Analysis Of Variance Sample Assignment

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1. In a one-way ANOVA, if the computed F statistic exceeds the critical F value we may

reject H₀ since there is evidence all the means differ.
reject H₀ since there is evidence of a treatment effect.
not reject H₀ since there is no evidence of a difference.
not reject H₀ because a mistake has been made.

ANSWER:

TYPE: MC DIFFICULTY: Easy

KEYWORDS: one-way analysis of variance, decision

2. Which of the following components in an ANOVA table are not additive?

Sum of squares.
Degrees of freedom.
Mean squares.
It is not possible to tell.

ANSWER:

TYPE: MC DIFFICULTY: Easy

KEYWORDS: one-way analysis of variance, sum of squares, properties

3. Why would you use the Tukey-Kramer procedure?

To test for normality.
To test for homogeneity of variance.
To test independence of errors.
To test for differences in pairwise means.

ANSWER:

TYPE: MC DIFFICULTY: Easy

KEYWORDS: Tukey-Kramer procedure

4. A completely randomized design

has only one factor with several treatment groups.
can have more than one factor, each with several treatment groups.
has one factor and one block.
has one factor and one block and multiple values.

ANSWER:

TYPE: MC DIFFICULTY: Moderate

KEYWORDS: completely randomized design

5. The F test statistic in a one-way ANOVA is

MSW/MSA.
SSW/SSA.
MSA/MSW.
SSA/SSW.

ANSWER:

TYPE: MC DIFFICULTY: Easy

KEYWORDS: one-way analysis of variance, F test for factor

6. The degrees of freedom for the F test in a one-way ANOVA are

(n – c) and (c – 1).
(c – 1) and (n – c).
(c – n) and (n – 1).
(n – 1) and (c – n).

ANSWER:

TYPE: MC DIFFICULTY: Easy

KEYWORDS: one-way analysis of variance, degrees of freedom

7. In a one-way ANOVA, the null hypothesis is always

there is no treatment effect.
there is some treatment effect.
all the population means are different.
some of the population means are different.

ANSWER:

TYPE: MC DIFFICULTY: Easy

KEYWORDS: one-way analysis of variance, form of hypothesis

8. In a one-way ANOVA

an interaction term is present.
an interaction effect can be tested.
there is no interaction term.
the interaction term has (c – 1)(n – 1) degrees of freedom.

ANSWER:

TYPE: MC DIFFICULTY: Moderate

KEYWORDS: one-way analysis of variance, properties, interaction

9. Interaction in an experimental design can be tested in

a completely randomized model.
a randomized block model.
a two-factor model.
all ANOVA models.

ANSWER:

TYPE: MC DIFFICULTY: Easy

KEYWORDS: two-factor analysis of variance, interaction, properties

10. In a two-way ANOVA the degrees of freedom for the interaction term is

(r – 1)(c – 1).
rc(n – 1).
(r – 1).
rcn + 1.

ANSWER:

TYPE: MC DIFFICULTY: Easy

KEYWORDS: two-factor analysis of variance, interaction, degrees of freedom

11. In a two-way ANOVA the degrees of freedom for the "error" term is

(r – 1)(c – 1).
rc(n’ – 1).
(r – 1).
rcn' + 1.

ANSWER:

TYPE: MC DIFFICULTY: Easy

KEYWORDS: two-factor analysis of variance, degrees of freedom

TABLE 11-1

Psychologists have found that people are generally reluctant to transmit bad news to their peers. This phenomenon has been termed the “MUM effect.” To investigate the cause of the MUM effect, 40 undergraduates at Duke University participated in an experiment. Each subject was asked to administer an IQ test to another student and then provide the test taker with his or her percentile score. Unknown to the subject, the test taker was a bogus student who was working with the researchers. The experimenters manipulated two factors: subject visibility and success of test taker, each at two levels. Subject visibility was either visible or not visible to the test taker. Success of the test taker was either top 20% or bottom20%. Ten subjects were randomly assigned to each of the 2 x 2 = 4 experimental conditions, then the time (in seconds) between the end of the test and the delivery of the percentile score from the subject to the test taker was measured. (This variable is called the latency to feedback.) The data were subjected to appropriate analyses with the following results.

Source	df	SS	MS	F	PR > F
Subject visibility	1	1380.24	1380.24	4.26	0.043
Test taker success	1	1325.16	1325.16	4.09	0.050
Interaction	1	3385.80	3385.80	10.45	0.002
Error	36	11,664.00	324.00
Total	39	17,755.20

12. Referring to Table 11-1, what type of experimental design was employed in this study?

Completely randomized design with 4 treatments
Randomized block design with four treatments and 10 blocks
2 x 2 factorial design with 10 replications
None of the above

ANSWER:

TYPE: MC DIFFICULTY: Moderate

KEYWORDS: two-factor analysis of variance, two-factor factorial design

13. Referring to Table 11-1, at the 0.01 level, what conclusions can you draw from the analyses?

At the 0.01 level, subject visibility and test taker success are significant predictors of latency feedback.
At the 0.01 level, the model is not useful for predicting latency to feedback.
At the 0.01 level, there is evidence to indicate that subject visibility and test taker success interact.
At the 0.01 level, there is no evidence of interaction between subject visibility and test taker success.

