Option C (Defines the number of iterations with no improvement to stop training.) matches our derived result under Purpose of n_iter_no_change in SGDRegressor. Do not confuse `tol` (the minimum improvement threshold) with `n_iter_no_change` (the patience count of epochs without improvement). D: **Incorrect:** Option D (Sets the learning rate schedule.       **BA** **Section Id :** 64065369403 **Section Number :** 4 **Section type :** Online **Mandatory or Optional :** Mandatory **Number of Que...) does not satisfy the conditions derived in the step-by-step analysis for Purpose of n_iter_no_change in SGDRegressor (correct choice: C).

**Answer: D** **Topic:** Hypothesis Testing — *Interpreting P-Value in a Normality Goodness-of-Fit Test* ### Step-by-Step Solution **Step 1:** In a goodness-of-fit test for normality: - **Null Hypothesis ($H_0$)**: The data follows a normal distribution. - **Alternative Hypothesis ($H_1$)**: The data does not follow a normal distribution. **Step 2:** Given $p\text{-value} = 0.67$, compare against the significance level $\alpha = 0.05$ ($5\%$): since $0.67 > 0.05$, we **DO NOT REJECT** the null hypothesis $H_0$ and conclude that the data follows a normal distribution (**D is true**, while A, B, C are false). **Step 3:** Note that Options E, F, G, and H phrase the decision in terms of *'rejecting / not rejecting the alternative hypothesis'*, which is statistically invalid because hypothesis tests always act on the **null** hypothesis. **Step 4:** Therefore, **D** is the only correct statement. **Key Takeaway / Tip:** In classical hypothesis testing, we always either *reject* or *do not reject* the **null hypothesis ($H_0$)**—never the alternative hypothesis. A: **Incorrect:** Option A (At a 5% level of significance **REJECT** the null hypothesis and conclude that the data does not follow a normal distribution) does not satisfy the conditions derived in the step-by-step analysis for Interpreting P-Value in a Normality Goodness-of-Fit Test (correct choice: D). B: **Incorrect:** Option B (At a 5% level of significance **DO NOT REJECT** the null hypothesis and conclude that the data does not follow a normal distribution) does not satisfy the conditions derived in the step-by-step analysis for Interpreting P-Value in a Normality Goodness-of-Fit Test (correct choice: D). C: **Incorrect:** Option C (At a 5% level of significance **REJECT** the null hypothesis and conclude that the data follows a normal distribution) does not satisfy the conditions derived in the step-by-step analysis for Interpreting P-Value in a Normality Goodness-of-Fit Test (correct choice: D). D: **Correct:** Option D (At a 5% level of significance **DO NOT REJECT** the null hypothesis and conclude that the data follows a normal distribution) matches our derived result under Interpreting P-Value in a Normality Goodness-of-Fit Test. In classical hypothesis testing, we always either *reject* or *do not reject* the **null hypothesis ($H_0$)**—never the alternative hypothesis. E: **Incorrect:** Option E (At a 10% level of significance **REJECT** the alternative hypothesis and conclude that the data does not follow a normal distribution) does not satisfy the conditions derived in the step-by-step analysis for Interpreting P-Value in a Normality Goodness-of-Fit Test (correct choice: D). F: **Incorrect:** Option F (At a 10% level of significance **DO NOT REJECT** the alternative hypothesis and conclude that the data does not follow a normal distribution) does not satisfy the conditions derived in the step-by-step analysis for Interpreting P-Value in a Normality Goodness-of-Fit Test (correct choice: D). G: **Incorrect:** Option G (At a 10% level of significance **REJECT** the alternative hypothesis and conclude that the data follows a normal distribution) does not satisfy the conditions derived in the step-by-step analysis for Interpreting P-Value in a Normality Goodness-of-Fit Test (correct choice: D). H: **Incorrect:** Option H (At a 10% level of significance **DO NOT REJECT** the alternative hypothesis and conclude that the data follows a normal distribution) does not satisfy the conditions derived in the step-by-step analysis for Interpreting P-Value in a Normality Goodness-of-Fit Test (correct choice: D).

