HLT-362V ANOVA and Chi-Square Exercises: Examples, Answers, Solved Problems & Guide

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If you are working through HLT-362V Applied Statistics for Health Care Professionals at Grand Canyon University, ANOVA and chi-square are usually the two topics that slow students down the most. Both tests show up in weekly assignments, in the topic quizzes, and later in the article-analysis assignments where you have to read a published study and correctly identify which statistical test the researchers used and why.

This guide walks through what ANOVA and chi-square actually test, how to calculate each one by hand, how to read the output, and how to avoid the mistakes that cost the most points on HLT-362V assignments. It includes two fully solved chi-square problems, a worked ANOVA example, a decision table for choosing between a t-test, ANOVA, and chi-square, and a checklist you can use to check your own work before you submit it. GCU’s current course description for HLT-362V frames it as an introductory statistics course focused on applying appropriate statistical methods to healthcare research, so the goal here is the same: understand the test well enough to apply it correctly, not just memorize a formula.

This assignment is part of HLT-362V Applied Statistics for Health Care at GCU. For a complete guide to every HLT-362V assignment in sequence, see the HLT-362V Applied Statistics for Health Care guide.

HLT-362V ANOVA and Chi-Square Exercises at a Glance

Start here if you are short on time and need to know which test applies to your assignment before you start calculating anything.

Use this table to identify the correct test before starting the calculation. If your outcome variable is a number (blood pressure, pain score, recovery time in days) and you are comparing three or more groups, you want ANOVA. If your outcome variable is a category (yes/no, diagnosis type, unit assignment) and you are asking whether two categorical variables are related, you want chi-square.

What Is ANOVA?

Analysis of variance (ANOVA) is a statistical test that compares the means of three or more groups to determine whether at least one group differs significantly from the others. It is called “analysis of variance” because the test works by comparing the variance between group means to the variance within each group. If the between-group variance is large relative to the within-group variance, that is evidence the groups do not all come from the same underlying population.

In healthcare research, ANOVA answers questions such as:

• Do mean pain scores differ across three treatment groups?
• Does mean systolic blood pressure differ among three patient groups receiving different interventions?
• Do average recovery times differ among multiple treatment approaches?

Key terms you need for HLT-362V:

• Independent variable: the categorical grouping variable (for example, treatment type: A, B, or C)
• Dependent variable: the continuous outcome being measured (for example, pain score)
• F-statistic: the ratio of between-group variance to within-group variance; a larger F suggests a larger difference between group means relative to variability within groups
• p-value: the probability of observing a result this extreme if the null hypothesis were true
• Null hypothesis (H0): all group means are equal
• Alternative hypothesis (Ha): at least one group mean differs from the others
• Statistical significance: typically declared when p < 0.05, meaning the result is unlikely to have occurred by chance alone

A statistically significant ANOVA tells you that at least one group mean differs from the others. It does not tell you which groups differ. That is what post hoc tests, such as Tukey’s Honestly Significant Difference (HSD) test, are for. Tukey HSD compares every pair of group means while controlling the overall error rate, so you can say specifically, for example, that Group A differed from Group C but Group B did not differ from either (Lane, n.d., as archived in statistical methods references).

One more distinction that matters for HLT-362V: a statistically significant result is not automatically a clinically meaningful one. A large sample can produce a very small p-value for a mean difference that is too small to matter at the bedside. Current guidance in the health research literature recommends reporting effect sizes and confidence intervals alongside p-values rather than treating “p < 0.05” as the whole story (Ciapponi et al., 2021).

What Are the Three Types of ANOVA?

Terminology can vary slightly across textbooks and statistical software, so pay attention to how your specific assignment phrases the question.

1. One-way ANOVA

Used when there is one categorical independent variable with three or more groups and one continuous dependent variable.

Healthcare example: Comparing mean recovery time (in days) among patients assigned to three different post-surgical mobility protocols.

2. Two-way ANOVA

Used when examining the effects of two categorical independent variables on a continuous dependent variable. Two-way ANOVA can also test for an interaction effect between the two independent variables, meaning the effect of one variable may depend on the level of the other.

Healthcare example: Comparing mean blood glucose levels across three medication types and two age groups, while also checking whether medication effect differs by age group.

3. Repeated-measures ANOVA

Used when the same participants are measured multiple times or under multiple conditions, rather than comparing separate independent groups.

Healthcare example: Comparing a single group of patients’ pain scores measured before surgery, immediately after surgery, and at a two-week follow-up.