ANSWER:

TYPE: MC DIFFICULTY: Moderate

KEYWORDS: two-factor analysis of variance, F test for interaction, decision, conclusion, interaction

14. Referring to Table 11-1, in the context of this study, interpret the statement: “Subject visibility and test taker success interact.”

The difference between the mean feedback time for visible and nonvisible subjects depends on the success of the test taker.
The difference between the mean feedback time for test takers scoring in the top 20% and bottom 20% depends on the visibility of the subject.
The relationship between feedback time and subject visibility depends on the success of the test taker.
All of the above are correct interpretations.

ANSWER:

TYPE: MC DIFFICULTY: Difficult

KEYWORDS: two-factor analysis of variance, interaction, conclusion

15. An airline wants to select a computer software package for its reservation system. Four software packages (1, 2, 3, and 4) are commercially available. The airline will choose the package that bumps as few passengers, on the average, as possible during a month. An experiment is set up in which each package is used to make reservations for 5 randomly selected weeks. (A total of 20 weeks was included in the experiment.) The number of passengers bumped each week is given below. How should the data be analyzed?

Package 1: 12, 14, 9, 11, 16

Package 2: 2, 4, 7, 3, 1

Package 3: 10, 9, 6, 10, 12

Package 4: 7, 6, 6, 15, 12

F test for differences in variances.
One-way ANOVA F
t test for the differences in means.
t test for the mean difference.

ANSWER:

TYPE: MC DIFFICULTY: Easy

KEYWORDS: one-way analysis of variance, F test for factor

TABLE 11-2

An airline wants to select a computer software package for its reservation system. Four software packages (1, 2, 3, and 4) are commercially available. The airline will choose the package that bumps as few passengers, on the average, as possible during a month. An experiment is set up in which each package is used to make reservations for 5 randomly selected weeks. (A total of 20 weeks was included in the experiment.) The number of passengers bumped each week is obtained, which gives rise to the following Excel output:

ANOVA
Source of Variation	SS	df	MS	F	P-value	F crit
Between Groups	212.4	3	8.304985	0.001474	3.238867
Within Groups	136.4	8.525
Total	348.8

16. Referring to Table 11-2, the within groups degrees of freedom is

ANSWER:

TYPE: MC DIFFICULTY: Easy

KEYWORDS: one-way analysis of variance, degrees of freedom

17. Referring to Table 11-2, the total degrees of freedom is

ANSWER:

TYPE: MC DIFFICULTY: Easy

KEYWORDS: one-way analysis of variance, degrees of freedom

18. Referring to Table 11-2, the among-group (between-group) mean squares is

ANSWER:

TYPE: MC DIFFICULTY: Easy

KEYWORDS: one-way analysis of variance, mean squares

19. Referring to Table 11-2, at a significance level of 1%,

there is insufficient evidence to conclude that the average numbers of customers bumped by the 4 packages are not all the same.
there is insufficient evidence to conclude that the average numbers of customers bumped by the 4 packages are all the same.
there is sufficient evidence to conclude that the average numbers of customers bumped by the 4 packages are not all the same.
there is sufficient evidence to conclude that the average numbers of customers bumped by the 4 packages are all the same.

ANSWER:

TYPE: MC DIFFICULTY: Easy

KEYWORDS: one-way analysis of variance, F test for factor, decision, conclusion

TABLE 11-3

A realtor wants to compare the average sales-to-appraisal ratios of residential properties sold in four neighborhoods (A, B, C, and D). Four properties are randomly selected from each neighborhood and the ratios recorded for each, as shown below.

A: 1.2, 1.1, 0.9, 0.4 C: 1.0, 1.5, 1.1, 1.3

B: 2.5, 2.1, 1.9, 1.6 D: 0.8, 1.3, 1.1, 0.7

Interpret the results of the analysis summarized in the following table:

Source	df	SS	MS	F	PR > F
Neighborhoods	3.1819	1.0606	10.76	0.001
Error	12
Total	4.3644

20. Referring to Table 11-3, the among group degrees of freedom is

ANSWER:

TYPE: MC DIFFICULTY: Easy

KEYWORDS: one-way analysis of variance, degrees of freedom

21. Referring to Table 11-3, the within group sum of squares is

0606
1825
1819
3644

ANSWER:

TYPE: MC DIFFICULTY: Easy

KEYWORDS: one-way analysis of variance, sum of squares

22. Referring to Table 11-3, the within group mean squares is

ANSWER:

TYPE: MC DIFFICULTY: Easy

KEYWORDS: one-way analysis of variance, mean squares

23. Referring to Table 11-3,

at the 0.05 level of significance, the mean ratios for the 4 neighborhoods are not all the same.
at the 0.01 level of significance, the mean ratios for the 4 neighborhoods are all the same.
at the 0.10 level of significance, the mean ratios for the 4 neighborhoods are not significantly different.
at the 0.05 level of significance, the mean ratios for the 4 neighborhoods are not significantly different from 0.

