Solution:

a) Number of comparisons:

b) Overall Type I error rate:

With 15 tests, there's more than a 50% chance of making at least one Type I error!

c) Why ANOVA is better:

• ANOVA performs a single test, maintaining α = 0.05
• Controls the family-wise error rate
• Avoids the multiple comparisons problem
• More powerful and statistically sound approach

Solution:

a) F = 0.8:

• Between-group variance is LESS than within-group variance
• Group means are very similar relative to natural variation
• Very little evidence of group differences
• Will definitely fail to reject H₀

b) F = 1.2:

• Between-group variance is slightly larger than within-group variance
• Weak evidence of group differences
• Likely not statistically significant
• Probably fail to reject H₀

c) F = 15.6:

• Between-group variance is MUCH larger than within-group variance
• Group means differ substantially
• Strong evidence that at least one group differs
• Very likely to reject H₀ (highly significant)

Solution:

a) Two-sample t-test

• Only 2 groups (two departments)
• Comparing means
• Quantitative variable (salary)

b) One-way ANOVA

• More than 2 groups (four majors)
• Comparing means
• One categorical factor (major)
• Quantitative variable (GPA)

c) Chi-square test of independence

• Both variables are categorical
• Testing for association/relationship
• Not comparing means

d) Kruskal-Wallis test

• More than 2 groups (5 cities)
• Highly skewed data (normality violated)
• Non-parametric alternative to ANOVA
• Compares medians instead of means

Solution:

a) Degrees of freedom:

• df₁ = 3 = df_between = k - 1, so k = 4 groups
• df₂ = 76 = df_within = N - k, so N = 80 total observations
• This means 4 locations with 80 total customers (average 20 per location)

b) Conclusion:

• p = 0.002 < 0.05, so reject H₀
• At the 0.05 significance level, there is sufficient evidence to conclude that at least one store location has a different mean customer satisfaction score than the others

c) What this does NOT tell us:

• Which specific locations differ from each other
• Which location has the highest/lowest satisfaction
• How many locations differ significantly
• Need post-hoc tests to answer these questions!

Solution:

Between-group variation:

• Measures how much the group means differ from each other
• Reflects the effect of the factor/treatment
• If groups truly differ, this should be large
• Example: Differences in average test scores across different teaching methods

Within-group variation:

• Measures how much individuals vary within each group
• Represents natural variability/"noise" in the data
• Exists regardless of group membership
• Example: Students in the same class have different scores due to individual differences

Why we compare them:

• If between-group variation is much larger than within-group variation, it suggests real group differences (not just random chance)
• The F-statistic is this ratio: F = MSB/MSW
• Large F → signal (group differences) >> noise (random variation) → significant result
• Small F → signal ≈ noise → no significant difference

Solution:

a) k = 3, N = 3 × 12 = 36

b) k = 5, N = 8 + 10 + 12 + 10 + 10 = 50

c) k = 4, N = 60

Solution:

Step 1: Find df_between

Step 2: Find N and df_total

Step 3: Calculate SST

Step 4: Calculate mean squares

Step 5: Calculate F

Complete table:

Source	SS	df	MS	F
Between	240	3	80	6.00
Within	360	27	13.33
Total	600	30

Solution:

a) Calculate means:

b) Calculate SSB:

c) Calculate SSW:

Group A:

Group B:

Group C:

Solution:

a) Calculate mean squares:

First find degrees of freedom:

Then calculate MS:

b) Calculate F:

c) Decision:

Conclusion: At the 0.05 significance level, there is sufficient evidence to conclude that at least one group mean is different from the others.

Solution:

a) Find MSB:

b) Find df_within and N:

c) Find SSW:

d) Find SST and verify:

Verification (alternatively: SST = MST × df_total, but MST isn't typically calculated)

Solution:

a) Calculate HSD:

b) Compare all pairs:

Comparison	|Difference|	vs HSD = 4.24	Significant?
Group 1 vs 2	|45 - 50| = 5	5 > 4.24	Yes
Group 1 vs 3	|45 - 53| = 8	8 > 4.24	Yes
Group 2 vs 3	|50 - 53| = 3	3 < 4.24	No

Conclusion: Group 1 differs significantly from both Groups 2 and 3, but Groups 2 and 3 do not differ significantly from each other.

Solution:

a) Number of comparisons:

b) Bonferroni-adjusted α:

c) Is p = 0.02 significant?

Explanation: While 0.02 < 0.05 (would be significant without correction), it's not less than the Bonferroni-adjusted 0.005, so we fail to reject H₀ for this comparison. This is why Bonferroni is considered conservative—it makes it harder to find significance.

