Business Statistics Quiz: One Way Anova
20 questions · exam conditions
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One Way AnovaQuestion 1 of 20

An analyst is comparing the daily website traffic for three different marketing campaigns. The following data represents the number of unique visitors for three days for each campaign:

  • Campaign A: 4, 5, 6
  • Campaign B: 7, 8, 9
  • Campaign C: 1, 2, 3

What is the Sum of Squares Within groups (SSW) for this dataset?

6.0
8.0
54.0
60.0
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Business Statistics Quiz

Business Statistics Quiz: One Way Anova

Practice One Way Anova in Business Statistics with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on One Way Anova, giving you a quick way to practice the rules, question types, and explanations that matter most for Business Statistics.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

All questions

Question 1

An analyst is comparing the daily website traffic for three different marketing campaigns. The following data represents the number of unique visitors for three days for each campaign:

  • Campaign A: 4, 5, 6
  • Campaign B: 7, 8, 9
  • Campaign C: 1, 2, 3

What is the Sum of Squares Within groups (SSW) for this dataset?

  1. 6.0 (correct answer)
  2. 8.0
  3. 54.0
  4. 60.0
Explanation: The Sum of Squares Within groups (SSW), or Sum of Squares Error (SSE), is the sum of the squared deviations of each observation from its own group's mean.
  1. Calculate group means: Mean_A = (4+5+6)/3 = 5; Mean_B = (7+8+9)/3 = 8; Mean_C = (1+2+3)/3 = 2.
  2. Calculate sum of squares for each group: SS_A = (4-5)² + (5-5)² + (6-5)² = 1 + 0 + 1 = 2. SS_B = (7-8)² + (8-8)² + (9-8)² = 1 + 0 + 1 = 2. SS_C = (1-2)² + (2-2)² + (3-2)² = 1 + 0 + 1 = 2.
  3. Sum the group sums of squares: SSW = SS_A + SS_B + SS_C = 2 + 2 + 2 = 6.0. Distractor C (54.0) is the Sum of Squares Between groups (SSB). Distractor D (60.0) is the Total Sum of Squares (SST).

Question 2

A research study compares a process outcome across four groups (k=4) with a total of 64 participants (N=64). The analysis finds a Total Sum of Squares (SST) of 8,000 and a Mean Square Error (MSE), also known as Mean Square Within, of 100.

Using this information, what is the calculated F-statistic?

  1. 5.00
  2. 6.67 (correct answer)
  3. 20.00
  4. 26.67
Explanation: This is a multi-step calculation.
  1. Find the degrees of freedom within: df_within = N - k = 64 - 4 = 60.
  2. Calculate the Sum of Squares Within (SSW): SSW = MSE * df_within = 100 * 60 = 6,000.
  3. Calculate the Sum of Squares Between (SSB): SSB = SST - SSW = 8,000 - 6,000 = 2,000.
  4. Find the degrees of freedom between: df_between = k - 1 = 4 - 1 = 3.
  5. Calculate the Mean Square Between (MSB): MSB = SSB / df_between = 2,000 / 3 ≈ 666.67.
  6. Calculate the F-statistic: F = MSB / MSE = 666.67 / 100 ≈ 6.67.

Question 3

A retail chain analyzes customer satisfaction scores across five store locations. The ANOVA yields MSBetween = 45.6, MSWithin = 12.8, with sample sizes of 8, 12, 10, 15, and 10 customers per location. What is the F-statistic and the appropriate degrees of freedom for this analysis?

  1. F = 3.56; df₁ = 4, df₂ = 50, providing moderate evidence of differences between store locations (correct answer)
  2. F = 3.56; df₁ = 5, df₂ = 55, indicating significant variation in customer satisfaction across all locations
  3. F = 0.28; df₁ = 4, df₂ = 50, suggesting no meaningful differences in satisfaction between store locations
  4. F = 0.28; df₁ = 5, df₂ = 55, indicating homogeneous satisfaction levels across the retail chain
Explanation: F = MSBetween/MSWithin = 45.6/12.8 = 3.56. With 5 groups, df₁ = k-1 = 4. Total n = 8+12+10+15+10 = 55, so df₂ = n-k = 55-5 = 50. Choice A correctly calculates both F and degrees of freedom. Choice B has incorrect degrees of freedom (df₁ should be 4, not 5; df₂ should be 50, not 55). Choices C and D incorrectly calculate F = 0.28 (which would be MSWithin/MSBetween, the reciprocal). The interpretation about evidence strength requires comparing F to critical values.