ANSWER:

TYPE: MC DIFFICULTY: Easy

KEYWORDS: one-way analysis of variance, F test for factor, decision, conclusion

24. Referring to Table 11-3, the null hypothesis for Levene’s test for homogeneity of variances is

ANSWER:

TYPE: MC DIFFICULTY: Easy

KEYWORDS: Levene’s test, form of hypothesis

25. Referring to Table 11-3, the value of the test statistic for Levene’s test for homogeneity of variances is

ANSWER:

TYPE: MC DIFFICULTY: Moderate

KEYWORDS: Levene’s test, test statistic

26. Referring to Table 11-3, the numerator and denominator degrees of freedom for Levene’s test for homogeneity of variances at a 5% level of significance are, respective,

3, 12
12, 3
3, 15
15, 3

ANSWER:

TYPE: MC DIFFICULTY: Moderate

KEYWORDS: Levene’s test, degrees of freedom

27. Referring to Table 11-3, the critical value of Levene’s test for homogeneity of variances at a 5% level of significance is

ANSWER:

TYPE: MC DIFFICULTY: Moderate

KEYWORDS: Levene’s test, critical value

28. Referring to Table 11-3, the p-value of the test statistic for Levene’s test for homogeneity of variances is

ANSWER:

TYPE: MC DIFFICULTY: Moderate

KEYWORDS: Levene’s test, p-value

29. Referring to Table 11-3, what should be the decision for the Levene’s test for homogeneity of variances at a 5% level of significance?

Reject the null hypothesis because the p-value is smaller than the level of significance.
Reject the null hypothesis because the p-value is larger than the level of significance.
Do not eject the null hypothesis because the p-value is smaller than the level of significance.
Do not reject the null hypothesis because the p-value is larger than the level of significance.

ANSWER:

TYPE: MC DIFFICULTY: Easy

KEYWORDS: Levene’s test, decision

30. Referring to Table 11-3, what should be the conclusion for the Levene’s test for homogeneity of variances at a 5% level of significance?

There is not sufficient evidence that the variances are all the same.
There is sufficient evidence that the variances are all the same.
There is not sufficient evidence that the variances are not all the same.
There is sufficient evidence that the variances are not all the same.

ANSWER:

TYPE: MC DIFFICULTY: Easy

KEYWORDS: Levene’s test, conclusion

31. A campus researcher wanted to investigate the factors that affect visitor travel time in a complex, multilevel building on campus. Specifically, he wanted to determine whether different building signs (building maps versus wall signage) affect the total amount of time visitors require to reach their destination and whether that time depends on whether the starting location is inside or outside the building. Three subjects were assigned to each of the combinations of signs and starting locations, and travel time in seconds from beginning to destination was recorded. How should the data be analyzed?

Starting Room
Interior	Exterior
Wall Signs	141, 119, 238	224, 339, 139
Map	85, 94, 126	226, 129, 130

Completely randomized design
Randomized block design
2 x 2 factorial design
Levene’s test

ANSWER:

TYPE: MC DIFFICULTY: Moderate

KEYWORDS: two-factor factorial design

TABLE 11-4

A campus researcher wanted to investigate the factors that affect visitor travel time in a complex, multilevel building on campus. Specifically, he wanted to determine whether different building signs (building maps versus wall signage) affect the total amount of time visitors require to reach their destination and whether that time depends on whether the starting location is inside or outside the building. Three subjects were assigned to each of the combinations of signs and starting locations, and travel time in seconds from beginning to destination was recorded. An Excel output of the appropriate analysis is given below:

ANOVA
Source of Variation	SS	df	MS	F	P-value	F crit
Signs	14008.33	14008.33	0.11267	5.317645
Starting Location	12288	2.784395	0.13374	5.317645
Interaction	48	48	0.919506	5.317645
Within	35305.33	4413.167
Total	61649.67	11

32. Referring to Table 11-4, the degrees of freedom for the different building signs (factor A) is

ANSWER:

TYPE: MC DIFFICULTY: Easy

KEYWORDS: two-factor analysis of variance, degrees of freedom

33. Referring to Table 11-4, the within (error) degrees of freedom is

ANSWER:

TYPE: MC DIFFICULTY: Easy

KEYWORDS: two-factor analysis of variance, degrees of freedom

34. Referring to Table 11-4, the mean squares for starting location (factor B) is

48
4,413.17
12,288
14,008.3

ANSWER:

TYPE: MC DIFFICULTY: Easy

KEYWORDS: two-factor analysis of variance, mean squares

35. Referring to Table 11-4, the F test statistic for testing the main effect of types of signs is

0109
7844
1742
3176

ANSWER:

