Question 1

When the effect of a level for one factor depends on which level of another factor is present, the most appropriate ANOVA design to use in this situation is the:
A) one-way ANOVA.
B) two-way ANOVA.
C) randomised block design.
D) matched pairs design.

Accepted Answer

A two-way ANOVA is appropriate in situations where there are two factors, and the effect of one factor depends on the level of the other factor. It allows us to examine the main effects of each factor as well as their interaction. The other designs listed (one-way ANOVA, randomised block design, and matched pairs design) are not suitable for examining the interaction between two factors.

Question 2

Which of the following best describes the between-treatments in single-factor analysis of variance?
A) Sum of squares for error.
B) Total sum of squares.
C) Sum of squares for treatments.
D) All of these choices are correct.

Accepted Answer

The between-treatments in single-factor analysis of variance is represented by the sum of squares for treatments, which measures the variation between the means of the different treatments or groups. The sum of squares for error measures the variation within each group, while the total sum of squares is the sum of the variation between and within groups. Therefore, choice C is the best answer.

Question 3

In the one-way ANOVA where k is the number of treatments and n is the number of observations in all samples, the number of degrees of freedom for treatments is given by:
A) k - 1.
B) n - k.
C) n - 1.
D) n - k + 1.

Accepted Answer

The degrees of freedom for treatments is given by k - 1. This is because we need to estimate the variability due to treatments by comparing the means of different treatment groups. Once we know the means of k groups, we can calculate a grand mean, and then we need to estimate variability between the means and variability within the means. Since the total variability has n - 1 degrees of freedom (total sum of squares), and the within-group variability has n - k degrees of freedom (error sum of squares), we can obtain the degrees of freedom for treatments by subtracting the degrees of freedom for error from the total degrees of freedom, which is k - 1.

Question 4

A professor of statistics at Wayne State University in the US wants to determine whether the average starting salaries among graduates of the 15 universities in Michigan are equal. A sample of 25 recent graduates from each university is randomly taken. The appropriate critical value for the ANOVA test is obtained from the F-distribution with number so of degrees of freedom equal to:
A) 15 and 25.
B) 14 and 360.
C) 360 and 14.
D) 25 and 15.

Accepted Answer

The degrees of freedom for the ANOVA test are calculated as follows: the numerator degrees of freedom is the number of groups minus one (15 - 1 = 14), and the denominator degrees of freedom is the total number of observations minus the number of groups (15*25 - 15 = 360).

Question 5

In single-factor analysis of variance, if large differences exist among the sample means, it is then reasonable to:
A) reject the null hypothesis.
B) reject the alternative hypothesis.
C) fail to reject the null hypothesis.
D) None of these choices are correct.

Accepted Answer

If large differences exist among the sample means, it indicates that there is a significant difference between at least two groups. This means that the null hypothesis, which states that all group means are equal, can be rejected in favor of the alternative hypothesis, which states that at least one group mean is different from the others. Therefore, the best choice is to reject the null hypothesis (choice A).

Question 6

A survey will be conducted to compare the grade point averages of US high-school students from four different school districts. Students are to be randomly selected from each of the four districts and their grade point averages recorded. The ANOVA model most likely to fit this situation is:
A) one-way ANOVA.
B) two-way ANOVA.
C) randomised block design.
D) complete 4 * 4 factorial design.

Accepted Answer

Since we are comparing grade point averages from different school districts, the most appropriate model is a one-way ANOVA. This model will allow us to compare the means of each district and determine if there is a significant difference between them. A two-way ANOVA would be appropriate if we were also interested in the effect of another variable, such as gender or socioeconomic status, on the grade point averages. A randomised block design and a complete 4 factorial design would be appropriate for experiments with multiple treatments or interventions, but not for this situation.

Question 7

One-way ANOVA is applied to independent samples taken from four normally distributed populations with equal variances. If the null hypothesis is rejected, then we can infer that:
A) all population means are equal.
B) all population means differ.
C) at least two population means are equal.
D) at least two population means differ.

Accepted Answer

Rejecting the null hypothesis in a one-way ANOVA with independent samples implies that at least two population means differ.

Question 8

Which of the following is not a required condition for one-way ANOVA?
A) The populations are normally distributed.
B) The population variances are equal.
C) The samples are selected independently of each other.
D) The population means are equal.

Accepted Answer

One-way ANOVA tests the hypothesis that the means of multiple groups are equal. Therefore, one of the required conditions is that the population means are equal. The other required conditions are that populations are normally distributed, population variances are equal, and samples are selected independently.

Question 9

In ANOVA, the F-test is the ratio of two sample variances. In the one-way ANOVA (completely randomised design), the variance used as the denominator of the ratio is the:
A) mean square for treatments.
B) mean square for error.
C) mean square for blocks.
D) total sum of squares.