**Answer: 1** **Topic:** Chi-Square Test of Independence — *Degrees of Freedom for a 2x2 Contingency Table* ### Step-by-Step Solution **Step 1:** The test compares two categorical variables: 1. **Hero** ($r = 2$ categories: `MK` and `YV`) 2. **Part of Week** ($c = 2$ categories: `Weekends [Fri, Sat, Sun]` and `Weekdays [Mon–Thu]`) **Step 2:** Apply the degrees of freedom formula for an $r \times c$ contingency table: $$df = (r - 1)(c - 1) = (2 - 1)(2 - 1) = 1$$ **Key Takeaway / Tip:** Even though Table-1 has 15 daily rows, aggregating into `Weekends` vs. `Weekdays` across the 2 heroes creates a $2 \times 2$ contingency table with $df = 1$.

**Answer: 104.78** **Topic:** Chi-Square Test of Independence — *Expected Collection for MK on Weekends* ### Step-by-Step Solution **Step 1:** Sum the box-office collections from Table-1 into a $2 \times 2$ contingency table (`Weekends` = Jan 1, 6, 7, 8, 13, 14, 15; `Weekdays` = Jan 2, 3, 4, 5, 9, 10, 11, 12): - **MK on Weekends**: $11 + 16 + 19 + 8 + 15 + 17 + 15 = 101$ - **MK on Weekdays**: $9 + 15 + 8 + 18 + 10 + 19 + 19 + 19 = 117$ - **YV on Weekends**: $10 + 10 + 7 + 6 + 8 + 5 + 2 = 48$ - **YV on Weekdays**: $10 + 9 + 6 + 2 + 1 + 3 + 5 + 8 = 44$ **Step 2:** Calculate the marginal totals and grand total: - Total for MK: $R_{\text{MK}} = 101 + 117 = 218$ - Total for YV: $R_{\text{YV}} = 48 + 44 = 92$ - Total on Weekends: $C_{\text{Weekend}} = 101 + 48 = 149$ - Total on Weekdays: $C_{\text{Weekday}} = 117 + 44 = 161$ - Grand Total: $N = 218 + 92 = 310$ **Step 3:** Compute the expected box-office collection for **MK on Weekends**: $$E_{\text{MK, Weekend}} = \frac{R_{\text{MK}} \times C_{\text{Weekend}}}{N} = \frac{218 \times 149}{310} = \frac{32482}{310} \approx 104.7806 \approx 104.78$$ **Key Takeaway / Tip:** Be sure to include **Friday** along with Saturday and Sunday in the `Weekends` bucket, as explicitly defined in the question prompt.

**Answer: 47.78** **Topic:** Chi-Square Test of Independence — *Expected Collection for YV on Weekdays* ### Step-by-Step Solution **Step 1:** Using the marginal totals computed in Question 54: - Total collection for YV: $R_{\text{YV}} = 48 + 44 = 92$ - Total collection on Weekdays: $C_{\text{Weekday}} = 117 + 44 = 161$ - Grand Total collection: $N = 310$ **Step 2:** Compute the expected box-office collection for **YV on Weekdays**: $$E_{\text{YV, Weekday}} = \frac{R_{\text{YV}} \times C_{\text{Weekday}}}{N} = \frac{92 \times 161}{310} = \frac{14812}{310} \approx 47.7806 \approx 47.78$$ **Key Takeaway / Tip:** Notice that in a $2 \times 2$ table, the difference $|O_{ij} - E_{ij}|$ is identical for all 4 cells: $|101 - 104.78| = |44 - 47.78| = 3.78$.