TYPE: MC DIFFICULTY: Easy

KEYWORDS: two-factor analysis of variance, F test for factor

36. Referring to Table 11-4, the F test statistic for testing the interaction effect between the types of signs and the starting location is

0109
7844
1742
3176

ANSWER:

TYPE: MC DIFFICULTY: Easy

KEYWORDS: two-factor analysis of variance, F test for interaction, interaction

37. Referring to Table 11-4, at 1% level of significance,

there is insufficient evidence to conclude that the difference between the average traveling time for the different starting locations depends on the types of signs.
there is insufficient evidence to conclude that the difference between the average traveling time for the different types of signs depends on the starting locations.
there is insufficient evidence to conclude that the relationship between traveling time and the types of signs depends on the starting locations.
All of the above.

ANSWER:

TYPE: MC DIFFICULTY: Easy

KEYWORDS: two-factor analysis of variance, F test for factor, F test for interaction, decision, conclusion

38. Referring to Table 11-4, at 10% level of significance,

there is sufficient evidence to conclude that the difference between the average traveling time for the different starting locations depends on the types of signs.
there is insufficient evidence to conclude that the difference between the average traveling time for the different types of signs depends on the starting locations.
there is sufficient evidence to conclude that the difference between the average traveling time for the different starting locations does not depend on the types of signs.
None of the above.

ANSWER:

TYPE: MC DIFFICULTY: Easy

KEYWORDS: two-factor analysis of variance, F test for factor, decision, conclusion

TABLE 11-5

A physician and president of a Tampa Health Maintenance Organization (HMO) are attempting to show the benefits of managed health care to an insurance company. The physician believes that certain types of doctors are more cost-effective than others. One theory is that Primary Specialty is an important factor in measuring the cost-effectiveness of physicians. To investigate this, the president obtained independent random samples of 20 HMO physicians from each of 4 primary specialties - General Practice (GP), Internal Medicine (IM), Pediatrics (PED), and Family Physicians (FP) - and recorded the total charges per member per month for each. A second factor which the president believes influences total charges per member per month is whether the doctor is a foreign or USA medical school graduate. The president theorizes that foreign graduates will have higher mean charges than USA graduates. To investigate this, the president also collected data on 20 foreign medical school graduates in each of the 4 primary specialty types described above. So information on charges for 40 doctors (20 foreign and 20 USA medical school graduates) was obtained for each of the 4 specialties. The results for the ANOVA are summarized in the following table.

Source	df	SS	MS	F	PR > F
Specialty	3	22,855	7,618	60.94	0.0001
Med school	1	105	105	0.84	0.6744
Interaction	3	890	297	2.38	0.1348
Error	152	18,950
Total	159	42,800

39. Referring to Table 11-5, what was the total number of doctors included in the study?

ANSWER:

TYPE: MC DIFFICULTY: Easy

KEYWORDS: two-factor analysis of variance, properties

40. Referring to Table 11-5, what degrees of freedom should be used to determine the critical value of the F ratio against which to test for interaction between the two factors?

numerator df = 1, denominator df = 159
numerator df = 3, denominator df = 159
numerator df = 1, denominator df = 152
numerator df = 3, denominator df = 152

ANSWER:

TYPE: MC DIFFICULTY: Easy

KEYWORDS: two-factor analysis of variance, F test for interaction, degrees of freedom

41. Referring to Table 11-5, interpret the test for interaction.

There is insufficient evidence to say at the 0.10 level of significance that the difference between the mean charges for foreign and USA graduates depends on primary specialty.
There is sufficient evidence to say at the 0.10 level of significance that the difference between the mean charges for foreign and USA graduates depends on primary specialty.
There is sufficient evidence at the 0.10 level of significance of a difference between the mean charges for foreign and USA medical graduates.
There is sufficient evidence to say at the 0.10 level of significance that mean charges depend on both primary specialty and medical school.

ANSWER:

TYPE: MC DIFFICULTY: Difficult

KEYWORDS: two-factor analysis of variance, interaction, interpretation

42. Referring to Table 11-5, what degrees of freedom should be used to determine the critical value of the F ratio against which to test for differences in the mean charges for doctors among the four primary specialty areas?

numerator df = 1, denominator df = 159
numerator df = 3, denominator df = 159
numerator df = 1, denominator df = 152
numerator df = 3, denominator df = 152

ANSWER:

TYPE: MC DIFFICULTY: Easy

KEYWORDS: two-factor analysis of variance, F test for factor, degrees of freedom

43. Referring to Table 11-5, what degrees of freedom should be used to determine the critical value of the F ratio against which to test for differences between the mean charges of foreign and USA medical school graduates?

numerator df = 1, denominator df = 159
numerator df = 3, denominator df = 159
numerator df = 1, denominator df = 152
numerator df = 3, denominator df = 152

ANSWER:

TYPE: MC DIFFICULTY: Easy

KEYWORDS: two-factor analysis of variance, F test for factor, degrees of freedom

44. Referring to Table 11-5, is there evidence of a difference between the mean charges of foreign and USA medical school graduates?