Solution:

a) NO, do not conduct post-hoc tests

• p = 0.18 > 0.05, so ANOVA is not significant
• Failed to reject H₀
• No evidence of group differences
• Post-hoc tests only make sense after significant ANOVA

b) YES, conduct post-hoc tests

• p = 0.001 < 0.05, so ANOVA is significant
• df₁ = 3 means k = 4 groups (more than 2)
• Know at least one group differs, but not which ones
• Post-hoc tests will identify specific group differences

c) NO, do not conduct post-hoc tests

• Even though p < 0.001 (significant)
• df₁ = 1 means k = 2 groups only
• With only 2 groups, ANOVA is equivalent to a t-test
• Already know which groups differ (the only two!)
• Post-hoc tests are only needed with 3+ groups

Solution:

Pairwise comparisons:

Comparison	Difference	vs HSD	Significant?
A vs B	|48-52| = 4	4 > 3.5	Yes
A vs C	|48-45| = 3	3 < 3.5	No
A vs D	|48-51| = 3	3 < 3.5	No
B vs C	|52-45| = 7	7 > 3.5	Yes
B vs D	|52-51| = 1	1 < 3.5	No
C vs D	|45-51| = 6	6 > 3.5	Yes

Interpretation:

• Brand C has significantly shorter lifespan than Brands B and D
• Brand B has significantly longer lifespan than Brands A and C
• Brands B and D perform similarly (no significant difference)
• Brands A and D perform similarly
• Brands A and C perform similarly

Ranking (from longest to shortest): B ≈ D > A ≈ C

Recommendation: Brands B or D are best choices; avoid Brand C.

Solution:

a) Use Bonferroni

• Sample sizes are unequal (8, 12, 10, 15)
• Tukey's HSD assumes equal sample sizes
• Bonferroni works well with unequal sample sizes
• Alternatively, could use modified Tukey with harmonic mean, but Bonferroni is simpler

b) Bonferroni-adjusted α:

Number of comparisons:

Adjusted α:

Use α = 0.0083 (or approximately 0.008) for each of the 6 pairwise t-tests.

Solution:

Step 1: Identify max and min variances

Step 2: Calculate ratio

Step 3: Apply rule of thumb

Conclusion: Yes, the equal variance assumption is reasonable. The ratio of 1.59 is less than 2, indicating that the variances are similar enough for ANOVA.

Solution:

a) Violation: Independence

• Same 30 students measured three times (repeated measures)
• Observations are NOT independent
• Alternative: Use Repeated Measures ANOVA or Friedman test (non-parametric)

b) Violation: Normality

• Highly right-skewed data violates normality assumption
• Small sample sizes (n = 15) means CLT may not apply
• Alternatives:
  • Use Kruskal-Wallis test (compares medians, no normality assumption)
  • Log-transform the data to reduce skewness, then use ANOVA

c) Violation: Independence

• Students within same classroom may not be independent (clustered data)
• Classroom effects could influence results
• Alternatives:
  • Use multilevel/hierarchical models to account for classroom nesting
  • Randomly sample students from each classroom (not all students)
  • Consider classroom as a random effect

Solution:

a) Two-sample t-test (or independent samples t-test)

• 2 groups (men vs women)
• Comparing means
• Quantitative variable (commute time)

b) One-way ANOVA

• More than 2 groups (5 brands)
• Comparing means
• One categorical factor (brand)
• Quantitative variable (satisfaction score)

c) Chi-square test of independence

• Two categorical variables (party preference and education level)
• Testing for association/relationship
• Not comparing means

d) Kruskal-Wallis test

• More than 2 groups (3 sectors)
• Comparing medians (not means due to extreme skew)
• Normality assumption violated
• Non-parametric alternative to ANOVA

Solution:

a) Yes, ANOVA is reasonable

• Large sample sizes: All groups have n ≥ 30, so Central Limit Theorem applies
• Moderate skewness: With large n, ANOVA is robust to moderate departures from normality
• Equal sample sizes: Groups are nearly equal (45, 50, 48), making ANOVA robust to unequal variances
• Variance ratio: While 2.3 is slightly above the ideal < 2 threshold, it's not severely violated, and equal sample sizes compensate

b) Features promoting robustness:

1. Large samples (n ≥ 30): Sampling distribution of means is approximately normal regardless of population shape
2. Balanced design: Similar sample sizes across groups reduce impact of variance inequality
3. Total sample size (N = 143): Large overall N increases power and stability

Note: If concerned, could also:

• Check with Levene's test for formal variance equality test
• Use Welch's ANOVA as a sensitivity check
• Examine residual plots to assess normality

Solution:

a) Check equal variance:

Equal variance assumption is reasonable.

b) ANOVA conclusion:

• p = 0.001 < 0.05 → Reject H₀
• At the 0.05 significance level, there is sufficient evidence to conclude that at least one diet produces a different mean weight loss than the others

c) Tukey's HSD comparisons:

Comparison	Difference	vs HSD = 2.8	Significant?
Diet A vs B	|8.2 - 12.5| = 4.3	4.3 > 2.8	Yes
Diet A vs C	|8.2 - 10.8| = 2.6	2.6 < 2.8	No
Diet B vs C	|12.5 - 10.8| = 1.7	1.7 < 2.8	No

d) Complete interpretation:

A one-way ANOVA was conducted to compare the effectiveness of three diets on weight loss over an 8-week period. Thirty-six participants were randomly assigned to one of three diet groups (n = 12 per group). The equal variance assumption was met (variance ratio = 1.27).

The ANOVA revealed a statistically significant difference in mean weight loss among the three diets, F(2, 33) = 8.45, p = 0.001. Post-hoc comparisons using Tukey's HSD indicated that Diet B (M = 12.5 lbs) produced significantly more weight loss than Diet A (M = 8.2 lbs), with a difference of 4.3 pounds. However, Diet C (M = 10.8 lbs) was not significantly different from either Diet A or Diet B.

Practical conclusion: Diet B appears to be the most effective for weight loss, producing an average of 4.3 more pounds of weight loss than Diet A over 8 weeks. Diet C shows intermediate results that are not statistically distinguishable from the other two diets.