**Answer: 0.89** **Topic:** Chi-Square Test of Independence — *Chi-Square Statistic for 2x2 Contingency Table* ### Step-by-Step Solution **Step 1:** List the observed ($O_{ij}$) and expected ($E_{ij}$) values for all 4 cells of the $2 \times 2$ table: - $(\text{MK, Weekend})$: $O_{11} = 101$, $E_{11} = \frac{218 \times 149}{310} \approx 104.7806$ - $(\text{MK, Weekday})$: $O_{12} = 117$, $E_{12} = \frac{218 \times 161}{310} \approx 113.2194$ - $(\text{YV, Weekend})$: $O_{21} = 48$, $E_{21} = \frac{92 \times 149}{310} \approx 44.2194$ - $(\text{YV, Weekday})$: $O_{22} = 44$, $E_{22} = \frac{92 \times 161}{310} \approx 47.7806$ **Step 2:** Every cell has squared deviation $(O_{ij} - E_{ij})^2 = (3.780645)^2 \approx 14.29328$. Compute the sum of $\frac{(O_{ij} - E_{ij})^2}{E_{ij}}$: $$\chi^2_{\text{calc}} = 14.29328 \left(\frac{1}{104.7806} + \frac{1}{113.2194} + \frac{1}{44.2194} + \frac{1}{47.7806}\right)$$ **Step 3:** Evaluate the sum: $$\chi^2_{\text{calc}} \approx 0.1364 + 0.1262 + 0.3232 + 0.2991 = 0.8850 \approx 0.89$$ **Key Takeaway / Tip:** You can also use the direct $2 \times 2$ formula $\chi^2 = \frac{N(ad - bc)^2}{R_1 R_2 C_1 C_2} = \frac{310(101 \times 44 - 117 \times 48)^2}{218 \times 92 \times 149 \times 161} \approx 0.885$.

**Answer: A** **Topic:** Chi-Square Test of Independence — *Hypothesis Test Conclusion Using Chi-Square Table* ### Step-by-Step Solution **Step 1:** At a $95\%$ confidence level, the significance level is $\alpha = 1 - 0.95 = 0.05$, and from Question 53 the degrees of freedom is $\nu = df = 1$. **Step 2:** Look up the critical value in Table-2 for $\nu = 1$ and upper-tail area $\alpha = .050$: $$\chi^2_{\text{crit}}(1, 0.05) = 3.84$$ **Step 3:** Compare the computed test statistic $\chi^2_{\text{calc}} \approx 0.89$ against $\chi^2_{\text{crit}} = 3.84$: since $0.89 < 3.84$, we **Do Not Reject the Null** (**A**). **Key Takeaway / Tip:** In formal statistical terminology, we say *'Do Not Reject the Null'* rather than *'Accept the Null'*. A: **Correct:** Option A (Do Not Reject the Null) matches our derived result under Hypothesis Test Conclusion Using Chi-Square Table. In formal statistical terminology, we say *'Do Not Reject the Null'* rather than *'Accept the Null'*. B: **Incorrect:** Option B (Reject the Null) does not satisfy the conditions derived in the step-by-step analysis for Hypothesis Test Conclusion Using Chi-Square Table (correct choice: A). C: **Incorrect:** Option C (Accept the Null) does not satisfy the conditions derived in the step-by-step analysis for Hypothesis Test Conclusion Using Chi-Square Table (correct choice: A). D: **Incorrect:** Option D (Do Not Accept the Null) does not satisfy the conditions derived in the step-by-step analysis for Hypothesis Test Conclusion Using Chi-Square Table (correct choice: A). E: **Incorrect:** Option E (None of these) does not satisfy the conditions derived in the step-by-step analysis for Hypothesis Test Conclusion Using Chi-Square Table (correct choice: A).