Yes, the test for the main effect for primary specialty is significant at = 0.10.
No, the test for the main effect for medical school is not significant at = 0.10.
No, the test for the interaction is not significant at = 0.10.
Maybe, but we need information on the -estimates to fully answer the question.

ANSWER:

TYPE: MC DIFFICULTY: Easy

KEYWORDS: two-factor analysis of variance, F test for factor, decision, conclusion

45. Referring to Table 11-5, what assumption(s) need(s) to be made in order to conduct the test for differences between the mean charges of foreign and USA medical school graduates?

There is no significant interaction effect between the area of primary specialty and the medical school on the doctors’ mean charges.
The charges in each group of doctors sampled are drawn from normally distributed populations.
The charges in each group of doctors sampled are drawn from populations with equal variances.
All of the above are necessary assumptions.

ANSWER:

TYPE: MC DIFFICULTY: Moderate

KEYWORDS: two-factor analysis of variance, assumptions

46. True or False: The analysis of variance (ANOVA) tests hypotheses about the population variance.

ANSWER: False

TYPE: TF DIFFICULTY: Easy

KEYWORDS: one-way analysis of variance

47. True or False: The F test in a completely randomized model is just an expansion of the t test for independent samples.

ANSWER: True

TYPE: TF DIFFICULTY: Easy

KEYWORDS: completely randomized design, F test for factor

48. True or False: When the F test is used for ANOVA, the rejection region is always in the right tail.

ANSWER: True

TYPE: TF DIFFICULTY: Easy

KEYWORDS: F test for factor, rejection region

49. True or False: A completely randomized design with 4 groups would have 6 possible pairwise comparisons.

ANSWER: True

TYPE: TF DIFFICULTY: Moderate

KEYWORDS: completely randomized design, properties

50. True or False: If you are comparing the average sales among 3 different brands you are dealing with a three-way ANOVA design.

ANSWER: False

TYPE: TF DIFFICULTY: Easy

KEYWORDS: one-way analysis of variance, properties

51. True or False: The MSE must always be positive.

ANSWER: True

TYPE: TF DIFFICULTY: Easy

KEYWORDS: mean squares, properties

52. True or False: In a two-way ANOVA, it is easier to interpret main effects when the interaction component is not significant.

ANSWER: True

TYPE: TF DIFFICULTY: Moderate

KEYWORDS: two-factor analysis of variance, interpretation

53. True or False: In a one-factor ANOVA analysis, the among sum of squares and within sum of squares must add up to the total sum of squares.

ANSWER: True

TYPE: TF DIFFICULTY: Easy

KEYWORDS: one-way analysis of variance, sum of squares, properties

54. True or False: In a two-factor ANOVA analysis, the sum of squares due to both factors, the interaction sum of squares and the within sum of squares must add up to the total sum of squares.

ANSWER: True

TYPE: TF DIFFICULTY: Easy

KEYWORDS: two-factor analysis of variance, sum of squares, properties

TABLE 11-6

As part of an evaluation program, a sporting goods retailer wanted to compare the downhill coasting speeds of 4 brands of bicycles. She took 3 of each brand and determined their maximum downhill speeds. The results are presented in miles per hour in the table below.

Trial	Barth	Tornado	Reiser	Shaw
1	43	37	41	43
2	46	38	45	45
3	43	39	42	46

55. Referring to Table 11-6, the sporting goods retailer decided to perform an ANOVA F test. The amount of total variation or SST is __________.

ANSWER: 102.67

TYPE: FI DIFFICULTY: Moderate

KEYWORDS: one-way analysis of variance, sum of squares, interpretation

56. Referring to Table 11-6, the among group variation or SSA is __________.

ANSWER: 81.33

TYPE: FI DIFFICULTY: Moderate

KEYWORDS: one-way analysis of variance, sum of squares

57. Referring to Table 11-6, the within group variation or SSW is __________.

ANSWER: 21.33

TYPE: FI DIFFICULTY: Moderate

KEYWORDS: one-way analysis of variance, sum of squares

58. Referring to Table 11-6, the value of MSA is __________, while MSW is __________.

ANSWER: 27.11; 2.67

TYPE: FI DIFFICULTY: Moderate

KEYWORDS: one-way analysis of variance, mean squares

59. Referring to Table 11-6, the null hypothesis that the average downhill coasting speeds of the 4 brands of bicycles are equal will be rejected at a level of significance of 0.05 if the value of the test statistic is greater than __________.

ANSWER: 4.07

TYPE: FI DIFFICULTY: Easy

KEYWORDS: one-way analysis of variance, F test for factor, decision

60. Referring to Table 11-6, in testing the null hypothesis that the average downhill coasting speeds of the 4 brands of bicycles are equal, the value of the test statistic is __________.

