DA0-002 Data Analysis Practice Question
A data analyst is performing a chi-square test for independence between two categorical variables. Which THREE of the following are necessary conditions for the test to be valid?
⚠ Common exam trap
DA0-002 often tests the confusion between parametric assumptions (normality, equal variance) and chi-square's non-parametric requirements — candidates incorrectly apply t-test or ANOVA assumptions to chi-square.
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
✓
Sample is randomly selected
Option C is correct because a chi-square test for independence requires that the sample be randomly selected from the population, ensuring the results can be generalized and that the expected counts reflect the underlying distribution. Option D is correct because the test assumes observations are independent; each subject or case must contribute to only one cell of the contingency table, and correlated or repeated observations violate the chi-square model. Option E is correct because the chi-square approximation is valid only when the expected frequency in each cell is at least 5 (or, in larger tables, when no more than 20% of cells have expected counts below 5 and none below 1). Option A is not required because chi-square is a nonparametric test of frequencies and does not assume homogeneity of variances. Option B is not required because chi-square does not assume normally distributed data; it operates on counts of categorical outcomes.
Answer analysis
Option-by-option breakdown
For each option: why learners choose it and why it is or isn't the right answer here.
- ✗
Variances are equal across groups
Why it's wrong here
Chi-square tests counts in contingency tables, not group means, so equal variances are irrelevant; that assumption belongs to ANOVA or t-tests comparing continuous outcomes. It tempts analysts because variance homogeneity is drilled into parametric testing, and it would be required when comparing means across the same categories.
- ✗
Data is normally distributed
Why it's wrong here
Chi-square for independence assumes categorical data with expected cell frequencies of at least five, not normality; it is a distribution-free test. It is tempting because normality underpins t-tests and ANOVA, but those compare continuous means, not categorical association.
- ✓
Sample is randomly selected
Why this is correct
Random selection ensures the sample represents the population, so observed cell counts estimate population proportions without selection bias. This satisfies the chi-square validity condition that expected frequencies reflect genuine population distributions rather than a biased subset.
- ✓
Observations are independent
Why this is correct
Independent observations mean each subject contributes to exactly one cell, so counts are not duplicated or clustered. This satisfies the chi-square validity condition by ensuring the summed cell counts follow the assumed multinomial distribution underlying the test statistic.
- ✓
Expected frequency in each cell is at least 5
Why this is correct
The chi-square statistic approximates a chi-square distribution only when each cell's expected count reaches five, preventing sparse cells from distorting the test. This satisfies the stem's validity condition, since smaller expected frequencies inflate Type I error rates.
About these practice questions
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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.