Chap 011

Chapter 11Analysis of Variance

True / False Questions1.One-factor ANOVA is a procedure intended to compare the variances of c samples.TrueFalse

2.Analysis of variance is a procedure intended to compare the means of c samples.TrueFalse

3.If you have four factors (call them A, B, C, and D) in an ANOVA experiment with replication, you could have a maximum of four different two-factor interactions.TrueFalse

4.Hartley's test measures the equality of the means for several groups.TrueFalse

5.Hartley's test is to check for unequal variances for c groups.TrueFalse

6.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 .TrueFalse

7.ANOVA assumes equal variances within each treatment group.TrueFalse

8.Three-factor ANOVA is required if we have three treatment groups (i.e., three data columns).TrueFalse

9.ANOVA assumes normal populations.TrueFalse

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

11.Tukey's test is similar to a two-sample t-test except that it pools the variances for all c samples.TrueFalse

12.Tukey's test is not needed if we have the overall F statistic for the ANOVA.TrueFalse

13.Interaction plots that show crossing lines indicate likely interactions.TrueFalse

14.Interaction plots that show parallel lines would suggest interaction effects.TrueFalse

15.In a two-factor ANOVA with three columns and four rows, there can be more than two interaction effects.TrueFalse

16.Sample sizes must be equal in one-factor ANOVA.TrueFalse

17.In a 34 randomized block (two-factor unreplicated) ANOVA, we have 12 treatment groups.TrueFalse

18.One-factor ANOVA with two groups is equivalent to a two-tailed t-test.TrueFalse

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

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

21.Tukey's test for five groups would require 10 comparisons of means.TrueFalse

22.ANOVA is robust to violations of the equal-variance assumption as long as group sizes are equal.TrueFalse

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

24.Tukey's test with seven groups would entail 21 comparisons of means.TrueFalse

25.Tukey's test pools all the sample variances.TrueFalse

26.It is desirable, but not necessary, that sample sizes be equal in a one-factor ANOVA.TrueFalse

Multiple Choice Questions27.Which is the Excel function to find the critical value of F for = .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)

28.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)

29.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.

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.

31.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.

32.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.

33.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.

34.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.

35.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.

36.To test the null hypothesis H0: 1 = 2 = 3 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.

37.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.

38.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.

39.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.

40.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.

41.Using one-factor ANOVA with 30 observations we find at = .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.

42.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.

43.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.

44.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.

45.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

46.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

47.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

48.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

49.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

50.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.


Degrees of freedom for between groups variation are:

A.3.

B.4.

C.5.

D.Can't tell from given information.


SS for between groups variation will be:

A.129.99.

B.630.83.

C.1233.4.

D.Can't tell.


The number of treatment groups is:

A.4.

B.3.

C.2.

D.1.


The sample size is:

A.20.

B.23.

C.24.

D.21.


Assuming equal group sizes, the number of observations in each group is:

A.2.

B.3.

C.4.

D.6.


Degrees of freedom for the F-test are:

A.5, 22.

B.4, 21.

C.3, 20.

D.impossible to determine.


The critical value of F at = 0.05 is:

A.1.645.

B.2.84.

C.3.10.

D.4.28.


At = 0.05, the difference between group means is:

A.highly significant.

B.barely significant.

C.not quite significant.

D.clearly insignificant.

59.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 test to use to compare the means for all three groups would require:

A.three-factor ANOVA.

B.one-factor ANOVA.

C.repeated two-sample test of means.

D.two-factor ANOVA with replication.

60.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.


Degrees of freedom for the between-groups sum of squares in the ANOVA would be:

A.11.

B.2.

C.4.

D.6.

62.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.

63.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.two-factor ANOVA.

C.three-factor ANOVA.

D.four-factor ANOVA.

64.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.depends on .


What are the degrees of freedom for the error sum of squares?

A.3

B.19

C.16

D.It depends on .

66.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.

67.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.


ANOVA Table


A.5.

B.4.

C.3.

D.impossible to ascertain from given.


ANOVA Table

The F statistic is:

A.2.88.

B.4.87.

C.5.93.

D.6.91.


ANOVA Table

The number of observations in the original sample was:

A.59.

B.60.

C.58.

D.54.


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.


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.


The MS (mean square) for the treatments is:

A.239.13.

B.106.88.

C.1,130.8.

D.impossible to ascertain from the information given.


The F statistic is:

A.4.87.

B.3.38.

C.5.93.

D.6.91.


The number of observations in the entire sample is:

A.20.

B.19.

C.22.


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.


Our decision about the hypothesis of equal treatment means is that the null hypothesis:

A.cannot be rejected at = .05.

B.can be rejected at = .05.

C.can be rejected for any typical value of .

D.cannot be assessed from the given information.

78.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, degrees of freedom for the between-groups sum of squares will be:

A.11.

B.3.

C.2.

D.9.


In a one-factor ANOVA, degrees of freedom for the within-groups sum of squares will be:

A.11.

B.3.

C.9.

D.2.


Degrees of freedom for the total sum of squares in a one-factor ANOVA would be:

A.11.

B.8.

C.2.

D.9.

81.Refer to the following MegaStat output (some information is missing). The sample size was n = 65 in a one-factor ANOVA.

At = .05, which is the critical value of the test statistic for a two-tailed test for a significant difference in means that are to be compared simultaneously? Note: This question requires a Tukey table.

