CompTIA DataX Practice Test (DY0-001)

Because the sampling distribution of Pearson's r is skewed and its variance depends on the unknown population correlation (ρ), a direct calculation using normal theory is inappropriate. Fisher's z-transformation-z = atanh(r) = ½ ln[(1+r)/(1−r)]-is a variance-stabilizing transform that makes the resulting statistic, z, approximately normally distributed as N(atanh(ρ), 1/(n−3)). A 95% interval for this transformed value is therefore z ± 1.96 / √(n−3). Applying the inverse transform (tanh) to the interval's endpoints yields the confidence interval for ρ. The Wilson score interval is designed for binomial proportions. A Box-Cox transformation applies to the raw data, not the correlation coefficient r. The statistic r√(n−2)/√(1−r²) follows a t-distribution and is used for hypothesis testing, not interval estimation.

The correct answer is that the standard errors of the coefficients are biased, which renders hypothesis tests and confidence intervals unreliable. The fan-shaped pattern in the residual plot is a classic indicator of heteroskedasticity, which means the variance of the error term is not constant across all levels of the independent variables. In the presence of heteroskedasticity, Ordinary Least Squares (OLS) coefficient estimates remain unbiased, but they are no longer efficient (i.e., not BLUE - Best Linear Unbiased Estimators). The primary issue for statistical inference is that the formulas used to calculate the variance and standard errors of the coefficients, which assume homoskedasticity, become biased. This bias in the standard errors leads to unreliable t-statistics, p-values, and confidence intervals, potentially causing the analyst to draw incorrect conclusions about the statistical significance of the predictor variables.

The coefficient estimates themselves do not become biased due to heteroskedasticity in an OLS model. Multicollinearity is a separate issue related to high correlation between predictor variables, not the variance of the residuals. While the normality of residuals is another OLS assumption, the fan shape specifically points to non-constant variance (heteroskedasticity), not necessarily a deviation from a normal distribution.

Adjusted R² modifies the ordinary R² by incorporating both sample size and the number of predictors, so it rises only when additional variables reduce the residual variance more than would be expected by chance. It therefore rewards genuine improvement while penalizing unnecessary complexity. In the table, Model 2 attains the highest adjusted R², indicating the best trade-off between parsimony and predictive power. Plain R² and RMSE both improve (or stay nearly the same) as more predictors are added, so they cannot flag overfitting. The F-statistic p-value only tests whether each model outperforms an intercept-only model; because all three p-values are identical, it offers no guidance for choosing among the competing models.

The missingness depends on two fully observed variables-age and loyalty tier-but, conditional on them, it is not related to the spending values that are actually missing. This matches the definition of Missing at Random (MAR). Under MAR, the missing-data mechanism is considered ignorable for likelihood-based models or multiple imputation, provided the observed predictors that drive missingness are included in the analysis. The mechanism is not Missing Completely at Random (MCAR) because younger, high-tier customers have a higher propensity for missingness, and it is not Missing Not at Random (MNAR) because spending itself does not influence whether it is missing once age and tier are taken into account.

The correct answer is Recall. Recall, also known as sensitivity or the true positive rate, is calculated as TP / (TP + FN), where TP is True Positives and FN is False Negatives. In scenarios where the cost of a False Negative is very high, such as failing to predict a critical equipment failure, maximizing recall is the primary objective. This metric directly measures the model's ability to identify all actual positive instances.

• Precision, calculated as TP / (TP + FP), measures the accuracy of positive predictions. Prioritizing precision would aim to reduce False Positives, which is not the main concern in this scenario.
• Accuracy is not suitable for highly imbalanced datasets because a model can achieve a high accuracy score by simply predicting the majority class, while completely failing to identify the rare, critical events.
• The F1 score is the harmonic mean of precision and recall and seeks a balance between them. While useful, it does not specifically prioritize minimizing False Negatives, which is the explicit goal here.

For the overall significance test in multiple regression, the F statistic can be written in terms of R²:

F = (R² / p) ÷ [(1 - R²) / (n - p - 1)]

Substituting the values:

• Numerator term: R² / p = 0.37 / 6 ≈ 0.0617
• Denominator term: (1 - R²) / (n - p - 1) = 0.63 / 73 ≈ 0.00863

F = 0.0617 / 0.00863 ≈ 7.15

Because this value greatly exceeds typical critical values for the F(6, 73) distribution at α = 0.05 (≈ 2.20), the null hypothesis would be rejected.

The distractor values reflect common errors: computing R² / (1 - R²) alone (≈0.59), using (n - p - 1) / p alone (≈12.2), or omitting the intercept when calculating degrees of freedom (≈37).

The correct answer is the Cox Proportional Hazards model. This model is a semi-parametric regression model and is ideal for this scenario because it allows for the estimation of the effects of covariates (like subscription plan and spending) on the hazard rate without making any assumptions about the shape of the baseline hazard function. This directly addresses the requirement to quantify covariate effects while avoiding strong distributional assumptions.

