One-Way ANOVA - Business Statistics

1. 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?
   Answer: Reject H₀ because the calculated F-statistic of 4.50 is greater than the critical value. 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: $df_{between} = k - 1 = 3 - 1 = 2$ (where k = number of groups), and $df_{within} = N - k = 30 - 3 = 27$ (where N = total sample size). Next, calculate the mean squares: $MSB = \frac{SSB}{df_{between}} = \frac{600}{2} = 300$ and $MSW = \frac{SSW}{df_{within}} = \frac{1800}{27} = 66.67$ Therefore: $F = \frac{MSB}{MSW} = \frac{300}{66.67} = 4.50$ 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.
2. What is the Sum of Squares Within groups (SSW) for this dataset?
   Answer: 6.0 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).
3. Which of the following represents the correct null ($H_0$) and alternative ($H_a$) hypotheses for this analysis, where $\mu_1, \mu_2, \mu_3$ are the population mean purchase amounts for the three layouts?
   Answer: $H_0: \mu_1 = \mu_2 = \mu_3$ versus $H_a:$ At least one population mean is different. 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 ($\mu$), not sample statistics ($\bar{x}$). Choice D incorrectly reverses the null and alternative hypotheses.
4. An operations manager conducts one-way ANOVA to compare mean production times across three shifts. The results show F = 2.45 with df₁ = 2 and df₂ = 27. Given that the critical value is 3.35 at α = 0.05, which statement best describes the implications for management decision-making?
   Answer: Since F < critical value, there's insufficient statistical evidence of differences, but consider practical significance and larger sample sizes When you encounter one-way ANOVA questions, focus on both statistical significance and practical implications for business decisions. ANOVA tests whether group means differ significantly, but failing to reject the null hypothesis doesn't automatically mean "do nothing." Here, we have F = 2.45 with critical value = 3.35 at α = 0.05. Since F < 3.35, we fail to reject the null hypothesis - there's insufficient statistical evidence that the three shifts have different mean production times. However, smart managers look beyond just statistical significance when making operational decisions. Answer C correctly recognizes that while we lack statistical evidence of differences, this doesn't mean the shifts are truly identical. The F-value of 2.45 suggests some variation exists, just not enough to reach significance. Management should consider whether the observed differences have practical importance and whether a larger sample might reveal meaningful patterns. Answer A is wrong because failing to reject H₀ doesn't prove equality - it just means insufficient evidence of difference. Answer B incorrectly suggests "marginal evidence" requires immediate action, but approaching the critical value without reaching it still means no statistical significance. Answer D makes a fundamental error by claiming F > 2.00 provides "sufficient evidence" - statistical significance depends on the critical value (3.35), not arbitrary thresholds like 2.00. Study tip: In business statistics, always distinguish between statistical significance and practical significance. A non-significant result doesn't automatically mean "maintain status quo" - consider effect sizes, sample sizes, and business context before making operational decisions.
5. 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?
   Answer: F = 15.00; this provides moderate evidence, but interpretation depends on the chosen significance level and critical value When you encounter one-way ANOVA problems, you need to calculate the F-statistic using the formula $F = \frac{MSB}{MSW}$, where MSB and MSW are the mean squares between and within groups, respectively. First, calculate the degrees of freedom: $df_B = k - 1 = 3 - 1 = 2$ (where k = number of groups) and $df_W = n - k = 48 - 3 = 45$. Then find the mean squares: $MSB = \frac{SSB}{df_B} = \frac{240}{2} = 120$ and $MSW = \frac{SSW}{df_W} = \frac{360}{45} = 8$. Therefore, $F = \frac{120}{8} = 15.00$. 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.
6. 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?
   Answer: 5.40 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.
7. Using this information, what is the calculated F-statistic?
   Answer: 6.67 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.
8. 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?
   Answer: F = 3.56; df₁ = 4, df₂ = 50, providing moderate evidence of differences between store locations 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.
9. 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?
   Answer: The test fails to provide sufficient evidence of differences at the 0.01 level, but may show significance at α = 0.05 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.
10. Assuming the assumptions for ANOVA have been met, what is the most accurate interpretation of this p-value?
    Answer: 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. 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.