A.2.81

B.2.54

C.2.33

D.1.96


Which pairs of days differ significantly? Note: This question requires access to a Tukey table.

A.(Mon, Thu) and (Mon, Wed) only.

B.(Mon, Wed) only.

C.(Mon, Thu) only.

D.(Mon, Thu) and (Mon, Wed) and (Mon, Fri) and (Mon, Tue).


At = .05, what is the critical value of the Tukey test statistic for a two-tailed test for a significant difference in means that are to be compared simultaneously? Note: This question requires access to a Tukey table.

A.2.07

B.2.80

C.2.76

D.1.96


Which pairs of meds differ at = .05? Note: This question requires access to a Tukey table.

A.Med 1, Med 2

B.Med 2, Med 4

C.Med 3, Med 4

D.None of them.

85.What is the .05 critical value of Hartley's test statistic for a one-factor ANOVA with n1 = 5, n2 = 8, n3 = 7, n4 = 8, n5 = 6, n6 = 8? Note: This question requires access to a Hartley table.

A.10.8

B.11.8

C.13.7

D.15.0

86.What is the .05 critical value of Tukey's test statistic for a one-factor ANOVA with n1 = 6, n2 = 6, n3 = 6? Note: This question requires access to a Tukey table.

A.3.67

B.2.60

C.3.58

D.2.75

87.What are the degrees of freedom for Hartley's test statistic for a one-factor ANOVA with n1 = 5, n2 = 8, n3 = 7, n4 = 8, n5 = 6, n6 = 8?

A.7, 6

B.6, 6

C.6, 41

88.What are the degrees of freedom for Tukey's test statistic for a one-factor ANOVA with n1 = 6, n2 = 6, n3 = 6?

A.3, 6

B.6, 3

C.6, 15

D.3, 15

89.After performing a one-factor ANOVA test, John noticed that the sample standard deviations for his four groups were, respectively, 33, 24, 73, and 35. John should:

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.

90.Which statement is incorrect?

A.We need a Tukey test because ANOVA doesn't 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.

91.Which is not an assumption of unreplicated two-factor ANOVA (randomized block)?

A.Normality of the population

B.Homogeneous variances

C.Additive treatment effects

D.There is factor interaction.

92.Which is correct concerning a two-factor unreplicated (randomized block) ANOVA?

A.No interaction effect is estimated.

B.The interaction effect would have its own F statistic.

C.The interaction would be insignificant unless the main effects were significant.

93.In a two-factor unreplicated (randomized block) ANOVA, what is the F statistic for the treatment effect given that SSA (treatments) = 216, SSB (block) = 126, SSE (error) = 18?

A.12

B.1.71

C.7

D.Can't tell without more information

94.Three bottles of wine are tasted by three experts. Each rater assigns a rating (scale is from 1 = terrible to 10 = superb). Which test would you use for the most obvious hypothesis?

A.t-test for independent means

B.One-factor ANOVA

C.Two-factor ANOVA without replication

D.Two-factor ANOVA with replication

95.To compare the cost of three shipping methods, a firm ships material to each of four different destinations over a six-month period. The average cost per shipment is shown below.

Which test would be appropriate?

A.Independent samples t-test

B.Two-factor ANOVA with replication

C.Dependent (paired-samples) t-test

D.Two-factor ANOVA without replication


For the appropriate type of ANOVA, total degrees of freedom would be:

A.11.

B.3.

C.4.

D.12.

97.Here is an Excel ANOVA table that summarizes the results of an experiment to assess the effects of ambient noise level and plant location on worker productivity. The test used = 0.05.

Is the effect of plant location significant at = .05?

A.Yes

B.No

C.Need more information to say


Is the effect of noise level significant at = .01?

A.Yes

B.No



The experimental design and ANOVA appear to be:

A.replicated two factor.

B.unreplicated two-factor.

C.impossible to determine.


The sample size is:

A.15.

B.10.

C.16.


101.At the Seymour Clinic, the number of patients seen by three doctors over three days is as follows:

This data set would call for:

A.two-factor ANOVA without replication.

B.two-factor ANOVA with replication.


D.five-factor ANOVA.


Degrees of freedom for the error sum of squares would be:

A.6.

B.14.

C.8.

D.15.

103.Here is an Excel ANOVA table for an experiment that analyzed factors that may affect patients' blood pressure (some information is missing).

The number of medication types is:

A.1.

B.2.

C.3.

D.4.

104.Here is an Excel ANOVA table for an experiment that analyzed two factors that may affect patients' blood pressure (some information is missing).

The number of patient age groups is:

A.1.

B.2.

C.3.

D.4.


The number of patients per replication is:

A.1.

B.2.

C.3.

D.4.


The overall sample size is:

A.7.

B.25.

C.32.

D.impossible to determine as given.


At = .05 the effect of medication type is:

A.significant.

B.insignificant.

C.borderline.


At = .01 the effect of patient age is:

A.very clearly significant.

B.just barely significant.



At = .10 the interaction is:

A.significant.

B.insignificant.

C.borderline.

110.Three randomly chosen pieces of four types of PVC pipe of equal wall thickness are tested to determine the burst strength (in pounds per square inch) under three temperature conditions, yielding the results shown below.


A.One-factor ANOVA



D.Two-factor ANOVA with no replication

111.Three randomly chosen pieces of four types of PVC pipe of equal wall thickness are tested to determine the burst strength (in pounds per square inch) under three temperature conditions, yielding the results shown below.

