Deck 11: Analysis of Variance
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Deck 11: Analysis of Variance
1
ANOVA is robust to violations of the equal-variance assumption as long as group sizes are equal.
True
2
Hartley's test is to check for unequal variances for c groups.
True
3
Tukey's test with seven groups would entail 21 comparisons of means.
True
4
Tukey's test is not needed if we have the overall F statistic for the ANOVA.
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5
ANOVA assumes normal populations.
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6
It is desirable,but not necessary,that sample sizes be equal in a one-factor ANOVA.
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7
Sample sizes must be equal in one-factor ANOVA.
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8
Tukey's test compares pairs of treatment means in an ANOVA.
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9
Tukey's test pools all the sample variances.
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10
Hartley's test is the largest sample mean divided by the smallest sample mean.
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11
Comparison of c means in one-factor ANOVA can equivalently be done by using c individual t-tests on c pairs of means at the same α.
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12
ANOVA is a procedure intended to compare the means of several groups (treatments).
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13
One-factor ANOVA stacked data for five groups will be arranged in five separate columns.
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14
Levene's test for homogeneity of variance is attractive because it does not depend on the assumption of normality.
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15
ANOVA assumes equal variances within each treatment group.
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16
ANOVA is a procedure intended to compare the variances of several groups (treatments).
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17
One-factor ANOVA with two groups is equivalent to a two-tailed t-test.
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18
Tukey's test is similar to a two-sample t-test except that it pools the variances for all c samples.
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19
Tukey's test for five groups would require 10 comparisons of means.
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20
Hartley's test measures the equality of the means for several groups.
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21
Which is not an assumption of ANOVA?
A)Normality of the populations
B)Homogeneous variances
C)Equal treatment effects
D)Independent observations
A)Normality of the populations
B)Homogeneous variances
C)Equal treatment effects
D)Independent observations
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22
The Kruskal-Wallis test is equivalent to comparing medians in c groups.
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23
The Kruskal-Wallis test does not require normal populations,but it does require them to be of similar shape.
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24
ANOVA is used to compare
A)proportions of several groups.
B)variances of several groups.
C)means of several groups.
D)both means and variances.
A)proportions of several groups.
B)variances of several groups.
C)means of several groups.
D)both means and variances.
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25
Degrees of freedom for the between-group variation in a one-factor ANOVA with n1 = 5,n2 = 6,n3 = 7 would be
A)18
B)17
C)6
D)2
A)18
B)17
C)6
D)2
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26
Which of the following is not a characteristic of the F distribution?
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.
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.
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27
The Kruskal-Wallis test is useful in finance or marketing,because normal populations can rarely be assumed for financial or marketing data.
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28
In an ANOVA,when would the F-test statistic be zero?
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
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
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29
Degrees of freedom for the between-group variation in a one-factor ANOVA with n1 = 8,n2 = 5,n3 = 7,n4 = 9 would be
A)28
B)3
C)29
D)4
A)28
B)3
C)29
D)4
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30
Which is not an assumption of ANOVA?
A)Normality of the treatment populations
B)Homogeneous treatment variances
C)Independent sample observations
D)Equal population sizes for groups
A)Normality of the treatment populations
B)Homogeneous treatment variances
C)Independent sample observations
D)Equal population sizes for groups
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31
Which is not assumed in ANOVA?
A)Observations are independent.
B)Populations are normally distributed.
C)Variances of all treatment groups are the same.
D)Population variances are known.
A)Observations are independent.
B)Populations are normally distributed.
C)Variances of all treatment groups are the same.
D)Population variances are known.
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32
In a one-factor ANOVA,the computed value of F will be negative
A)when there is no difference in the treatment means.
B)when there is no difference within the treatments.
C)when the SST (total)is larger than SSE (error).
D)under no circumstances.
A)when there is no difference in the treatment means.
B)when there is no difference within the treatments.
C)when the SST (total)is larger than SSE (error).
D)under no circumstances.
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33
In an ANOVA,the SSE (error)sum of squares reflects
A)the effect of the combined factor(s).
B)the overall variation in Y that is to be explained.
C)the variation that is not explained by the factors.
D)the combined effect of treatments and sample size.
A)the effect of the combined factor(s).
B)the overall variation in Y that is to be explained.
C)the variation that is not explained by the factors.
D)the combined effect of treatments and sample size.
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34
Which is the Excel function to find the critical value of F for α = .05,df1 = 3,df2 = 25?
A)=F.DIST(.05,2,24)
B)=F.INV.RT(.05,3,25)
C)=F.DIST(.05,3,25)
D)=F.INV(.05,2,24)
A)=F.DIST(.05,2,24)
B)=F.INV.RT(.05,3,25)
C)=F.DIST(.05,3,25)
D)=F.INV(.05,2,24)
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35
To test the null hypothesis H0: μ1 = μ2 = μ3 using samples from normal populations with unknown but equal variances,we
A)cannot safely use ANOVA.
B)can safely employ ANOVA.
C)would prefer three separate t-tests.
D)would need three-factor ANOVA.
A)cannot safely use ANOVA.
B)can safely employ ANOVA.
C)would prefer three separate t-tests.
D)would need three-factor ANOVA.
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36
The Kruskal-Wallis test is analogous to the one-factor ANOVA.
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37
The Kruskal-Wallis test is often almost as powerful as one-factor ANOVA.
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38
Which Excel function gives the right-tail p-value for an ANOVA test with a test statistic Fcalc = 4.52,n = 29 observations,and c = 4 groups?
A)=F.DIST.RT(4.52,3,25)
B)=F.INV(4.52,4,28)
C)=F.DIST(4.52,4,28)
D)=F.INV(4.52,3,25)
A)=F.DIST.RT(4.52,3,25)
B)=F.INV(4.52,4,28)
C)=F.DIST(4.52,4,28)
D)=F.INV(4.52,3,25)
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39
Analysis of variance is a technique used to test for
A)equality of two or more variances.
B)equality of two or more means.
C)equality of a population mean and a given value.
D)equality of more than two variances.
A)equality of two or more variances.
B)equality of two or more means.
C)equality of a population mean and a given value.
D)equality of more than two variances.
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40
Variation "within" the ANOVA treatments represents
A)random variation.
B)differences between group means.
C)differences between group variances.
D)the effect of sample size.
A)random variation.
B)differences between group means.
C)differences between group variances.
D)the effect of sample size.
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41
Refer to the following partial ANOVA results from Excel (some information is missing).
The sample size is
A)20
B)23
C)24
D)21