ANSWER: 10.17

TYPE: FI DIFFICULTY: Moderate

KEYWORDS: one-way analysis of variance, F test for factor, test statistic

61. Referring to Table 11-6, construct the ANOVA table from the sample data.

ANSWER:

Analysis of Variance

Source	df	SS	MS	F	p
Bicycle Brands	3	81.33	27.11	10.17	0.004*
Error	8	21.33	2.67
Total	11	102.67

* or p < 0.005, tabular value

TYPE: PR DIFFICULTY: Moderate

KEYWORDS: one-way analysis of variance, properties

62. True or False: Referring to Table 11-6, the null hypothesis should be rejected at a 5% level of significance.

ANSWER:

True

TYPE: TF DIFFICULTY: Easy

KEYWORDS: one-way analysis of variance, decision

63. True or False: Referring to Table 11-6, the decision made implies that all 4 means are significantly different.

ANSWER:

False

TYPE: TF DIFFICULTY: Moderate

KEYWORDS: one-way analysis of variance, conclusion, interpretation

64. True or False: Referring to Table 11-6, the test is valid only if the population of speeds has the same variance for the 4 brands.

ANSWER:

True

TYPE: TF DIFFICULTY: Easy

KEYWORDS: one-way analysis of variance, assumption

65. True or False: Referring to Table 11-6, the test is less sensitive to the assumption that the population of speeds has the same variance for the 4 brands if the sample sizes of the 4 brands are equal.

ANSWER:

True

TYPE: TF DIFFICULTY: Easy

KEYWORDS: one-way analysis of variance, assumption

66. True or False: Referring to Table 11-6, the test is valid only if the population of speeds is normally distributed.

ANSWER:

True

TYPE: TF DIFFICULTY: Easy

KEYWORDS: one-way analysis of variance, assumption

67. True or False: Referring to Table 11-6, the test is robust to the violation of the assumption that the population of speeds is normally distributed.

ANSWER:

True

TYPE: TF DIFFICULTY: Easy

KEYWORDS: one-way analysis of variance, assumption

68. Referring to Table 11-6, the sporting goods retailer decided to compare the 4 treatment means by using the Tukey-Kramer procedure with an overall level of significance of 0.05. There are ________ pairwise comparisons that can be made.

ANSWER:

TYPE: FI DIFFICULTY: Moderate

KEYWORDS: one-way analysis of variance, Tukey-Kramer procedure

69. Referring to Table 11-6, using an overall level of significance of 0.05, the critical value of the Studentized range Q used in calculating the critical range for the Tukey-Kramer procedure is ________.

ANSWER:

4.53

TYPE: FI DIFFICULTY: Moderate

KEYWORDS: one-way analysis of variance, Tukey-Kramer procedure, critical value

70. Referring to Table 11-6, using an overall level of significance of 0.05, the critical range for the Tukey-Kramer procedure is ________.

ANSWER:

4.27

TYPE: FI DIFFICULTY: Easy

KEYWORDS: one-way analysis of variance,, Tukey-Kramer procedure, critical value

71. True or False: Referring to Table 11-6, based on the Tukey-Kramer procedure with an overall level of significance of 0.05, the retailer would decide that there is a significant difference between all pairs of mean speeds.

ANSWER:

False

TYPE: TF DIFFICULTY: Moderate

KEYWORDS: one-way analysis of variance, Tukey-Kramer procedure, decision, conclusion

72. True or False: Referring to Table 11-6, based on the Tukey-Kramer procedure with an overall level of significance of 0.05, the retailer would decide that there is no significant difference between any pair of mean speeds.

ANSWER:

False

TYPE: TF DIFFICULTY: Moderate

KEYWORDS: one-way analysis of variance, Tukey-Kramer procedure, decision, conclusion

73. True or False: Referring to Table 11-6, based on the Tukey-Kramer procedure with an overall level of significance of 0.05, the retailer would decide that the mean speed for the Tornado brand is significantly different from each of the mean speeds for other brands.

ANSWER:

True

TYPE: TF DIFFICULTY: Moderate

KEYWORDS: one-way analysis of variance, Tukey-Kramer procedure, decision, conclusion

74. True or False: Referring to Table 11-6, based on the Tukey-Kramer procedure with an overall level of significance of 0.05, the retailer would decide that the 3 means other than the mean for Tornado are not significantly different from each other.

ANSWER:

True

TYPE: TF DIFFICULTY: Moderate

KEYWORDS: one-way analysis of variance, Tukey-Kramer procedure, decision, conclusion

75. Referring to Table 11-6, the null hypothesis for Levene’s test for homogeneity of variances is

ANSWER:

TYPE: MC DIFFICULTY: Easy

KEYWORDS: Levene’s test, form of hypothesis

76. Referring to Table 11-6, what is the value of the test statistic for Levene’s test for homogeneity of variances?

ANSWER:

0.1333

TYPE: PR DIFFICULTY: Moderate

KEYWORDS: Levene’s test, test statistic

77. Referring to Table 11-6, what are the numerator and denominator degrees of freedom for Levene’s test for homogeneity of variances respectively?