Total degrees of freedom for the ANOVA would be"

A.19.

B.12.

C.35.

D.59.

112.A firm is studying the effect of work shift and parts supplier on its defect rate (dependent variable is defects per 1000). The resulting ANOVA results are shown below (some information is missing).

How many suppliers were there?

A.1

B.2

C.3

D.4


How many replications per cell were there?

A.2

B.3

C.4

D.5


At = 0.01, the effect of supplier is:

A.clearly significant.


C.almost but not quite significant.



The number of observations was:

A.37.

B.45.

C.44.

D.40.


At = 0.01, the interaction effect is:

A.strongly significant.



117.A firm is concerned with variability in hourly output at several factories and shifts. Here are the results of an ANOVA using output per hour as the dependent variable (some information is missing).

The original data matrix has how many treatments (rows columns)?

A.4

B.6

C.3

D.8


The number of observations in each treatment cell (row-column intersection) is:

A.1.

B.2.

C.3.



At = 0.01 the effect of factory is:


B.clearly insignificant.

C.of borderline significance.


The p-value for the interaction effect is going to be:

A.very small (near 0).

B.very large (near 1).

C.impossible to knowcould be either large or small.

121.Sound engineers studied factors that might affect the output (in decibels) of a rock concert speaker system. The results of their ANOVA tests are shown (some information is missing).

Which is the number of amplifiers and positions tested?

A.1, 3

B.2, 4

C.3, 5

D.4, 1

122.Sound engineers studied factors that might affect the output (in decibels) of a rock concert speaker system. The results of their ANOVA tests are shown (some information is missing).

The number of observations per cell was:

A.1.

B.2.

C.3.

D.4.

123.Sound engineers studied factors that might affect the output (in decibels) of a rock concert speaker system. The desired level of significance was = .05. The results of their ANOVA tests are shown (some information is missing).

The most reasonable conclusion at = .05 about the three sources of variation (amplifier, position, and interaction) would be that their effects are:

A.significant, significant, insignificant.

B.insignificant, significant, significant.

C.very significant, almost significant, insignificant.

124.Sound engineers studied factors that might affect the output, in decibels, of a rock concert speaker system. The results of their ANOVA tests are shown (some information is missing).

The F statistic for amplifier was:

A.9.90.

B.10.16.

C.5.72.

D.4.27.

125.A multinational firm manufactures several types of 1280 1024 LCD displays in several locations. They designed a sampling experiment to analyze the number of pixels per screen that have significant color degradation after 52,560 hours (six years of continuous use) using accelerated life testing. The Excel ANOVA table for their experiment is shown below. Some table entries have been obscured. The response variable (Y) is the number of degraded pixels in a given display.

Degrees freedom for display type will be:

A.1.

B.4.

C.3.

D.5.


How many display types were there?

A.1

B.2

C.3

D.5


How many countries were studied?

A.1

B.2

C.3

D.4


The F statistic for display effect is:

A.1.78.

B.3.16.

C.2.39.

D.2.94.


At = .05, the interaction effect is:






The numerator degrees of freedom for the interaction test would be:

A.2.

B.4.

C.8.

D.16.

131.A veterinarian notes the age (months) at which dogs are brought to the clinic to be neutered.

What kind of test would be used?

A.One-factor ANOVA



D.Three-factor ANOVA with replication.

132.A veterinarian notes the age (months) at which dogs are brought in to the clinic to be neutered.

Numerator degrees of freedom for the ANOVA interaction test would be:

A.2.

B.3.

C.6.

D.can't tell.

133.A veterinarian notes the age (months) at which dogs are brought in to the clinic to be neutered.

Total degrees of freedom for a two-factor replicated ANOVA would be:

A.6.

B.14.

C.17.

D.11.


How many nozzle settings were observed?

A.3

B.2

C.1

D.Can't tell.


Degrees of freedom for pressure level would be:

A.2.

B.3.

C.4.

D.6.


Error degrees of freedom would be:

A.24.

B.15.

C.12.

D.13.


The overall sample size was:

A.24.

B.23.

C.22.

D.18.


How many pressure levels were observed?

A.4

B.3

C.2

D.1


At = .05, the critical F value for nozzle setting is:

A.4.71.

B.4.75.

C.3.68.

D.3.02.


The form of the original data matrix is:

A.3 1 table.

B.1 2 table.

C.4 3 table.

D.2 3 table.


The number of replications per treatment was:

A.4.

B.3.

C.2.

D.1.


At = 0.05, the effect of nozzle setting is:





143.As shown below, a hospital recorded the number of minutes spent in post-op recovery by three randomly chosen knee-surgery patients in each category, based on age and type of surgery. Which is the most appropriate test?

A.One-factor ANOVA

B.Two-factor ANOVA without replication

C.Two-factor ANOVA with replication

D.Rimsky-Korsakov test

144.Refer to the following partial ANOVA results from Excel (some information is missing). The response variable was Y = maximum amount of water pumped from wells (gallons per minute).

The degrees of freedom for age of well is:

A.2.

B.3.

C.4.

D.5.


The F statistic for depth of well is:

A.25.23.

B.25.78.

C.25.31.

D.25.06.


The MS for interaction is:

A.7.25.

B.8.17.

C.8.37.

D.9.28.


The MS for age of well is:

A.185.23.

B.179.26.

C.180.25.

D.182.33.