A)20
B)23
C)24
D)21
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42
Refer to the following partial ANOVA results from Excel (some information is missing).
The critical value of F at α = .05 is
A)1.645
B)2.84
C)3.10
D)4.28

A)1.645
B)2.84
C)3.10
D)4.28
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43
Given the following ANOVA table (some information is missing),find the critical value of F.05. 
A)3.06
B)2.90
C)2.36
D)3.41

A)3.06
B)2.90
C)2.36
D)3.41
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44
One-factor analysis of variance
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.
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.
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45
In a one-factor ANOVA,the total sum of squares is equal to
A)the sum of squares within groups plus the sum of squares between groups.
B)the sum of squares within groups times the sum of squares between groups.
C)the sum of squares within groups divided by the sum of squares between groups.
D)the means of all the groups squared.
A)the sum of squares within groups plus the sum of squares between groups.
B)the sum of squares within groups times the sum of squares between groups.
C)the sum of squares within groups divided by the sum of squares between groups.
D)the means of all the groups squared.
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46
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?
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.
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.
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47
The Internal Revenue Service wishes to study the time required to process tax returns in three regional centers.A random sample of three tax returns is chosen from each of three centers.The time (in days)required to process each return is recorded as shown below.
The appropriate test to use to compare the means for all three groups would be
A)test for independent samples.
B)one-factor ANOVA.
C)repeated two-sample test of means.
D)test for outliers.

A)test for independent samples.
B)one-factor ANOVA.
C)repeated two-sample test of means.
D)test for outliers.
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48
Given the following ANOVA table (some information is missing),find the F statistic. 
A)3.71
B)0.99
C)0.497
D)4.02

A)3.71
B)0.99
C)0.497
D)4.02
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49
Identify the degrees of freedom for the treatment and error in this one-factor ANOVA (blanks indicate missing information). 
A)4,24
B)3,20
C)5,23

A)4,24
B)3,20
C)5,23
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50
Refer to the following partial ANOVA results from Excel (some information is missing).
At α = .05,the difference between group means is
A)highly significant.
B)barely significant.
C)not quite significant.
D)clearly insignificant.