How to Read an ANOVA Table

An ANOVA output table typically includes the following columns:

• SS (Sum of Squares): total variability, split into Between Groups and Within Groups
• df (degrees of freedom): Between Groups df = k − 1 (k = number of groups); Within Groups df = N − k (N = total sample size)
• MS (Mean Square): SS divided by its corresponding df
• F: the test statistic (MS Between ÷ MS Within)
• p-value: the probability associated with the F statistic

Example ANOVA table:

This is reported in APA style as F(2, 27) = 5.84, p = 0.006. The first number in parentheses is the Between Groups df, and the second is the Within Groups df.

How to write the conclusion:

• Statistically significant result (p < 0.05): “There was a statistically significant difference in mean [outcome] among the three groups, F(2, 27) = 5.84, p = 0.006. The null hypothesis was rejected.”
• Statistically non-significant result (p ≥ 0.05): “There was no statistically significant difference in mean [outcome] among the three groups, F(2, 24) = 1.12, p = 0.341. The researchers failed to reject the null hypothesis.”

Use “reject the null hypothesis” or “fail to reject the null hypothesis.” Avoid “accept the null hypothesis” — failing to find a significant difference is not the same as proving the groups are equal.

HLT-362V ANOVA Worked Example

Research question: Does mean post-operative pain score (0–10 scale) differ among patients receiving three different pain management protocols?

H0: μA = μB = μC (mean pain scores are equal across all three protocols) Ha: At least one protocol group mean differs from the others

Data (pain scores, n = 5 per group):

Step 1: Group means

• Mean A = 30 / 5 = 6.0
• Mean B = 24 / 5 = 4.8
• Mean C = 15 / 5 = 3.0

Step 2: Grand mean Grand mean = (30 + 24 + 15) / 15 = 69 / 15 = 4.6

Step 3: Between-group sum of squares (SSB) SSB = Σ n(group mean − grand mean)² SSB = 5(6.0 − 4.6)² + 5(4.8 − 4.6)² + 5(3.0 − 4.6)² SSB = 5(1.96) + 5(0.04) + 5(2.56) SSB = 9.80 + 0.20 + 12.80 = 22.80

Step 4: Within-group sum of squares (SSW) Calculate the squared deviation of each score from its own group mean, then sum:

• Group A deviations from 6.0: 0, 1, 1, 0, 0 → squared: 0, 1, 1, 0, 0 → sum = 2.0
• Group B deviations from 4.8: −0.8, 0.2, −0.8, 1.2, 0.2 → squared: 0.64, 0.04, 0.64, 1.44, 0.04 → sum = 2.80
• Group C deviations from 3.0: 0, −1, 1, 0, 0 → squared: 0, 1, 1, 0, 0 → sum = 2.0

SSW = 2.0 + 2.80 + 2.0 = 6.80

Step 5: Degrees of freedom

• df Between = k − 1 = 3 − 1 = 2
• df Within = N − k = 15 − 3 = 12
• df Total = N − 1 = 14

Step 6: Mean squares

• MS Between = SSB / df Between = 22.80 / 2 = 11.40
• MS Within = SSW / df Within = 6.80 / 12 = 0.567

Step 7: F statistic F = MS Between / MS Within = 11.40 / 0.567 = 20.11

Step 8: Interpretation With df(2, 12), F = 20.11 far exceeds the critical F value of approximately 3.89 at α = 0.05, so this result is statistically significant (p < 0.001).

Step 9: Conclusion There was a statistically significant difference in mean post-operative pain score among the three pain management protocols, F(2, 12) = 20.11, p < 0.001. The null hypothesis was rejected. A Tukey HSD post hoc test would be the next step to determine which specific protocol pairs differed.

Need help working through your HLT-362V statistics exercises? Send Gradevia a message on WhatsApp for a quick scope review if you are stuck applying this process to your own dataset.

What Is Chi-Square?

Pearson’s chi-square test (χ²) is used with categorical (count) data to determine whether observed frequencies differ meaningfully from what would be expected if there were no relationship or pattern.

Key terms:

• Observed frequency (O): the actual count recorded in each category
• Expected frequency (E): the count you would expect in each category if the null hypothesis (no association, or a specified distribution) were true
• Contingency table: a table that cross-tabulates two categorical variables
• Independence: the assumption tested in a chi-square test of independence, that two variables are unrelated
• Degrees of freedom: for a contingency table, df = (rows − 1)(columns − 1)
• Chi-square statistic (χ²): a measure of how far observed frequencies deviate from expected frequencies
• p-value: the probability of observing a deviation this large if there were truly no association

Formula:

χ² = Σ [(O − E)² / E]

Each cell in the table contributes (O − E)² / E, and these contributions are summed across all cells to produce the total χ² statistic.