Short Answer Questions148.The table below shows raw data on air pollutant levels (micrograms of particulate per liter of air) sampled at four different randomly chosen times of day on three different freeways. State the most reasonable hypotheses. What test would a statistician probably use? How many total degrees of freedom? How many degrees of freedom for the treatment(s)? How many error degrees of freedom? Explain. Do not do the F-test.

149.The table below shows six random observations on the number of airline tickets booked on Orbitz per hour in the five months bracketing the summer travel season. State the most reasonable hypotheses. What test would a statistician probably use? How many total degrees of freedom? How many degrees of freedom for the treatment(s)? How many error degrees of freedom? Explain. Do not do the F-test.

150.What is GLM, and why do we need it?

Chapter 11 Analysis of Variance Answer Key

True / False Questions1.One-factor ANOVA is a procedure intended to compare the variances of c samples.FALSEANOVA compares several means (although its test statistic is based on variances).

AACSB: AnalyticBlooms: RememberDifficulty: 1 EasyLearning Objective: 11-01 Use basic ANOVA terminology correctly.Topic: Overview of ANOVA

2.Analysis of variance is a procedure intended to compare the means of c samples.TRUEAlthough its test statistic is based on variances, ANOVA compares several means.


3.If you have four factors (call them A, B, C, and D) in an ANOVA experiment with replication, you could have a maximum of four different two-factor interactions.FALSEThere could be six two-way interactions: AB, AC, AD, BC, BD, CD.

AACSB: AnalyticBlooms: UnderstandDifficulty: 3 HardLearning Objective: 11-11 Recognize the need for experimental design and GLM (optional).Topic: Higher-Order ANOVA Models (Optional)

4.Hartley's test measures the equality of the means for several groups.FALSEHartley's test is designed to detect unequal population variances.

AACSB: AnalyticBlooms: RememberDifficulty: 2 MediumLearning Objective: 11-08 Use Hartley's test for equal variances in c treatment groups.Topic: Tests for Homogeneity of Variances

5.Hartley's test is to check for unequal variances for c groups.TRUEUnequal population variances would violate an ANOVA assumption.


6.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 .FALSEMultiple two-sample t-tests from the same data set would inflate the overall .

AACSB: AnalyticBlooms: RememberDifficulty: 2 MediumLearning Objective: 11-07 Understand and perform Tukey's test for paired means.Topic: Multiple Comparisons

7.ANOVA assumes equal variances within each treatment group.TRUEANOVA checks for unequal means, while assuming homogeneous variances.

AACSB: AnalyticBlooms: RememberDifficulty: 1 EasyLearning Objective: 11-06 Explain the assumptions of ANOVA and why they are important.Topic: Overview of ANOVA

8.Three-factor ANOVA is required if we have three treatment groups (i.e., three data columns).FALSEIf there are only three columns of data, we only have one factor (with three treatments). The hypothesis is whether the three treatment group means are the same.

AACSB: AnalyticBlooms: UnderstandDifficulty: 2 MediumLearning Objective: 11-11 Recognize the need for experimental design and GLM (optional).Topic: Higher-Order ANOVA Models (Optional)

9.ANOVA assumes normal populations.TRUEPopulations are assumed to be normally distributed and to have equal variances.


10.Tukey's test compares pairs of treatment means in an ANOVA.TRUETukey's test is a follow-up to ANOVA to detect which pairs of means differ (if any).

AACSB: AnalyticBlooms: ApplyDifficulty: 2 MediumLearning Objective: 11-07 Understand and perform Tukey's test for paired means.Topic: Multiple Comparisons

11.Tukey's test is similar to a two-sample t-test except that it pools the variances for all c samples.TRUEThere is a strong analogy with the two-sample t-test, except that we pool all the variances.


12.Tukey's test is not needed if we have the overall F statistic for the ANOVA.FALSETukey's test is a follow-up to ANOVA to detect which pairs of means differ (if any).


13.Interaction plots that show crossing lines indicate likely interactions.TRUEInteraction plots provide an intuitive visual way of seeing possible interactions.

AACSB: AnalyticBlooms: RememberDifficulty: 2 MediumLearning Objective: 11-10 Interpret main effects and interaction effects in two-factor ANOVA.Topic: Two-Factor ANOVA with Replication (Full Factorial Model)

14.Interaction plots that show parallel lines would suggest interaction effects.FALSEInteraction plots that show crossing lines indicate likely interactions.


15.In a two-factor ANOVA with three columns and four rows, there can be more than two interaction effects.FALSEThere can only be one interaction (row column).


16.Sample sizes must be equal in one-factor ANOVA.FALSESample sizes often are equal by design, but it is not necessary.

AACSB: AnalyticBlooms: RememberDifficulty: 1 EasyLearning Objective: 11-02 Recognize from data format when one-factor ANOVA is appropriate.Topic: One-Factor ANOVA (Completely Randomized Model)

17.In a 34 randomized block (two-factor unreplicated) ANOVA, we have 12 treatment groups.TRUEEach row/column combination is a treatment group.

AACSB: AnalyticBlooms: UnderstandDifficulty: 2 MediumLearning Objective: 11-10 Interpret main effects and interaction effects in two-factor ANOVA.Topic: Two-Factor ANOVA without Replication (Randomized Block Model)

18.One-factor ANOVA with two groups is equivalent to a two-tailed t-test.TRUEThe p-values will be the same in either test as long as the t-test is two-tailed.

