Statistics Exam Questions and Solutions: Outliers,
Probability, and Hypothesis Testing - Prof. Mumo
(2026/2027)
Applied Statistical Analysis | Key Domains: Detection & Treatment of Outliers, Conditional
Probability & Bayes' Theorem, Design of Experiments, Advanced Hypothesis Testing (Two Samples,
ANOVA, Chi-Square), Nonparametric Methods, and Power & Error Analysis | Expert-Aligned
Structure | Multiple-Choice Exam Format
Introduction
This structured Statistics Exam for 2026/2027 provides 60 multiple-choice questions with
correct answers and rationales. It covers intermediate to advanced topics, focusing on the
application of statistical reasoning to analyze data, test hypotheses, and draw valid conclusions in
research and real-world contexts.
Exam Structure:
● Applied Statistics Exam: (60 MULTIPLE-CHOICE QUESTIONS)
Answer Format
All correct answers and data-driven conclusions must appear in bold and cyan blue, accompanied
by concise rationales explaining the criterion for identifying an outlier (e.g., IQR method), the
application of a probability rule (e.g., calculating P(A|B) using Bayes' Theorem), the selection of the
appropriate hypothesis test based on study design and data type (e.g., paired t-test, Chi-square test
of independence), the interpretation of a test statistic in context, and why the alternative
multiple-choice options involve misuse of formulas, incorrect test selection, or logical fallacies in
statistical inference.
Applied Statistics Exam (60 Multiple-Choice Questions)
,1. Using the IQR method, which value is an outlier in the data set: 10, 12, 15, 18, 20, 22, 25,
30, 50?
A. 10
B. 25
C. 30
D. 50
D. 50
Rationale: Q1 = 15, Q3 = 25, IQR = 10. Upper fence = Q3 + 1.5×IQR = 25 + 15 = 40. Lower fence = 15 - 15
= 0. Values >40 or <0 are outliers. 50 > 40, so it is an outlier. Other values are within [0, 40].
2. In a medical test, P(Disease) = 0.01, P(Positive|Disease) = 0.95, and P(Positive|No Disease)
= 0.05. What is P(Disease|Positive)?
A. 0.01
B. 0.16
C. 0.95
D. 0.99
B. 0.16
Rationale: By Bayes’ Theorem: P(D|P) = [P(P|D)P(D)] / [P(P|D)P(D) + P(P|¬D)P(¬D)] = (0.95×0.01) /
(0.95×0.01 + 0.05×0.99) = 0.0095 / (0.0095 + 0.0495) = 0..059 ≈ 0.161. Despite high sensitivity,
low prevalence leads to many false positives.
3. Which experimental design controls for confounding by using matched pairs?
A. Completely randomized design
B. Randomized block design
C. Matched-pairs design
D. Observational study
C. Matched-pairs design
, Rationale: Matched-pairs design compares two treatments on similar units (e.g., twins, before-after),
controlling for individual variability. Randomized block (B) groups similar units into blocks;
completely randomized (A) assigns randomly without blocking; observational studies (D) cannot
control confounding.
4. A researcher compares pre- and post-test scores for the same group. Which test is
appropriate?
A. Two-sample t-test
B. Paired t-test
C. One-sample t-test
D. Chi-square test
B. Paired t-test
Rationale: Paired t-test is used for dependent samples (same subjects measured twice). Two-sample
t-test (A) is for independent groups; one-sample (C) compares to a known mean; chi-square (D) is for
categorical data.
5. In a chi-square test of independence, the null hypothesis states that:
A. The variables are independent
B. The variables are dependent
C. The proportions are equal
D. The means are equal
A. The variables are independent
Rationale: Chi-square test of independence assesses whether two categorical variables are related. H₀:
no association (independent); H₁: association exists (dependent). Options C and D relate to other tests
(ANOVA, z-test for proportions).