What Are the Three Main Types of Chi-Square Tests?

1. Chi-square goodness-of-fit test

Used to determine whether observed categorical frequencies differ from an expected distribution — one variable, compared against a hypothesized pattern.

Healthcare example: Testing whether the observed distribution of patient falls across three hospital shifts (day, evening, night) matches an expected even distribution.

2. Chi-square test of independence

Used to determine whether two categorical variables are associated within a single sample.

Nursing example: Testing whether smoking status (smoker/non-smoker) is associated with diagnosis category (respiratory illness/no respiratory illness) among a single group of patients.

3. Chi-square test of homogeneity

Used to determine whether different populations or groups have the same distribution across categorical outcomes. The math is identical to the test of independence, but the sampling design differs: homogeneity compares separately drawn samples (for example, patients from two different hospitals) across the same categorical outcome.

Independence vs. homogeneity, in plain terms: independence asks “within this one sample, are these two variables related?” Homogeneity asks “were these separate samples drawn from populations with the same distribution?” The calculation is the same; the research design and the wording of the conclusion differ.

Application of Chi Square Test in Health Care

The application of chi square test in nursing research is almost always about categorical relationships, not numerical averages. Some realistic examples:

A statistically significant chi-square result tells you the two categorical variables are associated. It does not tell you that one causes the other. A significant relationship between shift type and medication errors, for example, could reflect staffing ratios, patient acuity, or fatigue rather than the shift itself.

Solved Problems on Chi Square Test

Example 1: Chi-Square Test of Independence

Scenario: A nurse researcher wants to know whether patient education participation is associated with follow-up appointment attendance.

Observed frequencies:

Step 1: Expected frequencies E = (row total × column total) / grand total

• E(Participated, Attended) = (60 × 70) / 120 = 35.0
• E(Participated, Did not attend) = (60 × 50) / 120 = 25.0
• E(Did not participate, Attended) = (60 × 70) / 120 = 35.0
• E(Did not participate, Did not attend) = (60 × 50) / 120 = 25.0

Step 2: Chi-square contribution per cell

Step 3: Total χ² χ² = 2.857 + 4.000 + 2.857 + 4.000 = 13.71

Step 4: Degrees of freedom df = (rows − 1)(columns − 1) = (2 − 1)(2 − 1) = 1

Step 5: Interpretation At df = 1, the critical χ² value at α = 0.05 is 3.84. Since 13.71 exceeds 3.84, the result is statistically significant (p < 0.001).

Step 6: Plain-English conclusion There is a statistically significant association between patient education participation and follow-up appointment attendance, χ²(1, N = 120) = 13.71, p < 0.001. Patients who participated in education were more likely to attend follow-up than those who did not. This is an association, not proof that education participation directly causes attendance; other factors, such as health literacy or motivation, could contribute to both.

Note: this example uses one internally consistent χ² value throughout (χ² = 13.71, df = 1). If your own HLT-362V worksheet uses a different 2 × 2 dataset, follow this same process and make sure the χ² value you report in your write-up matches the number your calculation actually produces — a common point deduction is stating one χ² value in the calculation and a different one in the narrative conclusion.

Example 2: Chi-Square Goodness-of-Fit

Scenario: A hospital expects patient falls to be evenly distributed across three shifts (day, evening, night) if shift has no effect on fall risk. Over one month, the following falls were recorded.

Observed and expected frequencies (expected = equal distribution of 90 total falls across 3 shifts):

Step 1: Chi-square contribution per category

Step 2: Total χ² χ² = 4.80 + 0.30 + 7.50 = 12.60

Step 3: Degrees of freedom df = k − 1 = 3 − 1 = 2

Step 4: Interpretation At df = 2, the critical χ² value at α = 0.05 is 5.99. Since 12.60 exceeds 5.99, the result is statistically significant (p < 0.01).

Step 5: Conclusion Observed fall frequency across shifts differed significantly from an even distribution, χ²(2, N = 90) = 12.60, p < 0.01. Falls were not evenly distributed across shifts; the night shift accounted for a disproportionately high share of falls.

Why this is goodness-of-fit, not independence: this example tests one categorical variable (shift) against a hypothesized expected pattern (even distribution). There is no second categorical variable being cross-tabulated, which is what distinguishes it from the test of independence in Example 1.

Stuck interpreting your own chi-square output or unsure whether your assignment calls for independence, homogeneity, or goodness-of-fit? Message Gradevia on WhatsApp and describe what your dataset looks like.

Chi Square Test Simulation

A chi-square test simulation is a useful way to build intuition before you calculate anything by hand. The core idea is simple: expected frequencies stay fixed (they are determined by your null hypothesis or by the row/column totals), while observed frequencies are what you actually recorded.