Question 4

A marketing analyst is comparing the average daily sales generated by three different in-store display strategies across a sample of 30 stores. A one-way ANOVA is conducted to analyze the data.

The analysis yields a Sum of Squares Between groups (SSB) of 1,200 and a Total Sum of Squares (SST) of 4,200. Based on this information, what is the calculated F-statistic for the test?

  1. 0.40
  2. 2.50
  3. 5.40 (correct answer)
  4. 6.75
Explanation: To calculate the F-statistic, several steps are needed. First, find the Sum of Squares Within groups (SSW): SSW = SST - SSB = 4,200 - 1,200 = 3,000. Second, determine the degrees of freedom: df_between = k - 1 = 3 - 1 = 2; df_within = N - k = 30 - 3 = 27. Third, calculate the Mean Squares: MSB = SSB / df_between = 1,200 / 2 = 600; MSW = SSW / df_within = 3,000 / 27 ≈ 111.11. Finally, calculate the F-statistic: F = MSB / MSW = 600 / 111.11 ≈ 5.40.

Question 5

A restaurant chain wants to determine if there is a significant difference in the average weekly sales of four new menu items. A manager suggests that instead of running an ANOVA, they should simply perform six separate two-sample t-tests to compare every possible pair of menu items (Item A vs. B, A vs. C, A vs. D, B vs. C, B vs. D, C vs. D) at a significance level of α=0.05\alpha = 0.05 for each test.

What is the primary statistical reason that this approach of conducting multiple t-tests is considered inappropriate?

  1. The power of each individual t-test would be lower than the power of a single ANOVA test, increasing the chance of a Type II error.
  2. It inflates the family-wise Type I error rate, substantially increasing the probability of incorrectly finding a significant difference. (correct answer)
  3. Two-sample t-tests are only valid when the sample sizes for all groups are exactly equal, which may not be the case.
  4. The process of running six separate tests is less efficient and more computationally intensive than a single ANOVA.
Explanation: The main issue with conducting multiple t-tests is the inflation of the family-wise (or experiment-wise) Type I error rate. If each test is conducted at α=0.05\alpha = 0.05, the probability of making at least one Type I error across all six tests is significantly higher than 0.05. The formula for the family-wise error rate is 1(1α)c1 - (1 - \alpha)^c, where c is the number of comparisons. In this case, it would be 1(0.95)60.2651 - (0.95)^6 \approx 0.265, or a 26.5% chance of a false positive. ANOVA is designed to test for differences across all groups simultaneously while maintaining the specified Type I error rate.

Question 6

An ANOVA was conducted to assess the effect of three different training programs on employee performance scores. The analysis resulted in a Sum of Squares Between groups (SSB) of 500 and a Sum of Squares Within groups (SSW) of 2000.

Based on these results, what is the value of eta-squared (η2\eta^2) and what is its correct interpretation?

  1. 0.20; The training programs explain 20% of the variance in employee performance scores. (correct answer)
  2. 0.25; The training programs explain 25% of the variance in employee performance scores.
  3. 4.00; The variance between training programs is 4 times the variance within the programs.
  4. 0.80; 80% of the variance in performance scores is due to factors other than the training programs.
Explanation: Eta-squared (η2\eta^2) is a measure of effect size in ANOVA, representing the proportion of variance in the dependent variable that is explained by the independent variable (the groups). It is calculated as η2=SSBSST\eta^2 = \frac{SSB}{SST}. First, we must find the Total Sum of Squares (SST), which is SST = SSB + SSW = 500 + 2000 = 2500. Then, η2=5002500=0.20\eta^2 = \frac{500}{2500} = 0.20. This means that 20% of the total variability in employee performance scores can be attributed to the different training programs.