6. Which nonparametric test is the analog of the paired t-test?
A. Mann-Whitney U test
B. Wilcoxon signed-rank test
C. Kruskal-Wallis test
Probability, and Hypothesis Testing - Prof. Mumo
(2026/2027)
Applied Statistical Analysis | Key Domains: Detection & Treatment of Outliers, Conditional
Probability & Bayes' Theorem, Design of Experiments, Advanced Hypothesis Testing (Two Samples,
ANOVA, Chi-Square), Nonparametric Methods, and Power & Error Analysis | Expert-Aligned
Structure | Multiple-Choice Exam Format
Introduction
This structured Statistics Exam for 2026/2027 provides 60 multiple-choice questions with
correct answers and rationales. It covers intermediate to advanced topics, focusing on the
application of statistical reasoning to analyze data, test hypotheses, and draw valid conclusions in
research and real-world contexts.
Exam Structure:
● Applied Statistics Exam: (60 MULTIPLE-CHOICE QUESTIONS)
Answer Format
All correct answers and data-driven conclusions must appear in bold and cyan blue, accompanied
by concise rationales explaining the criterion for identifying an outlier (e.g., IQR method), the
application of a probability rule (e.g., calculating P(A|B) using Bayes' Theorem), the selection of the
appropriate hypothesis test based on study design and data type (e.g., paired t-test, Chi-square test
of independence), the interpretation of a test statistic in context, and why the alternative
multiple-choice options involve misuse of formulas, incorrect test selection, or logical fallacies in
statistical inference.
Applied Statistics Exam (60 Multiple-Choice Questions)
,1. Using the IQR method, which value is an outlier in the data set: 10, 12, 15, 18, 20, 22, 25,
30, 50?
A. 10
B. 25
C. 30
D. 50
D. 50
Rationale: Q1 = 15, Q3 = 25, IQR = 10. Upper fence = Q3 + 1.5×IQR = 25 + 15 = 40. Lower fence = 15 - 15
= 0. Values >40 or <0 are outliers. 50 > 40, so it is an outlier. Other values are within [0, 40].
2. In a medical test, P(Disease) = 0.01, P(Positive|Disease) = 0.95, and P(Positive|No Disease)
= 0.05. What is P(Disease|Positive)?
A. 0.01
B. 0.16
C. 0.95
D. 0.99
B. 0.16
Rationale: By Bayes’ Theorem: P(D|P) = [P(P|D)P(D)] / [P(P|D)P(D) + P(P|¬D)P(¬D)] = (0.95×0.01) /
(0.95×0.01 + 0.05×0.99) = 0.0095 / (0.0095 + 0.0495) = 0..059 ≈ 0.161. Despite high sensitivity,
low prevalence leads to many false positives.
3. Which experimental design controls for confounding by using matched pairs?
A. Completely randomized design
B. Randomized block design
C. Matched-pairs design
D. Observational study
C. Matched-pairs design
, Rationale: Matched-pairs design compares two treatments on similar units (e.g., twins, before-after),
controlling for individual variability. Randomized block (B) groups similar units into blocks;
completely randomized (A) assigns randomly without blocking; observational studies (D) cannot
control confounding.
4. A researcher compares pre- and post-test scores for the same group. Which test is
appropriate?
A. Two-sample t-test
B. Paired t-test
C. One-sample t-test
D. Chi-square test
B. Paired t-test
Rationale: Paired t-test is used for dependent samples (same subjects measured twice). Two-sample
t-test (A) is for independent groups; one-sample (C) compares to a known mean; chi-square (D) is for
categorical data.
5. In a chi-square test of independence, the null hypothesis states that:
A. The variables are independent
B. The variables are dependent
C. The proportions are equal
D. The means are equal
A. The variables are independent
Rationale: Chi-square test of independence assesses whether two categorical variables are related. H₀:
no association (independent); H₁: association exists (dependent). Options C and D relate to other tests
(ANOVA, z-test for proportions).
6. Which nonparametric test is the analog of the paired t-test?
A. Mann-Whitney U test
B. Wilcoxon signed-rank test
C. Kruskal-Wallis test