What happens as observed frequencies change:

• When observed counts are close to expected counts, each (O − E)² term stays small, so the total χ² stays small and the p-value stays large (not significant).
• When observed counts diverge substantially from expected counts, the (O − E)² terms grow, χ² increases, and the p-value shrinks (more likely to be significant).
• The p-value is what ultimately tells you whether the observed deviation is large enough, relative to sample size and degrees of freedom, to be considered statistically significant rather than random variation.

A simple way to visualize this on paper or in a spreadsheet:

Add a running total column at the bottom for total χ², and compare it against the relevant critical value for your degrees of freedom, or convert it to a p-value using a chi-square distribution calculator.

For a verified, reputable interactive resource to build intuition (not a substitute for doing your own HLT-362V calculations), the NIST/SEMATECH Engineering Statistics Handbook’s chi-square goodness-of-fit page walks through the same logic in more technical depth (NIST/SEMATECH, n.d.).

When to Use t-Test vs ANOVA vs Chi-Square?

Use a t-test when: you are comparing the means of two groups.

Use ANOVA when: you are comparing the means of three or more groups.

Use chi-square when: you are examining categorical frequencies or whether categorical variables are associated.

Five quick scenarios:

1. “Do average pain scores differ between patients receiving Treatment A and Treatment B?” Answer: independent-samples t-test (two groups, continuous outcome).
2. “Do average pain scores differ among three treatment groups?” Answer: one-way ANOVA (three groups, continuous outcome).
3. “Is medication adherence category associated with readmission status?” Answer: chi-square test of independence (two categorical variables).
4. “Does a single group of patients’ anxiety score change across three measurement points before and after an intervention?” Answer: repeated-measures ANOVA (same subjects, multiple time points, continuous outcome).
5. “Does the distribution of patient satisfaction ratings (poor/fair/good/excellent) match a hospital’s expected historical distribution?” Answer: chi-square goodness-of-fit test (one categorical variable against an expected distribution).

HLT-362V ANOVA and Chi-Square Exercises Answers

Search intent around “HLT-362V ANOVA and Chi-Square Exercises answers” usually comes from two different places: students who genuinely want to check their reasoning, and students hoping to find a number to copy. Only the first approach actually helps you pass HLT-362V, because instructors are checking your work, not just your final statistic, and assignment datasets rotate.

The most useful way to use worked examples like the ones above is to compare your process step by step, not just your final number.

How to Check Your HLT-362V ANOVA and Chi-Square Answers

Run through this checklist before you submit:

• Did I identify the correct statistical test (t-test, ANOVA, or chi-square) based on my variable types?
• Did I identify the independent and dependent variables correctly?
• Did I calculate the expected frequencies correctly (for chi-square)?
• Did I use the correct degrees of freedom for my test?
• Did I calculate the test statistic (F or χ²) correctly, and does it match everywhere I reference it in my write-up?
• Did I interpret the p-value correctly relative to α = 0.05?
• Did I state a clear statistical conclusion (reject or fail to reject the null hypothesis)?
• Did I explain the result in plain healthcare language, not just symbols?
• Did I avoid claiming causation from an association (chi-square) or from a significant ANOVA?
• Did I distinguish statistical significance from clinical significance in my discussion?

If you get through this checklist and you’re still unsure whether your interpretation holds up, that’s a reasonable point to get a second set of eyes on your work before you submit.

Common HLT-362V Mistakes

1. Using ANOVA when only two groups are being compared (a t-test is correct here).
2. Using a t-test when there are three or more groups (ANOVA is correct here).
3. Using chi-square for continuous measurements instead of categorical/count data.
4. Confusing observed and expected frequencies when setting up the chi-square formula.
5. Calculating expected frequencies incorrectly (remember: row total × column total ÷ grand total).
6. Using incorrect degrees of freedom, especially in contingency tables larger than 2 × 2.
7. Misinterpreting a large p-value as “proof” that groups are equal.
8. Saying a significant ANOVA tells you which specific groups differ (it doesn’t — that requires a post hoc test).
9. Forgetting post hoc testing after a significant ANOVA with three or more groups.
10. Interpreting a significant chi-square association as evidence of causation.
11. Confusing statistical significance with clinical importance.
12. Reporting only the test statistic (F or χ²) without a plain-language conclusion in healthcare context.