Question 7

A financial analyst is using one-way ANOVA to compare the mean return on investment (ROI) for stocks in four different industry sectors. After collecting sample data, the analyst discovers that the sample mean ROI is exactly the same for all four sectors.

Assuming the analyst proceeds with the ANOVA calculation, what will the value of the F-statistic be?

  1. Exactly 0 (correct answer)
  2. Exactly 1
  3. A value that depends on the total sample size
  4. A negative value, indicating no difference
Explanation: The F-statistic is the ratio of Mean Square Between groups (MSB) to Mean Square Within groups (MSW). MSB is derived from the Sum of Squares Between groups (SSB), which measures the variation of each group's sample mean from the overall grand mean. If all sample means are identical, they will also be identical to the grand mean. Therefore, the variation between the group means (SSB) will be zero. This makes MSB zero, and consequently, the F-statistic (MSB/MSW) will also be exactly 0. The F-statistic cannot be negative.

Question 8

A retail company is testing three new website layouts to see if they result in different average purchase amounts. The company plans to use a one-way ANOVA to analyze the data.

Which of the following represents the correct null (H0H_0) and alternative (HaH_a) hypotheses for this analysis, where μ1,μ2,μ3\mu_1, \mu_2, \mu_3 are the population mean purchase amounts for the three layouts?

  1. H0:μ1=μ2=μ3H_0: \mu_1 = \mu_2 = \mu_3 versus Ha:μ1μ2μ3H_a: \mu_1 \neq \mu_2 \neq \mu_3
  2. H0:xˉ1=xˉ2=xˉ3H_0: \bar{x}_1 = \bar{x}_2 = \bar{x}_3 versus Ha:xˉ1xˉ2xˉ3H_a: \bar{x}_1 \neq \bar{x}_2 \neq \bar{x}_3
  3. H0:μ1=μ2=μ3H_0: \mu_1 = \mu_2 = \mu_3 versus Ha:H_a: At least one population mean is different. (correct answer)
  4. H0:H_0: At least one population mean is different versus Ha:μ1=μ2=μ3H_a: \mu_1 = \mu_2 = \mu_3
Explanation: In one-way ANOVA, the null hypothesis (H₀) always states that there is no effect, meaning all population means are equal. The alternative hypothesis (Hₐ) states that there is an effect, which means that not all population means are equal, or in other words, at least one population mean is different from the others. Choice A is incorrect because the alternative hypothesis is that at least one mean is different, not necessarily that all are different. Choice B is incorrect because hypotheses are always stated in terms of population parameters (μ\mu), not sample statistics (xˉ\bar{x}). Choice D incorrectly reverses the null and alternative hypotheses.

Question 9

In a one-way ANOVA, how would a substantial decrease in the variability within each group affect the F-statistic, assuming the differences between the group means remain the same?

  1. The F-statistic would decrease, making a significant result less likely.
  2. The F-statistic would increase, making a significant result more likely. (correct answer)
  3. The F-statistic would remain unchanged because it is a ratio of variances.
  4. The effect on the F-statistic cannot be determined without knowing the sample sizes.
Explanation: The F-statistic is calculated as F = MSB / MSW. The Mean Square Between (MSB) reflects the variability between the group means, while the Mean Square Within (MSW) reflects the average variability within the groups. If the differences between the group means remain the same, MSB will be unchanged. A decrease in the variability within each group will cause MSW to decrease. Since MSW is in the denominator of the F-ratio, a smaller denominator will result in a larger F-statistic. A larger F-statistic is more likely to be statistically significant.

Question 10

A supply chain manager wants to know if the mean delivery times for three different shipping carriers are equal. A one-way ANOVA is planned. The manager collects data and calculates the Sum of Squares Between groups (SSB) as 600 and the Sum of Squares Within groups (SSW) as 1,800. There were 10 deliveries sampled for each carrier.

If the critical F-value at a 0.05 significance level for the appropriate degrees of freedom is 3.35, what is the conclusion of the hypothesis test?

