به انگلیسی

Many means, one test

Mathematics · Data & Statistics · ages 21-22

Name ______________________ Date ____________

An ANOVA table has total SS = 150 and between-group SS = 45. What is within-group SS?
Answer: ______________
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
Between group sum of squares is 60 and within group is 40. What is the total sum of squares?
Answer: ______________
Four groups are compared. How many pairwise t tests would that take?
- 4
- 3
- 6
An ANOVA compares three groups. What are the between-group degrees of freedom?
Answer: ______________
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
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
Rejecting the ANOVA null hypothesis establishes that every pair of population means differs. True or false?
Circle one: True False
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
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

Answer key

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

Many means, one test W1-mt_WYGK3Ktnqz-s1

105 · Within-group SS is the total minus the between-group component: 150 - 45 = 105.
Compare measured plant heights across three soil types · Soil type defines the groups, and plant height is numerical. The other studies have categorical responses.
100 · Add the two components to get the total: 60 plus 40.
6 · Four groups have six distinct pairs. Comparing every pair separately would take six tests.
2 · Between-group df = k - 1. With three groups, 3 - 1 = 2.
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.
Population variances are equal across groups · The classical model assumes equal population variances, independent observations and normal populations within groups.
False · Rejection gives evidence that not all population means are equal. It does not establish a difference for every pair.
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.
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.