DA0-002 Data Analysis Practice Question
A data analyst is performing a chi-square test of independence on a contingency table of customer satisfaction (satisfied vs. dissatisfied) and product type (A, B, C). The test yields a p-value of 0.04 with α = 0.05. What is the correct conclusion?
⚠ Common exam trap
DA0-002 often tests the interpretation of p-values versus α, and candidates frequently confuse 'fail to reject' with 'accept the null' or misinterpret a significant result as proving causation rather than association.
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
✓
There is a significant association between satisfaction and product type.
With a p-value of 0.04 and α = 0.05, the p-value is less than the significance level, so we reject the null hypothesis of independence. This means there is statistically significant evidence of an association between customer satisfaction and product type. The correct conclusion is that a significant association exists.
Answer analysis
Option-by-option breakdown
For each option: why learners choose it and why it is or isn't the right answer here.
- ✗
There is no evidence of an association between satisfaction and product type.
Why it's wrong here
A p-value of 0.04 falls below α = 0.05, so the null hypothesis of independence is rejected; concluding no association contradicts the computed result. It tempts because a small p-value is often misread as weak evidence, which would be the correct reading only had the p-value exceeded the significance threshold.
- ✓
There is a significant association between satisfaction and product type.
Why this is correct
With p = 0.04 below the α = 0.05 threshold, the null hypothesis of independence is rejected, so satisfaction and product type are statistically associated. The chi-square test of independence detects whether the two categorical variables' observed cell frequencies deviate from those expected under independence.
- ✗
The test is invalid because the expected counts are too low.
Why it's wrong here
Nothing in the stem reports expected counts below five, and chi-square remains valid when each cell's expected frequency meets that threshold. Low expected counts would instead call for Fisher's exact test or category pooling, so this objection does not apply to the given table.
- ✗
Satisfaction and product type are independent.
Why it's wrong here
A p-value of 0.04 falls below α = 0.05, so the null hypothesis of independence is rejected; the data indicate an association between satisfaction and product type. Independence would be the conclusion only when the p-value exceeds the significance threshold, meaning no evidence of association was detected.
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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.