  1. Reject H₀ because the calculated F-statistic of 6.00 is greater than the critical value.
  2. Fail to reject H₀ because the calculated F-statistic of 4.50 is greater than the critical value.
  3. Fail to reject H₀ because the calculated F-statistic of 0.33 is less than the critical value.
  4. Reject H₀ because the calculated F-statistic of 4.50 is greater than the critical value. (correct answer)
Explanation: When you encounter a one-way ANOVA problem, you're testing whether multiple group means are significantly different. The key is calculating the F-statistic and comparing it to the critical value to make your decision. To find the F-statistic, you need the Mean Square Between (MSB) and Mean Square Within (MSW). First, determine your degrees of freedom: dfbetween=k1=31=2df_{between} = k - 1 = 3 - 1 = 2 (where k = number of groups), and dfwithin=Nk=303=27df_{within} = N - k = 30 - 3 = 27 (where N = total sample size). Next, calculate the mean squares: MSB=SSBdfbetween=6002=300MSB = \frac{SSB}{df_{between}} = \frac{600}{2} = 300 and MSW=SSWdfwithin=180027=66.67MSW = \frac{SSW}{df_{within}} = \frac{1800}{27} = 66.67 Therefore: F=MSBMSW=30066.67=4.50F = \frac{MSB}{MSW} = \frac{300}{66.67} = 4.50 Since 4.50 > 3.35 (the critical value), you reject H₀, concluding the carrier means differ significantly. Option A incorrectly calculates F as 6.00, likely from computational error. Option B correctly calculates F = 4.50 but reaches the wrong conclusion—when your test statistic exceeds the critical value, you reject H₀, not fail to reject it. Option C calculates F as 0.33, probably by dividing SSW by SSB instead of the mean squares in proper order. Remember: In ANOVA, always calculate degrees of freedom first, then mean squares, then the F-ratio. When F exceeds the critical value, you have evidence that at least one group mean differs from the others.

Question 11

When interpreting the output of a one-way ANOVA, the p-value is determined by comparing the calculated F-statistic to its theoretical sampling distribution. Which of the following correctly describes how the p-value is found?

  1. The area under the F-distribution curve to the left of the calculated F-statistic.
  2. Twice the area under the F-distribution curve to the right of the calculated F-statistic, to account for a two-tailed test.
  3. The height of the F-distribution curve at the point corresponding to the calculated F-statistic.
  4. The area under the F-distribution curve to the right of the calculated F-statistic. (correct answer)
Explanation: When you encounter ANOVA p-value questions, remember that ANOVA is always a right-tailed test because we're testing whether group means differ significantly from each other. The F-statistic measures the ratio of between-group variance to within-group variance, and larger F-values provide stronger evidence against the null hypothesis of equal group means. The p-value represents the probability of observing an F-statistic as extreme or more extreme than what we calculated, assuming the null hypothesis is true. Since the F-distribution only takes positive values and we're looking for evidence of differences (not specific directions), we only care about the right tail. The p-value is the area under the F-distribution curve to the right of our calculated F-statistic. Option A is wrong because the left-tail area would give us the probability of getting a smaller F-statistic, which isn't what we want for hypothesis testing. Option B incorrectly applies two-tailed logic—while many statistical tests are two-tailed, ANOVA's F-test is inherently one-tailed because we're testing for any differences among groups, not directional hypotheses. Option C confuses the p-value with a probability density; the height of the curve at a point isn't a probability—areas under curves represent probabilities. Study tip: Remember that ANOVA F-tests are always right-tailed tests. The larger the F-statistic, the smaller the p-value, because more area falls to the left of larger F-values. This makes intuitive sense: bigger F-statistics indicate larger differences between groups, providing stronger evidence against the null hypothesis.

Question 12

An e-commerce company tests four different ad creatives to see if they lead to different mean click-through rates. After running a one-way ANOVA, the result is an F-statistic of 1.25 with a p-value of 0.31.

Using a significance level of α=0.05\alpha = 0.05, what is the correct conclusion?