AACSB: AnalyticBlooms: UnderstandDifficulty: 2 MediumLearning Objective: 11-02 Recognize from data format when one-factor ANOVA is appropriate.Topic: One-Factor ANOVA (Completely Randomized Model)

19.One factor ANOVA stacked data for five groups will be arranged in five separate columns.FALSEOne column will contain the data, while a second column names the group.

AACSB: AnalyticBlooms: UnderstandDifficulty: 2 MediumLearning Objective: 11-02 Recognize from data format when one-factor ANOVA is appropriate.Topic: One-Factor ANOVA (Completely Randomized Model)

20.Hartley's test is the largest sample mean divided by the smallest sample mean.FALSEHartley's test statistic is the ratio of s2max to s2min.


21.Tukey's test for five groups would require 10 comparisons of means.TRUEThe number of possible comparisons is c(c - 1)/2 = 5(4)/2 = 10.


22.ANOVA is robust to violations of the equal-variance assumption as long as group sizes are equal.TRUEStudies suggest that equal group sizes strengthen the ANOVA test.

AACSB: AnalyticBlooms: RememberDifficulty: 2 MediumLearning Objective: 11-06 Explain the assumptions of ANOVA and why they are important.Topic: One-Factor ANOVA (Completely Randomized Model)

23.Levene's test for homogeneity of variance is attractive because it does not depend on the assumption of normality.TRUEWhile Hartley's test is sensitive to nonnormality, Levene's test statistic is not.

AACSB: AnalyticBlooms: RememberDifficulty: 2 MediumLearning Objective: 11-06 Explain the assumptions of ANOVA and why they are important.Topic: Tests for Homogeneity of Variances

24.Tukey's test with seven groups would entail 21 comparisons of means.TRUEThe number of possible comparisons is c(c - 1)/2 = 7(6)/2 = 21.


25.Tukey's test pools all the sample variances.TRUEIn a Tukey test, all c sample variances are combined (weighted by their degrees of freedom).

AACSB: AnalyticBlooms: RememberDifficulty: 1 EasyLearning Objective: 11-07 Understand and perform Tukey's test for paired means.Topic: Multiple Comparisons

26.It is desirable, but not necessary, that sample sizes be equal in a one-factor ANOVA.TRUEStudies suggest that equal group sizes strengthen the ANOVA test.


Multiple Choice Questions27.Which is the Excel function to find the critical value of F for = .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)

The equivalent Excel 2007 function would be =FINV(.05, 3, 25).

AACSB: TechnologyBlooms: ApplyDifficulty: 2 MediumLearning Objective: 11-05 Use a table or Excel to find critical values for the F distribution.Topic: One-Factor ANOVA (Completely Randomized Model)

28.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)

The equivalent Excel 2007 function would be =FDIST(.05, 3, 25).

AACSB: TechnologyBlooms: ApplyDifficulty: 3 HardLearning Objective: 11-04 Use Excel or other software for ANOVA calculations.Topic: One-Factor ANOVA (Completely Randomized Model)

29.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.

Variation within groups is also called error variance or unexplained variance.


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.

It is desirable, but not necessary, that sample sizes be equal in a one-factor ANOVA.


31.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.

If each group mean equals the overall mean, then Fcalc could be zero (an unusual situation).

AACSB: AnalyticBlooms: RememberDifficulty: 2 MediumLearning Objective: 11-03 Interpret sums of squares and calculations in an ANOVA table.Topic: One-Factor ANOVA (Completely Randomized Model)

32.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.

Although its test statistic is based on variances, ANOVA compares several means.


33.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.

Although its test statistic is based on variances, ANOVA compares several means.


34.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.

The F distribution is the ratio of two mean squares, so it cannot be negative.

AACSB: AnalyticBlooms: RememberDifficulty: 2 MediumLearning Objective: 11-04 Use Excel or other software for ANOVA calculations.Topic: One-Factor ANOVA (Completely Randomized Model)

35.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.

The error variance or unexplained variance is variation within groups.

AACSB: AnalyticBlooms: RememberDifficulty: 1 EasyLearning Objective: 11-03 Interpret sums of squares and calculations in an ANOVA table.Topic: One-Factor ANOVA (Completely Randomized Model)

36.To test the null hypothesis H0: 1 = 2 = 3 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.

As long as the variances are equal, we can safely use ANOVA.


37.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.

Population variances are almost never known.


38.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.

The F distribution is the ratio of two mean squares, so it cannot be negative.

AACSB: AnalyticBlooms: RememberDifficulty: 1 EasyLearning Objective: 11-03 Interpret sums of squares and calculations in an ANOVA table.Topic: One-Factor ANOVA (Completely Randomized Model)

39.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.

For between-group variation, we have dfnumerator = c - 1 = 3 - 1 = 2.

AACSB: AnalyticBlooms: ApplyDifficulty: 2 MediumLearning Objective: 11-03 Interpret sums of squares and calculations in an ANOVA table.Topic: One-Factor ANOVA (Completely Randomized Model)

40.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.

For between group variation we have dfnumerator = c - 1 = 4 - 1 = 3.


41.Using one-factor ANOVA with 30 observations we find at = .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.

With more degrees of freedom, the critical value F.05 will be smaller, so we might reject.


42.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.

Studies suggest that equal group sizes strengthen the power of the ANOVA test.

AACSB: AnalyticBlooms: ApplyDifficulty: 2 MediumLearning Objective: 11-06 Explain the assumptions of ANOVA and why they are important.Topic: One-Factor ANOVA (Completely Randomized Model)

43.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.