• The Kaplan-Meier estimator is a non-parametric method used to estimate and visualize the survival function. While useful for initial analysis, it cannot incorporate multiple or continuous covariates into a regression framework to quantify their individual effects on the hazard rate.
• The Weibull AFT (Accelerated Failure Time) model is a fully parametric model. It requires the assumption that survival times follow a specific distribution (the Weibull distribution). This contradicts the data scientist's goal of avoiding strong assumptions about the underlying distribution.
• An ARIMA model is used for time series forecasting, which analyzes data points collected over time to predict future values (e.g., monthly sales). It is not designed for time-to-event analysis, which involves understanding the duration until an event occurs and must account for censored data and individual-level covariates.

Precision is defined as TP / (TP + FP), the proportion of positive predictions that are correct.

• Classifier X: 260 / (260 + 110) ≈ 0.703 (> 0.70).
• Classifier Y: 350 / (350 + 180) ≈ 0.660 (< 0.70).

Because only Classifier X achieves a precision of at least 0.70, it alone satisfies the compliance requirement. Classifier Y falls short despite having more true positives, because it also generates more false positives, lowering its precision.

The correct answer is that 'Contract Type' should be selected because its split results in a lower weighted average entropy. The goal of a decision tree split is to maximize Information Gain, which is equivalent to minimizing the weighted average entropy of the child nodes.

The calculation is as follows:

1. Calculate the entropy for each child node. The formula for entropy is: E = -p * log2(p) - (1-p) * log2(1-p).

   • E(Node A) (9/12 Churn): -( (9/12) * log2(9/12) + (3/12) * log2(3/12) ) ≈ 0.811
   • E(Node B) (1/8 Churn): -( (1/8) * log2(1/8) + (7/8) * log2(7/8) ) ≈ 0.544
   • E(Node C) (3/10 Churn): -( (3/10) * log2(3/10) + (7/10) * log2(7/10) ) ≈ 0.881
   • E(Node D) (7/10 Churn): -( (7/10) * log2(7/10) + (3/10) * log2(3/10) ) ≈ 0.881

2. Calculate the weighted average entropy for each split. The formula is the sum of (samples_in_child / total_samples) * entropy_of_child.

   • W_avg_entropy('Contract Type') = (12/20) * 0.811 + (8/20) * 0.544 = 0.6 * 0.811 + 0.4 * 0.544 ≈ 0.487 + 0.218 = 0.705
   • W_avg_entropy('Has Tech Support') = (10/20) * 0.881 + (10/20) * 0.881 = 0.5 * 0.881 + 0.5 * 0.881 = 0.881

3. Compare the results. The split on 'Contract Type' (0.705) has a lower weighted average entropy than the split on 'Has Tech Support' (0.881). Therefore, 'Contract Type' yields a higher information gain and is the better split.

The other options are incorrect. The 'Has Tech Support' split has a higher weighted entropy, making it the less desirable choice. The balance of sample sizes in the child nodes for 'Has Tech Support' does not guarantee higher information gain; the purity of the classes within those nodes is what matters. Finally, calculating the Gini index is an alternative to entropy, not a necessary tie-breaker.

The correct answer is Recall.

Recall (Sensitivity or True Positive Rate) is calculated as TP / (TP + FN). It measures the proportion of actual positive cases that the model correctly identified. In this scenario, Recall = 800 / (800 + 200) = 80%. This metric directly addresses the business objective of minimizing missed fraudulent transactions (False Negatives). A high recall indicates that the model is effective at identifying the vast majority of actual fraud cases.

Accuracy is incorrect because it is a misleading metric for datasets with severe class imbalance. It is calculated as (TP + TN) / Total, which in this case is (800 + 998,500) / 1,000,000 = 99.93%. While this number seems very high, a naive model that predicts "Not Fraud" for every transaction would achieve 99.9% accuracy, making it a poor indicator of the model's ability to detect the rare positive class.

Precision is incorrect in this context. Precision is calculated as TP / (TP + FP) and measures the proportion of positive predictions that were actually correct. Here, Precision = 800 / (800 + 500) = 61.5%. This metric is important when the cost of a False Positive is high. However, the business objective explicitly prioritizes minimizing False Negatives over False Positives, making Recall the more relevant metric.

Matthews Correlation Coefficient (MCC) is a sophisticated and generally robust metric for imbalanced datasets because it considers all four cells of the confusion matrix. However, the question asks for the metric that is most relevant to the specific business objective of minimizing False Negatives. While MCC provides a balanced, overall score, Recall is the most direct and explicit measure of the model's performance against that particular goal.

The Box-Muller transform maps two independent Uniform(0, 1) variables to two independent N(0, 1) variables by converting the uniform samples into polar coordinates. The correct transformation is:

• Z1 = √(-2 ln U1) cos(2π U2)
• Z2 = √(-2 ln U1) sin(2π U2)

This works by setting the squared radius R² = -2 ln U1 and the angle Θ = 2π U2. Because the Jacobian of this transformation exactly cancels the standard bivariate normal density, the resulting pair has the desired distribution. The other choices break one or more of the requirements:

• Transformations that use π instead of 2π for the angle or take logarithms of both U1 and U2 distort the output, so it is no longer standard normal.
• Mixing the two uniform variables inside the logarithm, such as in the expression √(-2 ln (U1 / U2)), changes the radial distribution and makes Z1 and Z2 dependent, so they are neither independent nor standard normal.