**Answer: 3.75** **Topic:** Goodness-of-Fit Test — *Expected Frequency for Uniform Distribution Across Equal-Width Bins* ### Step-by-Step Solution **Step 1:** There are $N = 15$ dates in Table-1, and the range $[11, 30]$ is partitioned into $k = 4$ equal-width bins: `[11 to 15]`, `[16 to 20]`, `[21 to 25]`, and `[26 to 30]`. **Step 2:** Under the null hypothesis of a uniform distribution across these 4 equal-width bins, each bin has equal probability $p_i = \frac{1}{4} = 0.25$. **Step 3:** Calculate the expected frequency in any given bin: $$E_i = \frac{N}{k} = \frac{15}{4} = 3.75$$ **Key Takeaway / Tip:** Expected frequencies do not have to be integers; keep the exact decimal $15 / 4 = 3.75$.

**Answer: 3** **Topic:** Goodness-of-Fit Test — *Degrees of Freedom for Uniform Goodness-of-Fit Test* ### Step-by-Step Solution **Step 1:** In this uniform goodness-of-fit test, the minimum ($11$) and maximum ($30$) are pre-specified in the problem statement, and the data is grouped into $k = 4$ bins. **Step 2:** Therefore, the degrees of freedom is: $$df = k - 1 = 4 - 1 = 3$$ **Key Takeaway / Tip:** With $k = 4$ bins and pre-specified uniform bounds $[11, 30]$, the degrees of freedom is $k - 1 = 3$.

**Answer: 1.80** **Topic:** Goodness-of-Fit Test — *Chi-Square Test Statistic for Uniform Distribution of Daily Totals* ### Step-by-Step Solution **Step 1:** Calculate the combined daily box-office collection ($\text{MK} + \text{YV}$) for each of the 15 dates in Table-1: - Jan 1: $11 + 10 = 21$ - Jan 2: $9 + 10 = 19$ - Jan 3: $15 + 9 = 24$ - Jan 4: $8 + 6 = 14$ - Jan 5: $18 + 2 = 20$ - Jan 6: $16 + 10 = 26$ - Jan 7: $19 + 7 = 26$ - Jan 8: $8 + 6 = 14$ - Jan 9: $10 + 1 = 11$ - Jan 10: $19 + 3 = 22$ - Jan 11: $19 + 5 = 24$ - Jan 12: $19 + 8 = 27$ - Jan 13: $15 + 8 = 23$ - Jan 14: $17 + 5 = 22$ - Jan 15: $15 + 2 = 17$ **Step 2:** Count the observed frequencies $O_i$ in each of the 4 bins: - `[11 to 15]`: `14, 14, 11` $\implies O_1 = 3$ - `[16 to 20]`: `19, 20, 17` $\implies O_2 = 3$ - `[21 to 25]`: `21, 24, 22, 24, 23, 22` $\implies O_3 = 6$ - `[26 to 30]`: `26, 26, 27` $\implies O_4 = 3$ **Step 3:** Compute $\chi^2_{\text{calc}} = \sum_{i=1}^4 \frac{(O_i - 3.75)^2}{3.75}$: $$\chi^2_{\text{calc}} = \frac{(3 - 3.75)^2 + (3 - 3.75)^2 + (6 - 3.75)^2 + (3 - 3.75)^2}{3.75} = \frac{0.5625 + 0.5625 + 5.0625 + 0.5625}{3.75} = \frac{6.75}{3.75} = 1.80$$ **Key Takeaway / Tip:** Verify that the observed bin counts sum to the total number of dates: $3 + 3 + 6 + 3 = 15$.

**Answer: 0.50** **Topic:** Demand Modeling — *Point Price Elasticity of Linear Demand* ### Step-by-Step Solution **Step 1:** Evaluate the demand $D(P) = 4500 - 30P$ at the given price $P = 50$: $$D(50) = 4500 - 30(50) = 4500 - 1500 = 3000$$ **Step 2:** Note that the slope of the linear demand curve is $\frac{dD}{dP} = -30$. **Step 3:** Compute the magnitude of the price elasticity of demand at $P = 50$: $$|E_p| = \left|\frac{dD}{dP}\right| \cdot \frac{P}{D(P)} = 30 \times \frac{50}{3000} = \frac{1500}{3000} = 0.50$$ **Key Takeaway / Tip:** At $P = 50$, $|E_p| = 0.50 < 1$, which means demand is price-inelastic at this price point.

