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HLT-362V ANOVA and Chi-Square Exercises: Guide + Example
The HLT-362V ANOVA and chi-square exercises from the Grove & Cipher workbook ask you to do two things: interpret ANOVA tables and chi-square output from published nursing studies (Exercises 18 and 19), and calculate ANOVA F-ratios and chi-square statistics by hand from raw data (Exercises 33 and 35 in the 3rd edition; 35 and 37 in the 4th edition). ANOVA tests whether the means of three or more groups differ significantly, while chi-square tests whether two categorical variables are associated. This guide explains both tests in plain language, walks through representative problems with full solutions, and shows you exactly how to read the output so you can answer the workbook questions with confidence.
What Is ANOVA and When Is It Used?
ANOVA (analysis of variance) is a statistical test that compares the means of three or more groups to determine whether at least one group mean differs significantly from the others. It is used when the dependent variable is continuous (e.g., pain scores, blood pressure) and the independent variable is categorical with three or more levels (e.g., three treatment groups).
ANOVA produces an F-statistic, which is the ratio of the variance between groups to the variance within groups. A large F-value means the groups differ more than you would expect by chance. If the p-value is less than 0.05, you reject the null hypothesis and conclude that the group means are not all equal.
When ANOVA is significant, you need a post hoc test (such as Tukey HSD) to determine which specific pairs of groups differ, because ANOVA alone only tells you that a difference exists somewhere.
If you need a refresher on the descriptive statistics (means and standard deviations) that ANOVA builds on, see our Summary and Descriptive Statistics guide.
How Do You Read an ANOVA Table?
You read an ANOVA table by locating five values: the between-groups and within-groups sums of squares (SS), the degrees of freedom (df), the mean squares (MS), the F-statistic, and the p-value.
Here is what each row and column means:
- Between Groups — the variability explained by the group differences.
- Within Groups — the variability left over (random error).
- MS = SS / df — converting sums of squares into averages.
- F = MS(between) / MS(within) — the test statistic.
- p — the probability of getting this F-value if the groups were truly equal.
A worked example: F(2, 27) = 5.84, p = 0.006. Because p < 0.05, at least one group mean differs significantly. Post hoc analysis would then tell you which pair(s) of groups drive the difference.
How Do You Calculate ANOVA by Hand?
You calculate a one-way ANOVA by computing the between-groups and within-groups sums of squares, converting them to mean squares, and dividing to get the F-ratio. The formulas are straightforward once you see them in action.
The steps are:
- Compute the grand mean — the overall average of all values.
- Compute SS(between) = n x sum of (group mean – grand mean) squared.
- Compute df(between) = k – 1 (where k = number of groups).
- Compute MS(between) = SS(between) / df(between).
- Compute SS(within) from the individual deviations within each group.
- Compute df(within) = N – k (total observations minus number of groups).
- Compute MS(within) = SS(within) / df(within).
- Compute F = MS(between) / MS(within).
- Compare F to the critical value from the F-distribution table at your alpha level.
Our worked sample walks through this calculation with three groups of 10, producing F = 4.58 with a critical value of 3.35 — significant at alpha = 0.05.
What Is Chi-Square and When Is It Used?
The Pearson chi-square test determines whether there is a statistically significant association between two categorical variables. Unlike ANOVA, which works with continuous data, chi-square works with counts — the number of observations that fall into each combination of categories.
You use chi-square when you have a contingency table (e.g., shift type vs. complication yes/no) and want to know whether the distribution of one variable depends on the other. The null hypothesis is that the variables are independent.
Chi-square compares observed frequencies (what actually happened) to expected frequencies (what would happen if the variables were unrelated). A large chi-square value means the observed pattern deviates substantially from what independence would predict.
How Do You Calculate Chi-Square by Hand?
You calculate chi-square by computing the expected frequency for each cell, then summing the squared differences between observed and expected, each divided by the expected frequency.
The formula is: chi-square = sum of [(O – E) squared / E]
The steps are:
- Build the contingency table with row and column totals.
- Calculate expected frequencies: E = (row total x column total) / grand total.
- For each cell, compute (O – E) squared / E.
- Sum all cells to get the chi-square statistic.
- Compute df = (rows – 1) x (columns – 1).