  1. The company can conclude with 95% confidence that the mean click-through rates for all four ad creatives are identical.
  2. There is insufficient evidence to conclude that there are any differences in the mean click-through rates among the four ad creatives. (correct answer)
  3. The test proves that there is no difference in the effectiveness of the ad creatives.
  4. A Type II error has occurred, and the company should re-run the test with a different statistical method.
Explanation: The p-value (0.31) is greater than the significance level (α=0.05\alpha = 0.05). Therefore, we fail to reject the null hypothesis. The correct interpretation is that there is not enough statistical evidence to claim that a difference exists among the mean click-through rates. Choice A is incorrect because failing to reject H₀ does not prove that H₀ is true; it simply means we lack evidence to say it's false. Choice C uses the word 'proves,' which is too strong; statistical tests provide evidence, not proof. Choice D is incorrect because we can never know for sure if a Type II error (failing to detect a real difference) has occurred.

Question 13

A one-way ANOVA is conducted to compare the means of 5 groups. The calculated F-statistic from the sample data is 4.15. The appropriate critical F-value from the F-distribution table for a significance level of α=0.05\alpha = 0.05 is 2.58.

Based on this information, what can be concluded about the p-value of the test?

  1. The p-value is greater than 0.05.
  2. The p-value cannot be estimated without knowing the exact sample size.
  3. The p-value is exactly 0.05.
  4. The p-value is less than 0.05. (correct answer)
Explanation: When you encounter ANOVA hypothesis testing questions, focus on the relationship between the calculated F-statistic, critical F-value, and p-value. This is fundamentally about understanding where your test statistic falls in the F-distribution. In hypothesis testing, the p-value represents the probability of obtaining your observed test statistic (or more extreme) assuming the null hypothesis is true. The critical value marks the boundary where p-value equals your significance level (α=0.05\alpha = 0.05). Since the calculated F-statistic (4.15) exceeds the critical F-value (2.58), your test statistic falls in the rejection region of the F-distribution. This means the correct answer is D) The p-value is less than 0.05. When your calculated statistic exceeds the critical value, you're further into the tail of the distribution, making your result less likely under the null hypothesis, thus producing a smaller p-value. Answer A is incorrect because when F-calculated > F-critical, the p-value must be smaller than α\alpha, not larger. Answer B is wrong because while sample size affects degrees of freedom (already incorporated into the given critical value), you can still determine the p-value's relationship to α\alpha from the comparison of calculated versus critical values. Answer C is incorrect because the p-value equals α\alpha only when the calculated statistic exactly equals the critical value, which isn't the case here. Study tip: Remember this key relationship: F-calculated > F-critical means p-value < α\alpha (reject null), while F-calculated < F-critical means p-value > α\alpha (fail to reject null).

Question 14

A researcher compares mean productivity scores across three departments using one-way ANOVA. The analysis yields SSB = 240, SSW = 360, with total sample size n = 48. What is the calculated F-statistic, and what does it suggest about the strength of evidence for group differences?

  1. F = 15.00; this provides very strong evidence for differences since it greatly exceeds typical critical values
  2. F = 30.00; this suggests possible differences, but the unequal sum of squares indicates assumption violations
  3. F = 30.00; this provides extremely strong evidence since it represents a very large effect size
  4. F = 15.00; this provides moderate evidence, but interpretation depends on the chosen significance level and critical value (correct answer)
Explanation: When you encounter one-way ANOVA problems, you need to calculate the F-statistic using the formula F=MSBMSWF = \frac{MSB}{MSW}, where MSB and MSW are the mean squares between and within groups, respectively. First, calculate the degrees of freedom: dfB=k1=31=2df_B = k - 1 = 3 - 1 = 2 (where k = number of groups) and dfW=nk=483=45df_W = n - k = 48 - 3 = 45. Then find the mean squares: MSB=SSBdfB=2402=120MSB = \frac{SSB}{df_B} = \frac{240}{2} = 120 and MSW=SSWdfW=36045=8MSW = \frac{SSW}{df_W} = \frac{360}{45} = 8. Therefore, F=1208=15.00F = \frac{120}{8} = 15.00. Answer D correctly identifies F = 15.00 and appropriately characterizes this as moderate evidence that requires comparison to critical values for proper interpretation. Answer A calculates F correctly but overstates the conclusion. While F = 15.00 is substantial, calling it "very strong evidence" without referencing critical values is premature statistical interpretation. Answer B makes a calculation error, arriving at F = 30.00, and incorrectly suggests that unequal sum of squares indicates assumption violations. Unequal SSB and SSW values are perfectly normal in ANOVA. Answer C also miscalculates F as 30.00 and conflates the F-statistic with effect size. The F-statistic tests significance, while effect size requires different measures like eta-squared. Remember: Always calculate F-statistics step-by-step using degrees of freedom, and avoid making definitive conclusions about significance without comparing your calculated F to the appropriate critical value at your chosen alpha level.