The basic identify is SSbetween + SSwithin = SStotal.

AACSB: AnalyticBlooms: UnderstandDifficulty: 1 EasyLearning Objective: 11-03 Interpret sums of squares and calculations in an ANOVA table.Topic: One-Factor ANOVA (Completely Randomized Model)

44.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.

Variation within groups is also called error variance or unexplained variance.

AACSB: AnalyticBlooms: UnderstandDifficulty: 2 MediumLearning Objective: 11-03 Interpret sums of squares and calculations in an ANOVA table.Topic: One-Factor ANOVA (Completely Randomized Model)

45.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

MStreatment = 744/4 = 186, MSerror = (751.5)/15 = 50.1, so F = 186/50.1 = 3.71.


46.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

For df = (4, 15) we use Appendix F to get F.05 = 3.06.

AACSB: AnalyticBlooms: ApplyDifficulty: 2 MediumLearning Objective: 11-05 Use a table or Excel to find critical values for the F distribution.Topic: One-Factor ANOVA (Completely Randomized Model)

47.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

Since SS/df = MS, we know that df = SS/MS. Hence, 993/331 = 3 and 1002/50.1 = 20.


48.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

Since SS/df = MS, we know that df = SS/MS and, hence, 654/218 = 3 = c - 1.


49.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

Fcalc = (MStreatment)/(MSerror) = 218/128 = 1.703.



The F-test statistic is:

A.2.84.

B.3.56.

C.2.80.

D.2.79.

Fcalc = (MSbetween)/(MSwithin) = (210.2788)/(74.15) = 2.836.



Degrees of freedom for between groups variation are:

A.3.

B.4.

C.5.

D.Can't tell from given information.

SSbetween = 2113.833 - 1483 = 630.833, so df = (630.833)/(210.2778) = 3.

AACSB: AnalyticBlooms: ApplyDifficulty: 3 HardLearning Objective: 11-03 Interpret sums of squares and calculations in an ANOVA table.Topic: One-Factor ANOVA (Completely Randomized Model)


SS for between groups variation will be:

A.129.99.

B.630.83.

C.1233.4.

D.Can't tell.

SSbetween = 2113.833 - 1483 = 630.833.

AACSB: AnalyticBlooms: ApplyDifficulty: 1 EasyLearning Objective: 11-03 Interpret sums of squares and calculations in an ANOVA table.Topic: One-Factor ANOVA (Completely Randomized Model)



A.4.

B.3.

C.2.

D.1.

SSbetween = 2113.833 - 1483 = 630.833, so df = (630.833)/(210.2778) = 3 = c - 1.



The sample size is:

A.20.

B.23.

C.24.

D.21.

(630.833)/(210.2778) = 3 and (1483)/(74.15) = 20, so 3 + 20 = 23 = n - 1.



Assuming equal group sizes, the number of observations in each group is:

A.2.

B.3.

C.4.

D.6.

(630.833)/(210.2778) = 3 and (1483)/(74.15) = 20, so 3 + 20 = 23 = n - 1 and n/c = 24/4 = 6.



Degrees of freedom for the F-test are:

A.5, 22.

B.4, 21.

C.3, 20.


(630.833)/(210.2778) = 3 and (1483)/(74.15) = 20.



The critical value of F at = 0.05 is:

A.1.645.

B.2.84.

C.3.10.

D.4.28.

(630.833)/(210.2778) = 3 and (1483)/(74.15) = 20, so F.05 = 3.10 for df = (3, 20).

AACSB: AnalyticBlooms: ApplyDifficulty: 3 HardLearning Objective: 11-05 Use a table or Excel to find critical values for the F distribution.Topic: One-Factor ANOVA (Completely Randomized Model)


At = 0.05, the difference between group means is:


B.barely significant.



The p-value is not less than .05 so we cannot reject the hypothesis of equal means.



The test to use to compare the means for all three groups would require:

A.three-factor ANOVA.

B.one-factor ANOVA.

C.repeated two-sample test of means.

D.two-factor ANOVA with replication.

One factor (three group means to be compared).

AACSB: AnalyticBlooms: ApplyDifficulty: 1 EasyLearning Objective: 11-02 Recognize from data format when one-factor ANOVA is appropriate.Topic: One-Factor ANOVA (Completely Randomized Model)

60.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.

Error df = n - c = 9 - 3 = 6.



Degrees of freedom for the between-groups sum of squares in the ANOVA would be:

A.11.

B.2.

C.4.

D.6.

Between groups df = c - 1= 3 - 1 = 2.


62.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.

As there are only two groups, either ANOVA or a two-tailed t-test will give the same p-value.

AACSB: AnalyticBlooms: ApplyDifficulty: 2 MediumLearning Objective: 11-02 Recognize from data format when one-factor ANOVA is appropriate.Topic: One-Factor ANOVA (Completely Randomized Model)

63.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.two-factor ANOVA.


D.four-factor ANOVA.

One factor (four group means to be compared).

AACSB: AnalyticBlooms: ApplyDifficulty: 1 EasyLearning Objective: 11-02 Recognize from data format when one-factor ANOVA is appropriate.Topic: One-Factor ANOVA (Completely Randomized Model)


Degrees of freedom for the between-treatments sum of squares would be:

A.3.

B.19.

C.17.

D.depends on .

Between-reatments df = c - 1 = 4 - 1 = 3.