Accepted Answer

In one-way ANOVA, the F-test compares the variance between group means (mean square for treatments) to the variance within groups (mean square for error). Therefore, the variance used as the denominator of the ratio is the mean square for error.

Question 10

In the one-way ANOVA where k is the number of treatments and n is the number of observations in all samples, the number of degrees of freedom for error is given by:
A) k - 1.
B) n - k.
C) n - 1.
D) n - k + 1.

Accepted Answer

The degrees of freedom for error in a one-way ANOVA is calculated as the total number of observations minus the number of treatments, which is n - k.

Question 11

In one-way analysis of variance, within-treatments variation stands for the:
A) sum of squares for error.
B) sum of squares for treatments.
C) total sum of squares.
D) None of these choices are correct.

Accepted Answer

Within-treatments variation represents the variability or error that is left unexplained by the treatments being tested. It is measured by the sum of squares for error, which is calculated as the total sum of squares minus the sum of squares for treatments. Therefore, choice A is the correct answer.

Question 12

One-way ANOVA is performed on independent samples taken from three normally distributed populations with equal variances. The following summary statistics were calculated: $n _ { 1 } =$ 7, $\bar { x } _ { 1 } =$ 65, $s _ { 1 } =$ 4.2. $n _ { 2 } =$ 8, $\bar { x } _ { 2 } =$ 59, $S _ { 2 } =$ 4.9. $n _ { 3 } =$ 9, $\bar { x } _ { 3 } =$ 63, $s _ { 3 } =$ 4.6. The value of the test statistics, F, equals:
A) 71.250.
B) 0.322.
C) 21.104.
D) 3.376.

Accepted Answer

The F-statistic is calculated using the formula for ANOVA: $ F = \frac{MS_{between}}{MS_{within}} $. After calculating the mean squares for between and within groups using the given data, the F-statistic is found to be approximately 3.376.

Question 13

Consider the following partial ANOVA table: $$\begin{array} { | l | c | c | c | c | } 
\hline \text { Source of Variation } & \text { SS } & d f & M S & F \
\hline \text { Treatments } & 75 & * & 25 & 6.67 \
\hline \text { Error } & 60 & * & 3.75 & \
\hline \text { Total } & 135 & 19 & & \
\hline
\end{array}$$ The numbers of degrees of freedom for numerator and denominator, respectively, (identified by asterisks) are:
A) 4 and 15.
B) 3 and 16.
C) 15 and 4.
D) 16 and 3.

Accepted Answer

The total degrees of freedom is always equal to the sum of the degrees of freedom for the treatments and error. 
Therefore, $df_{total} = df_{treatment} + df_{error}$. 
Substituting the given values, we get $19 = df_{treatment} + df_{error}$.
Also, the numerator degrees of freedom for treatments is equal to the number of treatment groups minus 1. Since this information is not given, we cannot determine the exact value of $df_{treatment}$.
However, we can use the fact that the degrees of freedom for error is equal to $df_{error} = df_{total} - df_{treatment}$. Substituting the given values, we get $df_{error} = 19 - df_{treatment}$.
From the ANOVA table, we see that $MS_{error} = \frac{SS_{error}}{df_{error}} = 3.75$. Substituting the given values, we get $60/df_{error} = 3.75$ which gives $df_{error} = 16$.
Using $df_{total} = df_{treatment} + df_{error}$ and $df_{error} = 16$, we get $19 = df_{treatment} + 16$ which gives $df_{treatment} = 3$.
Therefore, the answer is B) 3 and 16.

Question 14

Which of the following is a correct formulation for the null hypothesis in one-way ANOVA?
A) $\mu _ { 1 } + \mu _ { 2 } + \mu _ { 3 } = 0$ .
B) $\mu _ { 1 } + \mu _ { 2 } + \mu _ { 3 } \neq 0$ .
C) $\mu _ { 1 } = \mu _ { 2 } = \mu _ { 3 }$ .
D) $\mu _ { 1 } \neq \mu _ { 2 } \neq \mu _ { 3 }$ .

Accepted Answer

The null hypothesis in one-way ANOVA is always that there is no difference between the means of the groups being compared. This is expressed as $\mu _ { 1 } = \mu _ { 2 } = \mu _ { 3 }$, which means that the population means of all three groups are equal. Option A implies that the sum of the population means for all three groups is zero, which is not a valid null hypothesis. Option B implies that at least one of the population means is not equal to the others, which is a statement of the alternative hypothesis. Option D implies that all three population means are different, which is also a statement of the alternative hypothesis.

Question 15

Which of the following is not true of Tukey's multiple comparison method?
A) It is based on the studentised range statistic q to obtain the critical value needed to construct individual confidence intervals.
B) It requires that all sample sizes are equal, or at least similar.
C) It can be employed instead of the analysis of variance.
D) All of these choices are correct.