ANSWER:

3, 8

TYPE: PR DIFFICULTY: Moderate

KEYWORDS: Levene’s test, degrees of freedom

78. Referring to Table 11-6, what is the critical value of Levene’s test for homogeneity of variances at a 5% level of significance?

ANSWER:

4.07

TYPE: PR DIFFICULTY: Moderate

KEYWORDS: Levene’s test, critical value

79. Referring to Table 11-6, what is the p-value of the test statistic for Levene’s test for homogeneity of variances?

ANSWER:

0.94

TYPE: PR DIFFICULTY: Moderate

KEYWORDS: Levene’s test, p-value

80. Referring to Table 11-6, what should be the decision for the Levene’s test for homogeneity of variances at a 5% level of significance?

Reject the null hypothesis because the p-value is smaller than the level of significance.
Reject the null hypothesis because the p-value is larger than the level of significance.
Do not eject the null hypothesis because the p-value is smaller than the level of significance.
Do not reject the null hypothesis because the p-value is larger than the level of significance.

ANSWER:

TYPE: MC DIFFICULTY: Easy

KEYWORDS: Levene’s test, decision

81. Referring to Table 11-6, what should be the conclusion for the Levene’s test for homogeneity of variances at a 5% level of significance?

There is not sufficient evidence that the variances are all the same.
There is sufficient evidence that the variances are all the same.
There is not sufficient evidence that the variances are not all the same.
There is sufficient evidence that the variances are not all the same.

ANSWER:

TYPE: MC DIFFICULTY: Easy

KEYWORDS: Levene’s test, conclusion

TABLE 11-7

An agronomist wants to compare the crop yield of 3 varieties of chickpea seeds. She plants 15 fields, 5 with each variety. She then measures the crop yield in bushels per acre. Treating this as a completely randomized design, the results are presented in the table that follows.

Trial	Smith	Walsh	Trevor
1	11.1	19.0	14.6
2	13.5	18.0	15.7
3	15.3	19.8	16.8
4	14.6	19.6	16.7
5	9.8	16.6	15.2

82. Referring to Table 11-7, the agronomist decided to perform an ANOVA F test. The amount of total variation or SST is __________.

ANSWER:

114.82

TYPE: FI DIFFICULTY: Moderate

KEYWORDS: one-way analysis of variance, sum of squares

83. Referring to Table 11-7, the among-group variation or SSA is __________.

ANSWER:

82.39

TYPE: FI DIFFICULTY: Moderate

KEYWORDS: one-way analysis of variance, sum of squares

84. Referring to Table 11-7, the within-group variation or SSW is __________.

ANSWER:

32.43

TYPE: FI DIFFICULTY: Moderate

KEYWORDS: one-way analysis of variance, sum of squares

85. Referring to Table 11-7, the value of MSA is __________, while MSW is __________.

ANSWER:

41.19; 2.70

TYPE: FI DIFFICULTY: Moderate

KEYWORDS: one-way analysis of variance, mean squares

86. Referring to Table 11-7, the null hypothesis will be rejected at a level of significance of 0.01 if the value of the test statistic is greater than __________.

ANSWER:

6.93

TYPE: FI DIFFICULTY: Easy

KEYWORDS: one-way analysis of variance, F test for factor, critical value

87. Referring to Table 11-7, the value of the test statistic is __________.

ANSWER:

15.24

TYPE: FI DIFFICULTY: Moderate

KEYWORDS: one-way analysis of variance, F test for factor, test statistic

88. Referring to Table 11-7, construct the ANOVA table from the sample data.

ANSWER:

Analysis of Variance

Source	df	SS	MS	F	p
Seed Varieties	2	82.39	41.19	15.24	0.000508*
Error	12	32.43	2.70
Total	14	114.82

* or p < 0.005, tabular value

TYPE: PR DIFFICULTY: Moderate

KEYWORDS: one-way analysis of variance, F test for factor, properties

89. Referring to Table 11-7, state the null hypothesis that can be tested.

ANSWER:

H₀:

TYPE: PR DIFFICULTY: Easy

KEYWORDS: one-way analysis of variance, F test for factor, form of hypothesis

90. True or False: Referring to Table 11-7, the null hypothesis should be rejected at 0.005 level of significance.

ANSWER:

True

TYPE: TF DIFFICULTY: Moderate

KEYWORDS: one-way analysis of variance, F test for factor, decision

91. True or False: Referring to Table 11-7, the decision made at 0.005 level of significance implies that all 3 means are significantly different.

ANSWER:

False

TYPE: TF DIFFICULTY: Moderate

KEYWORDS: one-way analysis of variance, F test for factor, conclusion

92. True or False: Referring to Table 11-7, the test is valid only if the population of crop yields has the same variance for the 3 varieties.