A)highly significant.
B)barely significant.
C)not quite significant.
D)clearly insignificant.
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51
Refer to the following partial ANOVA results from Excel (some information is missing).
Degrees of freedom for the F-test are
A)5,22
B)4,21
C)3,20
D)impossible to determine.

A)5,22
B)4,21
C)3,20
D)impossible to determine.
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52
For this one-factor ANOVA (some information is missing),what is the F-test statistic? 
A)0.159
B)2.833
C)1.703
D)Cannot be determined

A)0.159
B)2.833
C)1.703
D)Cannot be determined
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53
For this one-factor ANOVA (some information is missing),how many treatment groups were there? 
A)Cannot be determined
B)3
C)4
D)2

A)Cannot be determined
B)3
C)4
D)2
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54
Refer to the following partial ANOVA results from Excel (some information is missing).
The F-test statistic is
A)2.84
B)3.56
C)2.80
D)2.79

A)2.84
B)3.56
C)2.80
D)2.79
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55
The within-treatment variation reflects
A)variation among individuals of the same group.
B)variation between individuals in different groups.
C)variation explained by factors included in the ANOVA model.
D)variation that is not part of the ANOVA model.
A)variation among individuals of the same group.
B)variation between individuals in different groups.
C)variation explained by factors included in the ANOVA model.
D)variation that is not part of the ANOVA model.
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56
Refer to the following partial ANOVA results from Excel (some information is missing).
SS for between-groups variation will be
A)129.99
B)630.83
C)1233.4
D)We cannot tell from given information.

A)129.99
B)630.83
C)1233.4
D)We cannot tell from given information.
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57
The Internal Revenue Service wishes to study the time required to process tax returns in three regional centers.A random sample of three tax returns is chosen from each of three centers.The time (in days)required to process each return is recorded as shown below.Subsequently,an ANOVA test was performed.
Degrees of freedom for the error sum of squares in the ANOVA would be
A)11
B)2
C)4
D)6

A)11
B)2
C)4
D)6
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58
Refer to the following partial ANOVA results from Excel (some information is missing).
The number of treatment groups is
A)4
B)3
C)2
D)1

A)4
B)3
C)2
D)1
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59
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
A)2
B)3
C)4
D)6

A)2
B)3
C)4
D)6
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60
Refer to the following partial ANOVA results from Excel (some information is missing).
Degrees of freedom for between-groups variation are
A)3
B)4
C)5
D)We cannot tell from given information.

A)3
B)4
C)5
D)We cannot tell from given information.
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61
Refer to the following partial ANOVA results from Excel (some information is missing). ANOVA Table
The number of treatment groups is
A)5
B)4
C)3
D)impossible to ascertain from given information.

A)5
B)4
C)3
D)impossible to ascertain from given information.
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62
Sound levels are measured at random moments under typical driving conditions for various full-size truck models.The Excel ANOVA results are shown below.
The test statistic to compare the five means simultaneously is
A)2.96
B)15.8
C)5.56
D)4.45


A)2.96
B)15.8
C)5.56
D)4.45
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63
To compare the cost of three shipping methods,a random sample of four shipments is taken for each of three firms.The cost per shipment is shown below.
In a one-factor ANOVA,the degrees of freedom for the between-groups sum of squares will be
A)11
B)3
C)2
D)9

A)11
B)3
C)2
D)9
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64
Refer to the following partial ANOVA results from Excel (some information is missing). ANOVA Table
Using Appendix F,the 5 percent critical value for the F-test is approximately
A)3.24
B)6.91
C)2.56
D)2.06

A)3.24
B)6.91
C)2.56
D)2.06
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65
To compare the cost of three shipping methods,a random sample of four shipments is taken for each of three firms.The cost per shipment is shown below.
The degrees of freedom for the total sum of squares in a one-factor ANOVA would be
A)11
B)8
C)2
D)9

A)11
B)8
C)2
D)9
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66
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
A)3
B)19
C)17
D)It depends on α.