Quick HLT-362V Statistics Cheat Sheet

HLT-362V ANOVA and Chi-Square FAQ

What is ANOVA and chi-square? ANOVA (analysis of variance) is a statistical test that compares the means of three or more groups on a continuous outcome. Chi-square is a statistical test used with categorical data to test whether observed frequencies differ from expected frequencies, either to test association between two categorical variables or to compare an observed distribution to a hypothesized one. The fundamental difference is the type of outcome data: ANOVA needs a continuous (numeric) outcome, while chi-square needs categorical (count) data.

What are the three main types of chi-square tests?

1. Goodness-of-fit — tests whether one categorical variable’s observed distribution matches an expected distribution (example: comparing observed patient fall counts by shift to an expected even split).
2. Test of independence — tests whether two categorical variables are associated within a single sample (example: smoking status and respiratory diagnosis).
3. Test of homogeneity — tests whether separate populations or groups share the same distribution across a categorical outcome (example: fall rates across different hospital units).

When to use t-test vs ANOVA vs chi-square? Use a t-test to compare means between two groups, ANOVA to compare means across three or more groups, and chi-square when your data is categorical and you’re testing association or distribution rather than comparing averages.

What are the three types of ANOVA? One-way ANOVA (one categorical independent variable, 3+ groups), two-way ANOVA (two categorical independent variables, possibly with an interaction effect), and repeated-measures ANOVA (same subjects measured across multiple time points or conditions). Note that terminology and classification can vary somewhat depending on the textbook or statistical software your course uses.

What does a significant ANOVA result mean? It means at least one group mean differs significantly from the others. It does not identify which specific group or groups differ — that requires a post hoc test such as Tukey HSD.

What does the F-value mean? The F-value is the ratio of between-group variance to within-group variance. A larger F suggests the differences between group means are large relative to the variability within each group.

What does p < 0.05 mean? It means that, assuming the null hypothesis is true, there is less than a 5% probability of observing a result this extreme purely by chance. By convention, p < 0.05 is treated as statistically significant, though current guidance encourages reporting exact p-values and effect sizes rather than relying only on the 0.05 cutoff (Ciapponi et al., 2021).

What is the difference between observed and expected frequency? Observed frequency is the actual count recorded in your data. Expected frequency is the count you would predict for that category if the null hypothesis were true (no association, or a specified distribution).

How do you calculate expected frequency in chi-square? For a contingency table: E = (row total × column total) / grand total. For a goodness-of-fit test against a specified distribution, E is typically the total sample size multiplied by the hypothesized proportion for that category.

What are the degrees of freedom for chi-square? For a test of independence or homogeneity: df = (number of rows − 1)(number of columns − 1). For a goodness-of-fit test: df = number of categories − 1.

Does chi-square prove causation? No. A significant chi-square result shows an association between two categorical variables. It does not establish that one variable causes the other; other explanations, including confounding variables, are always possible.

When is a post hoc test needed after ANOVA? Whenever your ANOVA result is statistically significant and you have three or more groups. The ANOVA alone tells you a difference exists somewhere; a post hoc test such as Tukey HSD tells you where.

Is HLT-362V statistics difficult? Most students find the biggest challenge isn’t the math itself but correctly identifying which test applies to a given scenario and interpreting the output in healthcare language. Working through several practice scenarios, like the decision table and worked examples above, tends to help more than re-reading formula definitions.

How can I study ANOVA and chi-square when I have limited time? Focus first on correctly identifying which test applies to a scenario (the decision table above), since that’s where most points are lost. Then practice one worked example of each test by hand so you understand where each number in the output comes from, rather than trying to memorize every formula in isolation.

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References

Ciapponi, A., Belizán, J. M., Piaggio, G., & Yaya, S. (2021). There is life beyond the statistical significance. Reproductive Health, 18, 78. https://doi.org/10.1186/s12978-021-01131-w

National Center for Health Statistics. (n.d.). Statistical significance. Centers for Disease Control and Prevention. https://www.cdc.gov/nchs/hus/sources-definitions/statistical-significance.htm

NIST/SEMATECH. (n.d.). Chi-square goodness-of-fit test. e-Handbook of Statistical Methods, National Institute of Standards and Technology. https://itl.nist.gov/div898/handbook/eda/section3/eda35f.htm

Tomaszewska, K., Majchrowicz, B., Snarska, K., & Telega, D. (2022). Stress and occupational burnout of nurses working with COVID-19 patients. International Journal of Environmental Research and Public Health, 19(19), 12688. https://doi.org/10.3390/ijerph191912688

Grand Canyon University. (n.d.). HLT-362V: Applied Statistics for Health Care Professionals — course description. Course catalog, Grand Canyon University. (Course-level description reflected in this article; consult your current course syllabus for assignment-specific requirements, as these are set by individual instructors and may change by term.)