Accepted Answer

Tukey's multiple comparison method requires that all sample sizes are equal, or at least similar. It cannot be employed instead of the analysis of variance.

Question 16

The randomised block design with exactly two treatments is equivalent to a two-tailed:
A) independent samples z-test.
B) independent-samples equal-variances t-test.
C) independent-samples unequal-variances t-test.
D) matched pairs t-test.

Accepted Answer

The randomised block design with exactly two treatments involves pairing each block (subject) with themselves and randomly assigning one treatment to one of the pairs and the other treatment to the other pair. Therefore, this design is equivalent to a matched pairs t-test, which is the only option that involves paired data. The other options involve independent samples, which is not the case in a randomised block design with exactly two treatments.

Question 17

One-way ANOVA is performed on independent samples taken from three normally distributed populations with equal variances. The following summary statistics are calculated: $n _ { 1 } =$ 6, $\bar { x } _ { 1 } =$ 50, $s _ { 1 } =$ 5.2. $n _ { 2 } =$ 8, $\bar { x } _ { 2 } =$ 55, $S _ { 2 } =$ 4.9 . $n _ { 3 } =$ 6, $\bar { x } _ { 3 } =$ 51, $s _ { 3 } =$ 5.4. The grand mean equals:
A) 50.0.
B) 52.0.
C) 52.3.
D) 53.0.

Accepted Answer

To find the grand mean, we need to take the average of all the sample means. Thus, the grand mean is (50 + 55 + 51)/3 = 52.0.

Question 18

Three tennis players, one a beginner, one intermediate and one advanced, have been randomly selected from the membership of a club in a large city. Using the same tennis ball, each player hits ten serves, one with each of three racquet models, with the three racquet models selected randomly. The speed of each serve is measured with a machine and the result recorded. Among the ANOVA models listed below, the most likely model to fit this situation is the:
A) one-way ANOVA.
B) two-way ANOVA.
C) randomised block design.
D) matched-pairs model.

Accepted Answer

The randomised block design is appropriate here as it accounts for variability among players (blocks) and racquet models (treatments).

Question 19

One-way ANOVA is performed on three independent samples with n₁= 10, n₂= 8 and n₃= 9. The critical value obtained from the F-table for this test at the 5% level of significance equals: A) 39.46. B) 3.40. C) 4.32. D) 19.45.

Accepted Answer

The degrees of freedom for the numerator (between groups) is k-1 = 3-1 = 2, and for the denominator (within groups) is N-k = 10+8+9-3 = 24. The critical F-value at 5% significance level for df1=2 and df2=24 is approximately 3.40.

Question 20

In the randomised block design ANOVA, the sum of squares for error equals:
A) SS(Total) - SST.
B) SS(Total) - SSB.
C) SS(Total) - SST - SSB.
D) SS(Total) - SS(A) - SS(B) - SS(AB).

Accepted Answer

In a randomized block design ANOVA, the total variation can be decomposed into three parts: the variation due to treatments (SSB), the variation due to blocks (SSA), and the variation due to error (SSE). Therefore, the sum of squares for error is equal to SS(Total) - SST - SSB.

Quiz 16: Analysis of Variance

When the effect of a level for one factor depends on which level of another factor is present, the most appropriate ANOVA design to use in this situation is the:

Which of the following best describes the between-treatments in single-factor analysis of variance?

In the one-way ANOVA where k is the number of treatments and n is the number of observations in all samples, the number of degrees of freedom for treatments is given by:

In single-factor analysis of variance, if large differences exist among the sample means, it is then reasonable to:

A survey will be conducted to compare the grade point averages of US high-school students from four different school districts. Students are to be randomly selected from each of the four districts and their grade point averages recorded. The ANOVA model most likely to fit this situation is:

One-way ANOVA is applied to independent samples taken from four normally distributed populations with equal variances. If the null hypothesis is rejected, then we can infer that:

Which of the following is not a required condition for one-way ANOVA?

In ANOVA, the F-test is the ratio of two sample variances. In the one-way ANOVA (completely randomised design) , the variance used as the denominator of the ratio is the:

In the one-way ANOVA where k is the number of treatments and n is the number of observations in all samples, the number of degrees of freedom for error is given by:

In one-way analysis of variance, within-treatments variation stands for the:

Which of the following is a correct formulation for the null hypothesis in one-way ANOVA?

Which of the following is not true of Tukey's multiple comparison method?

The randomised block design with exactly two treatments is equivalent to a two-tailed:

One-way ANOVA is performed on three independent samples with n₁= 10, n₂= 8 and n₃= 9. The critical value obtained from the F-table for this test at the 5% level of significance equals:

In the randomised block design ANOVA, the sum of squares for error equals: