One-Way Analysis of Variance · seed 1 · A4, ink-friendly. The answer key prints on its own page for grown-ups.

Many means, one test

Mathematics · Data & Statistics · ages 21-22
Name ______________________   Date ____________

  1. An ANOVA table has total SS = 150 and between-group SS = 45. What is within-group SS?

    Answer: ______________

  2. Which study has a categorical factor and a numerical response for one-way ANOVA?

    • Compare measured plant heights across three soil types
    • Compare favourite colours across three soil types
    • Compare preferred soil types across three colour groups
    • Compare flower colours across three seed brands

  3. Between group sum of squares is 60 and within group is 40. What is the total sum of squares?

    Answer: ______________

  4. Four groups are compared. How many pairwise t tests would that take?

    • 4
    • 3
    • 6

  5. An ANOVA compares three groups. What are the between-group degrees of freedom?

    Answer: ______________

  6. Why can six unadjusted pairwise t-tests at 5% give a higher overall false-rejection risk than one ANOVA at 5%?

    • Each pairwise test must use more observations than ANOVA
    • There are more chances for a false rejection
    • ANOVA proves which pairs have unequal population means

  7. Which condition belongs to the classical one-way ANOVA model for its usual F reference distribution?

    • All observations within each group are identical
    • Population variances increase with group means
    • Population variances are equal across groups
    • The response is a category rather than a number

  8. Rejecting the ANOVA null hypothesis establishes that every pair of population means differs. True or false?

    Circle one:   True   False

  9. ANOVA rejects equal population means for four groups. You want to compare every pair. Which plan controls error across those comparisons?

    • Run six unadjusted t-tests, each at the chosen family level
    • Declare every pair unequal because the joint null was rejected
    • Use pairwise tests with family-wise error control

  10. The ANOVA assumptions hold. With df (2, 12), observed F is 8.1 and the chosen right-tail critical value is 3.9. What is the decision?

    • Conclude that all population means are exactly equal
    • Do not reject the equal-means null
    • Use 8.1 as the cutoff and 3.9 as the statistic
    • Reject the equal-means null hypothesis
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Answer key

For grown-ups. Fold this page away before handing over the rest.

Many means, one test W1-mt_WYGK3Ktnqz-s1

  1. 105 · Within-group SS is the total minus the between-group component: 150 - 45 = 105.
  2. Compare measured plant heights across three soil types · Soil type defines the groups, and plant height is numerical. The other studies have categorical responses.
  3. 100 · Add the two components to get the total: 60 plus 40.
  4. 6 · Four groups have six distinct pairs. Comparing every pair separately would take six tests.
  5. 2 · Between-group df = k - 1. With three groups, 3 - 1 = 2.
  6. There are more chances for a false rejection · A 5% level applies to each pairwise test, not automatically to the whole family. Several tests can raise the chance of at least one false rejection.
  7. Population variances are equal across groups · The classical model assumes equal population variances, independent observations and normal populations within groups.
  8. False · Rejection gives evidence that not all population means are equal. It does not establish a difference for every pair.
  9. Use pairwise tests with family-wise error control · The joint test does not identify the differing pairs. An adjusted comparison procedure controls false-rejection risk across the follow-up family.
  10. Reject the equal-means null hypothesis · Observed F exceeds the stated right-tail cutoff: 8.1 > 3.9. Reject the joint equal-means null. This does not identify individual differing pairs.
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