Therefore, the transformation using √(-2 ln U1) for the radius and the trigonometric functions of 2π U2 for the angle is the only one that satisfies the specification.

The Poisson distribution is designed for modeling the number of independent events that occur in a fixed interval when those events are rare and occur at a constant average rate. A defining property of the Poisson distribution is that its mean and variance are both equal to the rate parameter λ. Because the observed data are non-negative integer counts, the mean (0.18) is very close to the variance (0.20), and outages are presumed independent from day to day, the Poisson distribution is the most appropriate first model.

The binomial distribution is inappropriate here because it requires a fixed number of identical trials (n) each day; in this context n would need to be the unknown number of "possible outage opportunities" within a day, and its variance is np(1 − p), which is strictly less than the mean np. Since the sample variance is slightly greater than the sample mean, the binomial is a poorer fit than the Poisson.

The Student's t-distribution is a continuous distribution used for inference on sample means when population variance is unknown; it cannot generate non-negative integer counts.

A power-law distribution is heavy-tailed and continuous or defined on positive integers with probabilities that decay polynomially; it is suited to modeling extreme events across many orders of magnitude, not tightly bounded low counts such as 0, 1, 2 outages. Therefore, only the Poisson model aligns with both the empirical moment relationship and the data-generation mechanism.

The correct answer is 0.417. The weighted Gini impurity is calculated by finding the Gini impurity for each child node and then computing their weighted average based on the number of samples in each node.

Step 1: Calculate Gini Impurity for Child Node 1

• The proportion of 'Churn' is p1 = 40/60 = 2/3.
• The proportion of 'No Churn' is p2 = 20/60 = 1/3.
• Gini(Node 1) = 1 - (p1^{2 + p2}2) = 1 - ((2/3)^2 + (1/3)^2) = 1 - (4/9 + 1/9) = 1 - 5/9 = 4/9 ≈ 0.444.

Step 2: Calculate Gini Impurity for Child Node 2

• The proportion of 'Churn' is p1 = 10/40 = 1/4.
• The proportion of 'No Churn' is p2 = 30/40 = 3/4.
• Gini(Node 2) = 1 - (p1^{2 + p2}2) = 1 - ((1/4)^2 + (3/4)^2) = 1 - (1/16 + 9/16) = 1 - 10/16 = 6/16 = 0.375.

Step 3: Calculate the Weighted Gini Impurity

• The weight for Node 1 is the number of samples in it divided by the total samples in the parent: w1 = 60/100 = 0.6.
• The weight for Node 2 is w2 = 40/100 = 0.4.
• Weighted Gini = (w1 * Gini(Node 1)) + (w2 * Gini(Node 2)) = (0.6 * 4/9) + (0.4 * 0.375) ≈ 0.267 + 0.150 = 0.417.

Incorrect Answer Analysis:

• 0.500 is the Gini impurity of the parent node (1 - (0.5^{2 + 0.5}2) = 0.5), not the weighted impurity of the split.
• 0.083 is the Information Gain (Gini Gain), which is calculated by subtracting the weighted Gini impurity from the parent node's Gini impurity (0.500 - 0.417 = 0.083).
• 0.410 is an incorrect calculation, likely resulting from taking a simple average of the two child node impurities instead of a weighted average.

The correct answer is Missing at Random (MAR). In this scenario, the missingness of the systolic_blood_pressure (SBP) is dependent on another observed variable, which is patient_age. This is the key characteristic of MAR: the probability of a value being missing is related to other observed information in the dataset but not to the unobserved value itself.

• Missing Completely at Random (MCAR) is incorrect because the missingness is not completely random; it has a systematic relationship with the patient_age variable. If the data were MCAR, the probability of a missing SBP value would be the same for all patients, regardless of their age or any other characteristic.

• Not Missing at Random (NMAR) is incorrect because the scenario explicitly states that the missingness is not related to the actual unobserved blood pressure level. NMAR would apply if, for example, patients with very high blood pressure were less likely to have their SBP recorded, meaning the missingness depends on the value of the missing variable itself.

• Structural Missingness is incorrect. This term typically refers to data that is missing for a logical reason inherent in the study's design. For instance, a question about the number of pregnancies would be structurally missing for male participants. The scenario described does not fit this definition.

The Pearson correlation for a sample is r = Σ(x_i − x̄)(y_i − ȳ) / [(n − 1)s_X s_Y]. Substituting the given values gives r = 36 / (6 × 4 × 3) = 0.50. To test significance, use the t statistic t = r √(n − 2) / √(1 − r²). With n = 7, t = 0.50 × √5 / √(1 − 0.25) ≈ 1.29. The critical value for a two-tailed test with df = 5 at α = 0.05 is about ±2.57, so |1.29| does not exceed the critical value. Therefore the analyst fails to reject H₀; the correlation is not statistically significant at the 0.05 level.