Question 15

In a one-way ANOVA comparing four groups, the calculated F-statistic is 7.5. The Mean Square Within groups (MSW), also known as Mean Square Error (MSE), is 20.

What is the value of the Sum of Squares Between groups (SSB)?

  1. 150
  2. 450 (correct answer)
  3. 600
  4. Cannot be determined without the total sample size.
Explanation: This problem requires working backward. The formula for the F-statistic is F = MSB / MSW. We are given F = 7.5 and MSW = 20. We can solve for the Mean Square Between groups (MSB): MSB = F * MSW = 7.5 * 20 = 150. The formula for MSB is MSB = SSB / df_between. The degrees of freedom between groups (df_between) is k - 1, where k is the number of groups. Here, k = 4, so df_between = 4 - 1 = 3. Now we can solve for SSB: SSB = MSB * df_between = 150 * 3 = 450.

Question 16

An operations manager plans to use a one-way ANOVA to compare the mean productivity (units per hour) of employees across five different production departments.

Which of the following scenarios would most seriously challenge the validity of the conclusions drawn from the F-test?

  1. The variance of productivity in one department is double the variance of another, but sample sizes are large and equal (n=100 per department).
  2. The sample means of the five departments are found to be very close to one another, resulting in a low F-statistic.
  3. Productivity data for two of the departments are heavily skewed, and the sample sizes are small (n=12 per department). (correct answer)
  4. The overall distribution of productivity scores from all 60 employees combined appears to be slightly bimodal.
Explanation: The F-test in ANOVA relies on three main assumptions: independence, normality (of the populations/residuals), and homoscedasticity (equal variances). The F-test is most sensitive to violations of the normality assumption when sample sizes are small. Heavily skewed data violates this assumption. While ANOVA is somewhat robust to violations, the combination of strong non-normality and small sample sizes is the most problematic scenario. Distractor A is less of a concern because ANOVA is robust to unequal variances when sample sizes are large and equal. Distractor B describes a potential outcome, not a violation of assumptions. Distractor D is irrelevant, as the normality assumption applies to the populations within each group, not the combined data.

Question 17

A quality control engineer uses one-way ANOVA to compare the mean breaking strength of a polymer produced by five different suppliers. The analysis yields an F-statistic of 8.12 with a p-value of 0.0001.

Given this statistically significant result, what is the most appropriate next step for the engineer to take in the data analysis process?

  1. Conclude that the supplier with the highest sample mean breaking strength is definitively the best.
  2. Re-run the analysis with a larger sample size to confirm the result's validity.
  3. Perform post-hoc tests, such as Tukey's HSD, to identify which specific pairs of suppliers have different mean breaking strengths. (correct answer)
  4. Accept the alternative hypothesis that all five suppliers have different mean breaking strengths from one another.
Explanation: A significant ANOVA result (a low p-value) indicates that there is a statistically significant difference among the group means, but it does not specify which means are different. The null hypothesis is that all means are equal, and rejecting it means only that at least one mean is different. The next logical step is to use post-hoc (multiple comparison) tests, like Tukey's HSD or Bonferroni correction, to conduct pairwise comparisons and determine which specific groups differ from each other. Concluding that all means are different (D) or that the highest sample mean is definitively best (A) are invalid inferences from the ANOVA result alone.

Question 18

An analyst performs a one-way ANOVA comparing the performance of several groups and calculates an F-statistic of 0.85. Without knowing the degrees of freedom or the p-value, what is the most likely conclusion?

  1. The test results are invalid because the F-statistic cannot be less than 1.
  2. The null hypothesis will be rejected, regardless of the significance level.
  3. The variance within groups is smaller than the variance between groups.
  4. The null hypothesis will not be rejected at any conventional significance level. (correct answer)
Explanation: When you encounter F-statistics in ANOVA problems, remember that the F-statistic represents the ratio of variance between groups to variance within groups. This ratio tells you whether group differences are likely due to real effects or just random variation. An F-statistic of 0.85 means the variance between groups is actually smaller than the variance within groups. Since F-statistics follow a distribution where values cluster around 1 when there are no real group differences, and larger values indicate stronger evidence of differences, a value below 1 suggests the groups are more similar than you'd expect by chance alone. In hypothesis testing, you reject the null hypothesis only when your test statistic exceeds the critical value. For F-tests, critical values are always greater than 1 at any conventional significance level (0.05, 0.01, etc.). Since 0.85 is below 1, it cannot possibly exceed any critical value, making option D correct. Option A is wrong because F-statistics can definitely be less than 1 - there's nothing invalid about this result. Option B is incorrect because an F-statistic of 0.85 provides no evidence for rejecting the null hypothesis at any significance level. Option C reverses the relationship - an F-statistic below 1 means variance within groups is larger than variance between groups, not smaller. Remember this pattern: F-statistics below 1 always mean you fail to reject the null hypothesis. You don't need to see degrees of freedom or p-values to know that values below 1 cannot reach statistical significance in ANOVA.

Question 19

A one-way ANOVA was performed to compare the mean customer satisfaction scores (on a 100-point scale) for four different customer service centers. The analysis produced a p-value of 0.023.

Assuming the assumptions for ANOVA have been met, what is the most accurate interpretation of this p-value?

  1. There is a 2.3% probability that the null hypothesis of equal means is true.
  2. There is a 2.3% chance of observing sample mean differences as large or larger than the ones found, if the service centers truly have the same mean satisfaction score. (correct answer)
  3. There is a 97.7% probability that there is a real difference in the mean satisfaction scores among the four service centers.
  4. The results prove that all four service centers have different mean satisfaction scores from one another.
Explanation: The p-value is the probability of obtaining test results at least as extreme as the results actually observed, under the assumption that the null hypothesis is correct. In this context, it means there is a 2.3% probability of seeing the observed differences (or greater differences) between the sample means if, in reality, there were no differences between the population means. Distractors represent common misinterpretations of the p-value: A confuses it with the probability of H0 being true; C incorrectly states the confidence (1-p); D incorrectly concludes that all means are different, whereas ANOVA only indicates that at least one is different.

Question 20

An analyst performs one-way ANOVA on sales data from five regions with sample sizes of 12, 15, 18, 10, and 20. The calculated F-statistic is 2.89. If the critical value at α = 0.01 is 3.83, which statement best describes the appropriate interpretation?

  1. The test provides strong evidence against equal means since the F-statistic exceeds the degrees of freedom ratio
  2. The test fails to provide sufficient evidence of differences at the 0.01 level, but may show significance at α = 0.05 (correct answer)
  3. The test is inconclusive because the sample sizes are unequal, violating a key ANOVA assumption
  4. The test shows marginal significance since the F-statistic is close to the critical value within one standard error
Explanation: Since F = 2.89 < 3.83 (critical value), we fail to reject H₀ at α = 0.01. However, this doesn't mean there's no effect—it might be detectable at a higher α level like 0.05. Choice B correctly identifies this. Choice A incorrectly interprets the F-statistic comparison. Choice C is wrong because unequal sample sizes don't violate ANOVA assumptions (equal variances and normality are the key assumptions). Choice D incorrectly discusses 'marginal significance' and 'standard error' in this context.