11. Given this statistically significant result, what is the most appropriate next step for the engineer to take in the data analysis process?
    Answer: Perform post-hoc tests, such as Tukey's HSD, to identify which specific pairs of suppliers have different mean breaking strengths. 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.
12. 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?
    Answer: The null hypothesis will not be rejected at any conventional significance level. 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.
13. 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?
    Answer: The F-statistic would increase, making a significant result more likely. 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.
14. What is the value of the Sum of Squares Between groups (SSB)?
    Answer: 450 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.
15. Based on this information, what can be concluded about the p-value of the test?
    Answer: The p-value is less than 0.05. 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 ($\alpha = 0.05$). 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 $\alpha$, 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 $\alpha$ from the comparison of calculated versus critical values. Answer C is incorrect because the p-value equals $\alpha$ 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 < $\alpha$ (reject null), while F-calculated < F-critical means p-value > $\alpha$ (fail to reject null).
16. Assuming the analyst proceeds with the ANOVA calculation, what will the value of the F-statistic be?
    Answer: Exactly 0 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.
17. The F-test in a one-way ANOVA is generally considered robust to violations of the homoscedasticity (equal variances) assumption under which specific condition?
    Answer: When the sample sizes for all groups are equal or very close to equal. One of the key assumptions of ANOVA is homoscedasticity, or the equality of variances across all groups. However, the F-test is known to be robust to violations of this assumption, meaning it still provides reliable results, provided that the sample sizes of the groups are equal (a balanced design) or nearly equal. When sample sizes are unequal, the F-test becomes much more sensitive to differences in variances. While normality is also an assumption, it does not compensate for heteroscedasticity. Large total N is not sufficient if group sizes are highly unbalanced.
18. Using a significance level of $\alpha = 0.05$, what is the correct conclusion?
    Answer: There is insufficient evidence to conclude that there are any differences in the mean click-through rates among the four ad creatives. The p-value (0.31) is greater than the significance level ($\alpha = 0.05$). 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.
19. A food scientist compares the mean protein content of three different grain varieties using one-way ANOVA. With 24 total observations (8 per group), she calculates F = 4.12. If the critical F-value at α = 0.01 is 5.61 and at α = 0.05 is 3.47, what is the most precise conclusion?
    Answer: The result is statistically significant at α = 0.05 but not at α = 0.01, suggesting moderate evidence of protein differences When analyzing one-way ANOVA results, you need to compare your calculated F-statistic against critical values at different significance levels to determine the strength of evidence for group differences. Here, the calculated F = 4.12 falls between the two critical values: it's greater than 3.47 (the α = 0.05 threshold) but less than 5.61 (the α = 0.01 threshold). This means you can reject the null hypothesis at the 0.05 level but not at the more stringent 0.01 level, indicating moderate evidence of protein differences among grain varieties. Looking at the incorrect options: Answer A is wrong because F = 4.12 < 5.61, so the result is not significant at α = 0.01. Answer B incorrectly concludes no significance exists, but since 4.12 > 3.47, we do have significance at α = 0.05. Answer D mischaracterizes the situation as "inconclusive" when standard statistical interpretation provides a clear conclusion about significance levels. The precision of statistical evidence depends on your chosen alpha level. Lower alpha levels (like 0.01) require stronger evidence to reject the null hypothesis, while higher levels (like 0.05) are more lenient. When your F-statistic falls between these thresholds, you've found moderate evidence—significant enough to detect a difference, but not strong enough to meet the most rigorous standards. Remember: Always compare your test statistic to the appropriate critical value for your chosen significance level, and recognize that falling between critical values indicates moderate rather than strong or no evidence.
20. What is the primary statistical reason that this approach of conducting multiple t-tests is considered inappropriate?
    Answer: It inflates the family-wise Type I error rate, substantially increasing the probability of incorrectly finding a significant difference. 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 $\alpha = 0.05$, 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 $1 - (1 - \alpha)^c$, where c is the number of comparisons. In this case, it would be $1 - (0.95)^6 \approx 0.265$, 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.

Start this quiz