DA0-002 Data Analysis Practice Question
A data analyst is examining a dataset of daily website visits collected over the past year. The analyst wants to determine whether the distribution of visits on weekdays differs significantly from the distribution on weekends, without assuming normality. The analyst has already separated the data into two independent groups. Which statistical test should the analyst use?
⚠ Common exam trap
The trap here is assuming that any nonparametric test for two groups works, but the paired t-test and one-sample t-test require different data structures, while the chi-square test is for categorical frequencies, not continuous distributions.
Answer choices
Why each option matters
Answer the question above first, then reveal the full breakdown to understand why each option is right or wrong.
Correct answer & explanation
✓
Mann-Whitney U test
The Mann-Whitney U test is a nonparametric method that compares two independent groups without requiring normality. It assesses whether one group tends to have larger values than the other by ranking all observations. Since the analyst wants to compare weekday and weekend website visits and cannot assume normality, this test is the correct choice. It is specifically designed for independent groups and continuous or ordinal data.
Answer analysis
Option-by-option breakdown
For each option: why learners choose it and why it is or isn't the right answer here.
- ✗
One-sample t-test
Why it's wrong here
A one-sample t-test compares a single sample mean to a known or hypothesized population mean. The analyst has two separate groups and wants to compare their distributions, not test one group against a fixed value. This test also assumes normality, which contradicts the requirement to avoid that assumption. Therefore, it does not fit the scenario of comparing weekday versus weekend visit patterns.
- ✓
Mann-Whitney U test
Why this is correct
The Mann-Whitney U test is a nonparametric alternative to the independent samples t-test, comparing whether two independent groups come from the same distribution. It does not assume normality, making it appropriate for skewed visit counts. It ranks all observations and assesses whether one group tends to have higher values. This directly addresses the analyst's question of whether weekday and weekend visit distributions differ without distributional assumptions.
- ✗
Paired samples t-test
Why it's wrong here
A paired samples t-test requires dependent groups, such as before-and-after measurements on the same subjects. Here, weekday and weekend visits are independent groups from different days, not matched pairs. Using a paired test would violate the independence assumption and produce misleading results. Additionally, the t-test assumes normality, which the analyst explicitly wants to avoid, making this choice doubly inappropriate.
- ✗
Chi-square goodness-of-fit test
Why it's wrong here
The chi-square goodness-of-fit test determines whether an observed frequency distribution matches an expected distribution for a single categorical variable. It is not designed to compare two independent groups' distributions of a continuous variable like visit counts. While it is nonparametric, it operates on counts in categories, not on ranks of continuous data. Thus, it does not answer whether weekday and weekend visit distributions differ.
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JA
Written and reviewed by Johnson Ajibi, MSc IT Security
Senior Network & Security Engineer · founder of Courseiva
Last reviewed September 2026 · checked against the official CompTIA exam blueprint
This DA0-002 practice question is part of Courseiva's free CompTIA certification practice question bank. Courseiva provides original exam-style practice questions with explanations, topic-based practice, mock exams, readiness tracking, and study analytics to help learners prepare for the DA0-002 exam.