Question 1

Refer to the following partial ANOVA results from Excel (some information is missing) . ANOVA Table
 The number of treatment groups is

Accepted Answer

A)  5
B)  4
C)  3
D)  impossible to ascertain from given information.

Question 2

Refer to the following partial ANOVA results from Excel (some information is missing) .  At α = .05,the difference between group means is

Accepted Answer

A)  highly significant.
B)  barely significant.
C)  not quite significant.
D)  clearly insignificant.

Question 3

Which statement is incorrect?

Accepted Answer

A)  We need a Tukey test because ANOVA does not tell which pairs of means differ.
B)  Hartley's test is needed to determine whether the means of the groups differ.
C)  ANOVA assumes equal variances in the c groups being compared.

Question 4

Hartley's test is the largest sample mean divided by the smallest sample mean.

Accepted Answer

A) True 
 B)False

Question 5

After performing a one-factor ANOVA test,John noticed that the sample standard deviations for his four groups were,respectively,33,24,79,and 35.John should

Accepted Answer

A)  feel confident in his ANOVA test.
B)  use Hartley's test to check his assumptions.
C)  use an independent samples t-test instead of ANOVA.
D)  use a paired t-test instead of ANOVA.

Question 6

The Kruskal-Wallis test is analogous to the one-factor ANOVA.

Accepted Answer

A) True 
 B)False

Question 7

Which is not an assumption of ANOVA?

Accepted Answer

A)  Normality of the populations
B)  Homogeneous variances
C)  Equal treatment effects
D)  Independent observations

Question 8

Which is a characteristic of the Kruskal-Wallis test?

Accepted Answer

A)  It is rarely used because it assumes normality.
B)  It assumes equal population means.
C)  It is useful in comparing the medians in c groups.
D)  It requires at least 5 groups of identical size.

Question 9

One-factor analysis of variance

Accepted Answer

A)  requires that the number of observations in each group be identical.
B)  has less power when the number of observations per group is not identical.
C)  is extremely sensitive to slight departures from normality.
D)  is a generalization of the t-test for paired observations.

Question 10

Refer to the following partial ANOVA results from Excel (some information is missing) .  SS for between-groups variation will be

Accepted Answer

A)  129.99
B)  630.83
C)  1233.4
D)  We cannot tell from given information.

Question 11

Using one-factor ANOVA with 30 observations,we find at α = .05 that we cannot reject the null hypothesis of equal means.We increase the sample size from 30 observations to 60 observations and obtain the same value for the sample F-test statistic.Which is correct?

Accepted Answer

A)  We might now be able to reject the null hypothesis.
B)  We surely must reject H0 for 60 observations
C)  We cannot reject H0 since we obtained the same F-value.
D)  It is impossible to get the same F-value for n = 60 as for n = 30.

Question 12

One-factor ANOVA stacked data for five groups will be arranged in five separate columns.

Accepted Answer

A) True 
 B)False

Question 13

Refer to the following partial ANOVA results from Excel (some information is missing) .  Assuming equal group sizes,the number of observations in each group is

Accepted Answer

A)  2
B)  3
C)  4
D)  6

Question 14

Levene's test for homogeneity of variance is attractive because it does not depend on the assumption of normality.

Accepted Answer

A) True 
 B)False

Question 15

Degrees of freedom for the between-group variation in a one-factor ANOVA with n1 = 8,n2 = 5,n3 = 7,n4 = 9 would be

Accepted Answer

A)  28
B)  3
C)  29
D)  4

Question 16

Systolic blood pressure of randomly selected HMO patients was recorded on a particular Wednesday,with the results shown here.An ANOVA test was performed using these data.  Degrees of freedom for the between-treatments sum of squares would be

Accepted Answer

A)  3
B)  19
C)  17
D)  It depends on α.

Question 17

In an ANOVA,when would the F-test statistic be zero?

Accepted Answer

A)  When there is no difference in the variances
B)  When the treatment means are the same
C)  When the observations are normally distributed
D)  The F-test statistic cannot ever be zero

Question 18

Refer to the following partial ANOVA results from Excel (some information is missing) .  The sample size is

Accepted Answer

A)  20
B)  23
C)  24
D)  21

Question 19

Which of the following is not a characteristic of the F distribution?

Accepted Answer

A)  It is always right-skewed.
B)  It describes the ratio of two variances.
C)  It is a family based on two sets of degrees of freedom.
D)  It is negative when s12 is smaller than s22.

Question 20

Tukey's test compares pairs of treatment means in an ANOVA.

Accepted Answer

A) True 
 B)False

Exam 11: Analysis of Variance

Refer to the following partial ANOVA results from Excel (some information is missing) . Assuming equal group sizes,the number of observations in each group is

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In an ANOVA,when would the F-test statistic be zero?

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Which is a characteristic of the Kruskal-Wallis test?

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Refer to the following partial ANOVA results from Excel (some information is missing) . The sample size is

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Refer to the following partial ANOVA results from Excel (some information is missing) . SS for between-groups variation will be

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Using one-factor ANOVA with 30 observations,we find at α = .05 that we cannot reject the null hypothesis of equal means.We increase the sample size from 30 observations to 60 observations and obtain the same value for the sample F-test statistic.Which is correct?

Correct AnswerverifiedShow Answer

Tukey's test compares pairs of treatment means in an ANOVA.

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Refer to the following partial ANOVA results from Excel (some information is missing) . At α = .05,the difference between group means is

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Levene's test for homogeneity of variance is attractive because it does not depend on the assumption of normality.

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One-factor analysis of variance

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After performing a one-factor ANOVA test,John noticed that the sample standard deviations for his four groups were,respectively,33,24,79,and 35.John should

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The Kruskal-Wallis test is analogous to the one-factor ANOVA.

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Which of the following is not a characteristic of the F distribution?

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Refer to the following partial ANOVA results from Excel (some information is missing) . ANOVA Table The number of treatment groups is

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One-factor ANOVA stacked data for five groups will be arranged in five separate columns.

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Which is not an assumption of ANOVA?

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Systolic blood pressure of randomly selected HMO patients was recorded on a particular Wednesday,with the results shown here.An ANOVA test was performed using these data. Degrees of freedom for the between-treatments sum of squares would be

Correct AnswerverifiedShow Answer

Degrees of freedom for the between-group variation in a one-factor ANOVA with n1 = 8,n2 = 5,n3 = 7,n4 = 9 would be

Correct AnswerverifiedShow Answer

Hartley's test is the largest sample mean divided by the smallest sample mean.

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Which statement is incorrect?

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