- Compare to the critical value from the chi-square distribution table.
Our worked sample uses a 2×2 table (shift type vs. complications) with N = 160, producing chi-square = 8.53 with df = 1 and a critical value of 3.841 — significant at alpha = 0.05.
HLT-362V ANOVA and Chi-Square Exercises Example
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ANOVA and Chi-Square Exercises
Grove & Cipher Workbook — Exercises 18, 19, 33 & 35
[Student Name]
College of Nursing and Health Care Professions, Grand Canyon University
HLT-362V: Applied Statistics for Health Care
[Instructor Name]
[Due Date]
ANOVA and Chi-Square: Worked Exercises
Part 1 — Understanding ANOVA (Exercise 18)
Analysis of variance (ANOVA) tests whether the means of three or more groups differ significantly. The null hypothesis states that all group means are equal; a significant F-statistic leads to its rejection. When ANOVA is significant, post hoc tests (such as Tukey HSD) identify which specific pairs of groups differ.
Sample ANOVA Table (pain scores across three treatment groups, n = 30):
| Source | SS | df | MS | F | p |
| Between Groups | 142.56 | 2 | 71.28 | 5.84 | 0.006 |
| Within Groups | 329.44 | 27 | 12.20 | ||
| Total | 472.00 | 29 |
Q1. Is the F-value statistically significant?
Yes. F(2, 27) = 5.84, p = 0.006. Because p < 0.05, the null hypothesis is rejected. At least one group mean differs significantly from the others.
Q2. What does the between-groups SS represent?
The between-groups sum of squares (142.56) represents the variability in pain scores that is attributable to the differences among the three treatment groups.
Q3. What does the within-groups MS represent?
The within-groups mean square (12.20) represents the average variability within each group—the random, unexplained variation among individual scores after group differences are removed.
Q4. If post hoc analysis (Tukey HSD) shows Group 1 vs. Group 3 is significant (p = 0.004) but Group 1 vs. Group 2 is not (p = 0.22), what does this mean?
It means the pain scores differ significantly only between Groups 1 and 3. Groups 1 and 2, and Groups 2 and 3, do not differ enough to reach statistical significance. Post hoc tests are necessary because ANOVA tells you that a difference exists somewhere but does not tell you where.
Part 2 — Calculating ANOVA (Exercise 33 / 35)
To calculate a one-way ANOVA by hand, you compute the between-groups and within-groups sums of squares, convert them to mean squares, and divide to get the F-ratio.
Given data: three groups (A, B, C), each with n = 10, grand mean = 15.0
Group A mean = 12.5, Group B mean = 15.3, Group C mean = 17.2
SS_B (between) = n × Σ(group mean − grand mean)² = 10 × [(12.5−15)² + (15.3−15)² + (17.2−15)²]
= 10 × [6.25 + 0.09 + 4.84] = 10 × 11.18 = 111.80
df_B = k − 1 = 3 − 1 = 2
MS_B = SS_B / df_B = 111.80 / 2 = 55.90
Given: SS_W (within) = 329.44, df_W = N − k = 30 − 3 = 27
MS_W = 329.44 / 27 = 12.20
F = MS_B / MS_W = 55.90 / 12.20 = 4.58
Compare F = 4.58 to the critical value of F at df(2, 27) and α = 0.05, which is approximately 3.35. Because 4.58 > 3.35, the result is statistically significant; reject the null hypothesis.
Part 3 — Understanding Chi-Square (Exercise 19)
The Pearson chi-square (χ²) test determines whether there is a statistically significant association between two categorical variables. It compares observed frequencies in a contingency table to the frequencies expected if the variables were independent.
Sample Contingency Table (shift type vs. postoperative complications, N = 160):
| Complication: Yes | Complication: No | Total | |
| Night Shift | 28 (E = 20) | 52 (E = 60) | 80 |
| Day Shift | 12 (E = 20) | 68 (E = 60) | 80 |
| Total | 40 | 120 | 160 |
E = expected frequency = (row total × column total) / grand total
Q1. What are the expected frequencies?
For Night Shift / Yes: (80 × 40) / 160 = 20. For Night Shift / No: (80 × 120) / 160 = 60. The expected frequencies for Day Shift are the same by symmetry (20 and 60).