What are the degrees of freedom for the error sum of squares?

A.3

B.19

C.16

D.It depends on .

Error df = n - c = 20 - 4 = 16.


66.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.

Fcalc = (154.1)/(34.6) = 4.45.


67.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.

Hartley's H = s2max/s2min = (8.944)2/(3.983)2 = 5.04.

AACSB: AnalyticBlooms: ApplyDifficulty: 2 MediumLearning Objective: 11-08 Use Hartley's test for equal variances in c treatment groups.Topic: Tests for Homogeneity of Variances


ANOVA Table


A.5.

B.4.

C.3.

D.impossible to ascertain from given.

59 - 55 = 4 = c - 1, so c = 5



ANOVA Table

The F statistic is:

A.2.88.

B.4.87.

C.5.93.

D.6.91.

Fcalc = 11,189/1619 = 6.91.



ANOVA Table

The number of observations in the original sample was:

A.59.

B.60.

C.58.

D.54.

n - 1 = 59, so n = 60.



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.

Treatment df = 59 - 55 = 4, so F.05 = 2.56 using df = (4, 50) in Appendix F.



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.

Fcalc = 11,189/1619 = 6.91 while F.05 = 2.56 using df = (4, 50) in Appendix F.



The MS (mean square) for the treatments is:

A.239.13.

B.106.88.

C.1,130.8.

D.impossible to ascertain from the information given.

(717.4)/3 = 239.133.



The F statistic is:

A.4.87.

B.3.38.

C.5.93.

D.6.91.

Between-groups MS = (717.4)/3 = 239.133, so Fcalc = (239.133)/(70.675) = 3.383.



The number of observations in the entire sample is:

A.20.

B.19.

C.22.

n - 1 = 19, so n = 20.



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.

Error df = 19 - 3 = 16, so F.05 = 3.24 using df = (3, 16) in Appendix F.

AACSB: AnalyticBlooms: ApplyDifficulty: 2 MediumLearning Objective: 11-05 Use a table or Excel to find critical values for the F distribution.Topic: One-Factor ANOVA (Completely Randomized Model)


Our decision about the hypothesis of equal treatment means is that the null hypothesis:

A.cannot be rejected at = .05.

B.can be rejected at = .05.

C.can be rejected for any typical value of .

D.cannot be assessed from the given information.

The p-value is less than .05, so we conclude unequal population means.



In a one-factor ANOVA, degrees of freedom for the between-groups sum of squares will be:

A.11.

B.3.

C.2.

D.9.

Between-groups df = c - 1 = 3 - 1 = 2.



In a one-factor ANOVA, degrees of freedom for the within-groups sum of squares will be:

A.11.

B.3.

C.9.

D.2.

Within-groups df = n - c = 12 - 3 = 9.



Degrees of freedom for the total sum of squares in a one-factor ANOVA would be:

A.11.

B.8.

C.2.

D.9.

Total df = n - 1 = 12 - 1 = 11.



At = .05, which is the critical value of the test statistic for a two-tailed test for a significant difference in means that are to be compared simultaneously? Note: This question requires a Tukey table.

A.2.81

B.2.54

C.2.33

D.1.96

T.05 = 2.81 for df = (c, n - c) with c = 5 and n = 65.

AACSB: AnalyticBlooms: ApplyDifficulty: 3 HardLearning Objective: 11-07 Understand and perform Tukey's test for paired means.Topic: Multiple Comparisons


Which pairs of days differ significantly? Note: This question requires access to a Tukey table.

A.(Mon, Thu) and (Mon, Wed) only.

B.(Mon, Wed) only.

C.(Mon, Thu) only.

D.(Mon, Thu) and (Mon, Wed) and (Mon, Fri) and (Mon, Tue).

Use T.05 = 2.81 for df = (c, n - c) with c = 5 and n = 65.

AACSB: AnalyticBlooms: ApplyDifficulty: 3 HardLearning Objective: 11-07 Understand and perform Tukey's test for paired means.Topic: Multiple Comparisons


At = .05, what is the critical value of the Tukey test statistic for a two-tailed test for a significant difference in means that are to be compared simultaneously? Note: This question requires access to a Tukey table.

A.2.07

B.2.80

C.2.76

D.1.96




Which pairs of meds differ at = .05? Note: This question requires access to a Tukey table.

A.Med 1, Med 2

B.Med 2, Med 4

C.Med 3, Med 4

D.None of them.



85.What is the .05 critical value of Hartley's test statistic for a one-factor ANOVA with n1 = 5, n2 = 8, n3 = 7, n4 = 8, n5 = 6, n6 = 8? Note: This question requires access to a Hartley table.

A.10.8

B.11.8

C.13.7

D.15.0

H.05 = 13.7 for df = (c, (n/c) - 1) where c = 6 and n = 42, so we use df = (6, 6).


86.What is the .05 critical value of Tukey's test statistic for a one-factor ANOVA with n1 = 6, n2 = 6, n3 = 6? Note: This question requires access to a Tukey table.

A.3.67

B.2.60

C.3.58

D.2.75



87.What are the degrees of freedom for Hartley's test statistic for a one-factor ANOVA with n1 = 5, n2 = 8, n3 = 7, n4 = 8, n5 = 6, n6 = 8?

A.7, 6

B.6, 6

C.6, 41

Use df = (c, (n/c) - 1) where c = 6 and n = 42, or df = (6, 6).