A)3
B)19
C)17
D)It depends on α.
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67
The Internal Revenue Service wishes to study the time required to process tax returns in three regional centers.A random sample of three tax returns is chosen from each of three centers.The time (in days)required to process each return is recorded as shown below.
Degrees of freedom for the between-groups sum of squares in the ANOVA would be
A)11
B)2
C)4
D)6

A)11
B)2
C)4
D)6
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68
Refer to the following partial ANOVA results from Excel (some information is missing). ANOVA Table
The F statistic is
A)2.88
B)4.87
C)5.93
D)6.91

A)2.88
B)4.87
C)5.93
D)6.91
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69
Refer to the following partial ANOVA results from Excel (some information is missing). ANOVA table
The 5 percent critical value for the F test is
A)2.46
B)3.24
C)3.38
D)impossible to ascertain from the given information.

A)2.46
B)3.24
C)3.38
D)impossible to ascertain from the given information.
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70
Refer to the following partial ANOVA results from Excel (some information is missing). ANOVA Table
The p-value for the F-test would be
A)much less than .05.
B)slightly less than .05.
C)slightly greater than .05.
D)much greater than .05.

A)much less than .05.
B)slightly less than .05.
C)slightly greater than .05.
D)much greater than .05.
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71
Refer to the following partial ANOVA results from Excel (some information is missing). ANOVA table
Our decision about the hypothesis of equal treatment means is that the null hypothesis
A)cannot be rejected at α = .05.
B)can be rejected at α = .05.
C)can be rejected for any typical value of α.
D)cannot be assessed from the given information.

A)cannot be rejected at α = .05.
B)can be rejected at α = .05.
C)can be rejected for any typical value of α.
D)cannot be assessed from the given information.
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72
Sound levels are measured at random moments under typical driving conditions for various full-size truck models.The ANOVA results are shown below.
The test statistic for Hartley's test for homogeneity of variance is
A)2.25
B)5.04
C)4.61
D)4.45


A)2.25
B)5.04
C)4.61
D)4.45
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73
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.
What are the degrees of freedom for the error sum of squares?
A)3
B)19
C)16
D)It depends on α.

A)3
B)19
C)16
D)It depends on α.
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74
Refer to the following partial ANOVA results from Excel (some information is missing). ANOVA table
The F statistic is
A)4.87
B)3.38
C)5.93
D)6.91

A)4.87
B)3.38
C)5.93
D)6.91
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75
Refer to the following partial ANOVA results from Excel (some information is missing). ANOVA table
The MS (mean square)for the treatments is
A)239.13
B)106.88
C)1130.8
D)impossible to ascertain from the information given.

A)239.13
B)106.88
C)1130.8
D)impossible to ascertain from the information given.
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76
Refer to the following partial ANOVA results from Excel (some information is missing). ANOVA Table
The number of observations in the original sample was
A)59
B)60
C)58
D)54

A)59
B)60
C)58
D)54
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77
Refer to the following partial ANOVA results from Excel (some information is missing). ANOVA table
The number of observations in the entire sample is
A)20
B)19
C)22

A)20
B)19
C)22
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78
Prof.Gristmill sampled exam scores for five randomly chosen students from each of his two sections of ACC 200.His sample results are shown.
He could test the population means for equality using
A)a t-test for two means from independent samples.
B)a t-test for two means from paired (related)samples.
C)a one-factor ANOVA.
D)either a one-factor ANOVA or a two-tailed t-test.

A)a t-test for two means from independent samples.
B)a t-test for two means from paired (related)samples.
C)a one-factor ANOVA.
D)either a one-factor ANOVA or a two-tailed t-test.
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79
Systolic blood pressure of randomly selected HMO patients was recorded on a particular Wednesday,with the results shown here:
The appropriate hypothesis test is
A)one-factor ANOVA.
B)repeated t-tests to compare means.
C)correlation test.
D)two-tailed test of proportions.

A)one-factor ANOVA.
B)repeated t-tests to compare means.
C)correlation test.
D)two-tailed test of proportions.
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80
To compare the cost of three shipping methods,a random sample of four shipments is taken for each of three firms.The cost per shipment is shown below.
In a one-factor ANOVA,the degrees of freedom for the within-groups sum of squares will be
A)11
B)3
C)9
D)2

A)11
B)3
C)9
D)2
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