Q2. What does it mean that observed > expected in the Night Shift / Yes cell?
It means more complications occurred during the night shift than would be expected if shift type and complications were independent. This suggests an association between night-shift work and complication rates.
Q3. If χ² = 8.89, df = 1, p = 0.003, is the result significant?
Yes. Because p = 0.003 < 0.05, the null hypothesis of independence is rejected. There is a statistically significant association between shift type and postoperative complications.
Part 4 — Calculating Chi-Square (Exercise 35 / 37)
The chi-square statistic is calculated as: χ² = Σ [(O − E)² / E], where O is the observed frequency and E is the expected frequency, summed across all cells.
Using the data above:
Cell 1 (Night/Yes): (28 − 20)² / 20 = 64 / 20 = 3.20
Cell 2 (Night/No): (52 − 60)² / 60 = 64 / 60 = 1.07
Cell 3 (Day/Yes): (12 − 20)² / 20 = 64 / 20 = 3.20
Cell 4 (Day/No): (68 − 60)² / 60 = 64 / 60 = 1.07
χ² = 3.20 + 1.07 + 3.20 + 1.07 = 8.53
df = (rows − 1)(columns − 1) = (2 − 1)(2 − 1) = 1
The critical value of χ² at df = 1 and α = 0.05 is 3.841. Because 8.53 > 3.841, the result is significant: shift type is significantly associated with complications.
How Do These Exercises Connect to the Rest of HLT-362V?
ANOVA and chi-square are the inferential tests that appear most often in the research articles you analyzed in Article Analysis 1 and Article Analysis 2. Understanding them here deepens your ability to evaluate published findings.
The concepts also connect directly to the Correlation and Causation paper: ANOVA tells you whether group differences are real (not just chance), while chi-square tells you whether two categorical variables are associated — but neither test, on its own, proves causation. That distinction is exactly what the Correlation and Causation assignment tests.
For the full course map and links to every assignment guide, see the HLT-362V Applied Statistics for Health Care pillar page.
Common Mistakes to Avoid
Most lost points on these exercises come from a few recurring errors:
- Confusing ANOVA with a t-test — ANOVA is for three or more groups; a t-test is for exactly two.
- Forgetting post hoc tests — a significant F only tells you some groups differ, not which ones.
- Mixing up observed and expected frequencies in chi-square.
- Using the wrong degrees of freedom — df(between) = k – 1 for ANOVA; df = (r – 1)(c – 1) for chi-square.
- Reporting the wrong conclusion — always state whether you reject or fail to reject the null hypothesis, and what that means in context.
HLT-362V ANOVA and Chi-Square FAQ
What is the difference between ANOVA and a t-test?
A t-test compares the means of exactly two groups, while ANOVA compares the means of three or more groups. ANOVA uses the F-statistic instead of the t-statistic, and a significant result requires post hoc testing to identify which groups differ.
What is a post hoc test and when do you use it?
A post hoc test (such as Tukey HSD) is used after a significant ANOVA result to determine which specific pairs of groups have significantly different means. You only run post hoc tests when the overall F-test is significant.
What does the chi-square test tell you?
The chi-square test tells you whether there is a statistically significant association between two categorical variables. A significant result means the observed distribution of counts differs from what would be expected if the variables were independent.
How do you calculate expected frequencies for chi-square?
Calculate the expected frequency for each cell by multiplying the row total by the column total and dividing by the grand total: E = (row total x column total) / N.
Which Grove and Cipher exercises cover ANOVA and chi-square?
In the 3rd edition, Exercise 18 covers understanding ANOVA, Exercise 33 covers calculating ANOVA, Exercise 19 covers understanding chi-square, and Exercise 35 covers calculating chi-square. In the 4th edition, the calculation exercises shift to 35 (ANOVA) and 37 (chi-square).
About the Author
This guide was prepared by the Gradevia academic team, specialists in nursing and health-sciences coursework support for students at GCU, WGU, Walden, and Liberty University. Our writers hold graduate degrees in nursing, public health, and applied statistics. We focus on helping busy working nurses understand the method, not just the answer.
Article Update Log
- June 18, 2026 — Initial publication
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