**Answer: 150.00** **Topic:** Demand Modeling — *Satiating (Choke) Price of Linear Demand* ### Step-by-Step Solution **Step 1:** The **satiating price** is the price $P$ at which the quantity demanded falls to zero ($D(P) = 0$). **Step 2:** Set $D(P) = 4500 - 30P = 0$ and solve for $P$: $$4500 - 30P = 0 \implies 30P = 4500 \implies P = \frac{4500}{30} = 150.00$$ **Key Takeaway / Tip:** At $P = 50$, notice that $|E_p| = \frac{P}{P_{\text{satiating}} - P} = \frac{50}{150 - 50} = \frac{50}{100} = 0.50$, which provides a fast cross-check between Questions 61 and 62!

**Answer: 4500** **Topic:** Demand Modeling — *Market Size (Maximum Potential Demand) of Linear Demand* ### Step-by-Step Solution **Step 1:** In a linear demand-response model $D(P) = a - bP$, the **market size** is the maximum potential demand achieved when the product is offered for free ($P = 0$). **Step 2:** Substitute $P = 0$ into $D(P) = 4500 - 30P$: $$D(0) = 4500 - 30(0) = 4500$$ **Key Takeaway / Tip:** The intercept $a = 4500$ of the linear demand function $D(P) = a - bP$ represents the total market size.

**Answer: 0.09** **Topic:** Probability and Bayes' Theorem — *Joint Probability of Centre A and Visa Granted* ### Step-by-Step Solution **Step 1:** Identify the prior probability of applying through Centre A and the conditional probability of a visa being granted given Centre A: - $P(A) = 15\% = 0.15$ - $P(\text{Granted} \mid A) = 60\% = 0.60$ **Step 2:** Apply the multiplication rule of probability to find the joint probability $P(A \cap \text{Granted})$: $$P(A \cap \text{Granted}) = P(A) \times P(\text{Granted} \mid A) = 0.15 \times 0.60 = 0.09$$ **Key Takeaway / Tip:** The phrase *'applying through centre A **and** it is granted'* refers to the joint probability $P(A \cap G) = P(A)P(G \mid A)$.

**Answer: 0.02** **Topic:** Probability and Bayes' Theorem — *Joint Probability of Centre B and Visa Not Granted* ### Step-by-Step Solution **Step 1:** Identify the prior probability of applying through Centre B and the conditional probability of a visa NOT being granted given Centre B: - $P(B) = 10\% = 0.10$ - $P(\text{Granted} \mid B) = 80\% = 0.80 \implies P(\text{Not Granted} \mid B) = 1 - 0.80 = 0.20$ **Step 2:** Multiply to find the joint probability $P(B \cap \text{Not Granted})$: $$P(B \cap \text{Not Granted}) = P(B) \times P(\text{Not Granted} \mid B) = 0.10 \times 0.20 = 0.02$$ **Key Takeaway / Tip:** Remember to use the complement $1 - 0.80 = 0.20$ since the question asks for the visa **not** being granted.

**Answer: 187.5** **Topic:** Probability and Bayes' Theorem — *Expected Number of Granted Visas from Centres C, D, and E* ### Step-by-Step Solution **Step 1:** Each of the three centres C, D, and E receives $25\%$ ($0.25$) of applications and grants a visa with probability $50\%$ ($0.50$). **Step 2:** Compute the combined probability of an application coming from Centres C, D, or E and being granted a visa: $$P((C \cup D \cup E) \cap \text{Granted}) = 3 \times (0.25 \times 0.50) = 3 \times 0.125 = 0.375$$ **Step 3:** Multiply by $500$ applications to find the expected number coming from Centres C, D, and E: $$500 \times 0.375 = 187.5$$ **Key Takeaway / Tip:** Each of C, D, and E contributes $500 \times 0.125 = 62.5$ granted applications, totaling $3 \times 62.5 = 187.5$.