88.What are the degrees of freedom for Tukey's test statistic for a one-factor ANOVA with n1 = 6, n2 = 6, n3 = 6?

A.3, 6

B.6, 3

C.6, 15

D.3, 15

Use df = (c, n - c) with c = 3 and n = 18, or df = (3, 15).


89.After performing a one-factor ANOVA test, John noticed that the sample standard deviations for his four groups were, respectively, 33, 24, 73, and 35. John should:

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.

The unusually large standard deviation for group 3 suggests unequal variances.


90.Which statement is incorrect?

A.We need a Tukey test because ANOVA doesn't 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.

Hartley's test compares variances (not means).


91.Which is not an assumption of unreplicated two-factor ANOVA (randomized block)?

A.Normality of the population

B.Homogeneous variances

C.Additive treatment effects

D.There is factor interaction.

The usual assumptions apply to a two-factor ANOVA (but no interaction estimate is possible without replication).

AACSB: AnalyticBlooms: RememberDifficulty: 2 MediumLearning Objective: 11-06 Explain the assumptions of ANOVA and why they are important.Topic: Two-Factor ANOVA without Replication (Randomized Block Model)

92.Which is correct concerning a two-factor unreplicated (randomized block) ANOVA?

A.No interaction effect is estimated.

B.The interaction effect would have its own F statistic.

C.The interaction would be insignificant unless the main effects were significant.

We cannot estimate the interaction effect without replication in a two-factor ANOVA.

AACSB: AnalyticBlooms: RememberDifficulty: 2 MediumLearning Objective: 11-10 Interpret main effects and interaction effects in two-factor ANOVA.Topic: Two-Factor ANOVA without Replication (Randomized Block Model)

93.In a two-factor unreplicated (randomized block) ANOVA, what is the F statistic for the treatment effect given that SSA (treatments) = 216, SSB (block) = 126, SSE (error) = 18?

A.12

B.1.71

C.7

D.Can't tell without more information

We cannot calculate the mean squares without knowing r, c, and n, so no F statistics.

AACSB: AnalyticBlooms: ApplyDifficulty: 2 MediumLearning Objective: 11-10 Interpret main effects and interaction effects in two-factor ANOVA.Topic: Two-Factor ANOVA without Replication (Randomized Block Model)

94.Three bottles of wine are tasted by three experts. Each rater assigns a rating (scale is from 1 = terrible to 10 = superb). Which test would you use for the most obvious hypothesis?

A.t-test for independent means

B.One-factor ANOVA


D.Two-factor ANOVA with replication

Only one observation per row/column cell (two factors but no replication).

AACSB: AnalyticBlooms: ApplyDifficulty: 1 EasyLearning Objective: 11-09 Recognize from data format when two-factor ANOVA is needed.Topic: Two-Factor ANOVA without Replication (Randomized Block Model)



A.Independent samples t-test



D.Two-factor ANOVA without replication




For the appropriate type of ANOVA, total degrees of freedom would be:

A.11.

B.3.

C.4.

D.12.

df = n - 1 = 12 - 1 = 11.



Is the effect of plant location significant at = .05?

A.Yes

B.No


The p-value is not less than .05, so plant location has no significant effect.



Is the effect of noise level significant at = .01?

A.Yes

B.No


The p-value is much less than .05, so noise level has a significant effect.



The experimental design and ANOVA appear to be:

A.replicated two factor.

B.unreplicated two-factor.

C.impossible to determine.

The absence of an interaction suggests an unreplicated two-factor model.



The sample size is:

A.15.

B.10.

C.16.


For unreplicated two-factor ANOVA, total df = 3 + 3 + 9 = 15 = n - 1, so n = 16.



This data set would call for:

A.two-factor ANOVA without replication.

B.two-factor ANOVA with replication.


D.five-factor ANOVA.




Degrees of freedom for the error sum of squares would be:

A.6.

B.14.

C.8.

D.15.

For unreplicated two-factor ANOVA, the error df = (r - 1)(c - 1) = (5 - 1)(3 - 1) = 8.

AACSB: AnalyticBlooms: ApplyDifficulty: 3 HardLearning Objective: 11-09 Recognize from data format when two-factor ANOVA is needed.Topic: Two-Factor ANOVA without Replication (Randomized Block Model)

103.Here is an Excel ANOVA table for an experiment that analyzed factors that may affect patients' blood pressure (some information is missing).

The number of medication types is:

A.1.

B.2.

C.3.

D.4.

df = 1 = (number of medications - 1), so there were 2 medications.

AACSB: AnalyticBlooms: ApplyDifficulty: 2 MediumLearning Objective: 11-10 Interpret main effects and interaction effects in two-factor ANOVA.Topic: Two-Factor ANOVA with Replication (Full Factorial Model)


The number of patient age groups is:

A.1.

B.2.

C.3.

D.4.

For patient age group, df = (25.0938)/(8.3646) = 3 = c - 1 (so 4 age groups).

AACSB: AnalyticBlooms: ApplyDifficulty: 3 HardLearning Objective: 11-10 Interpret main effects and interaction effects in two-factor ANOVA.Topic: Two-Factor ANOVA with Replication (Full Factorial Model)


The number of patients per replication is:

A.1.

B.2.

C.3.

D.4.

c - 1 = (25.0938)/(8.3646) = 3

Chap 011

Documents

anova test

factor anova

assumption of anova

c groups

anova treatments

anova experiment

hartleys test

sample ttest