ANSWER:

True

TYPE: TF DIFFICULTY: Easy

KEYWORDS: one-way analysis of variance, F test for factor, assumption

93. True or False: Referring to Table 11-7, the test is valid only if the population of crop yields is normally distributed for the 3 varieties.

ANSWER:

True

TYPE: TF DIFFICULTY: Easy

KEYWORDS: one-way analysis of variance, F test for factor, assumption

94. Referring to Table 11-7, the agronomist decided to compare the 3 treatment means by using the Tukey-Kramer procedure with an overall level of significance of 0.01. There are ________ pairwise comparisons that can be made.

ANSWER:

TYPE: FI DIFFICULTY: Moderate

KEYWORDS: one-way analysis of variance, Tukey-Kramer procedure, properties

95. Referring to Table 11-7, using an overall level of significance of 0.01, the critical value of the Studentized range Q used in calculating the critical range for the Tukey-Kramer procedure is ________.

ANSWER:

5.04

TYPE: FI DIFFICULTY: Easy

KEYWORDS: one-way analysis of variance, Tukey-Kramer procedure, critical value

96. Referring to Table 11-7, using an overall level of significance of 0.01, the critical range for the Tukey-Kramer procedure is ________.

ANSWER:

3.70

TYPE: FI DIFFICULTY: Moderate

KEYWORDS: one-way analysis of variance, Tukey-Kramer procedure, critical value

97. True or False: Referring to Table 11-7, based on the Tukey-Kramer procedure with an overall level of significance of 0.01, the agronomist would decide that there is a significant difference between the crop yield of Smith and Walsh seeds.

ANSWER:

True

TYPE: TF DIFFICULTY: Moderate

KEYWORDS: one-way analysis of variance, Tukey-Kramer procedure, decision, conclusion

98. True or False: Referring to Table 11-7, based on the Tukey-Kramer procedure with an overall level of significance of 0.01, the agronomist would decide that there is a significant difference between the crop yield of Smith and Trevor seeds.

ANSWER:

False

TYPE: TF DIFFICULTY: Moderate

KEYWORDS: one-way analysis of variance, Tukey-Kramer procedure, decision, conclusion

99. True or False: Referring to Table 11-7, based on the Tukey-Kramer procedure with an overall level of significance of 0.01, the agronomist would decide that there is a significant difference between the crop yield of Walsh and Trevor seeds.

ANSWER:

False

TYPE: TF DIFFICULTY: Moderate

KEYWORDS: one-way analysis of variance, Tukey-Kramer procedure, decision, conclusion

TABLE 11-8

A hotel chain has identically sized resorts in 5 locations. The data that follow resulted from analyzing the hotel occupancies on randomly selected days in the 5 locations.

ROW	Caymen	Pennkamp	California	Mayaguez	Maui
1	28	40	21	37	22
2	33	35	21	47	19
3	41	33	27	45	25

Analysis of Variance

Source	df	SS	MS	F	p
Location	4	963.6	11.47	0.001
Error	10	210.0
Total

100. True or False: Referring to Table 11-8, if a level of significance of 0.05 is chosen, the null hypothesis should be rejected.

ANSWER:

True

TYPE: TF DIFFICULTY: Easy

KEYWORDS: one-way analysis of variance, F test for factor, decision

101. True or False: Referring to Table 11-8, if a level of significance of 0.05 is chosen, the decision made indicates that all 5 locations have different mean occupancy rates.

ANSWER:

False

TYPE: TF DIFFICULTY: Easy

KEYWORDS: one-way analysis of variance, F test for factor, conclusion

102. True or False: Referring to Table 11-8, if a level of significance of 0.05 is chosen, the decision made indicates that at least 2 of the 5 locations have different mean occupancy rates.

ANSWER:

True

TYPE: TF DIFFICULTY: Easy

KEYWORDS: one-way analysis of variance, F test for factor, conclusion

103. Referring to Table 11-8, the among-group variation or SSA is _________.

ANSWER:

963.6

TYPE: FI DIFFICULTY: Easy

KEYWORDS: one-way analysis of variance, sum of squares

104. Referring to Table 11-8, the within-group variation or SSW is _________.

ANSWER:

210.0

TYPE: FI DIFFICULTY: Easy

KEYWORDS: one-way analysis of variance, sum of squares

105. Referring to Table 11-8, the total variation or SST is ________.

ANSWER:

1,173.6

TYPE: FI DIFFICULTY: Easy

KEYWORDS: one-way analysis of variance, sum of squares

106. Referring to Table 11-8, the value of MSA is ______ while MSW is _______.

ANSWER:

240.9; 21.0

TYPE: FI DIFFICULTY: Easy

KEYWORDS: one-way analysis of variance, mean squares

107. Referring to Table 11-8, the numerator and denominator degrees of freedom of the test ratio are ________ and ________, respectively.

ANSWER:

4 and 10

TYPE: FI DIFFICULTY: Easy

KEYWORDS: one-way analysis of variance, degrees of freedom

108. True or False: Referring to Table 11-8, the total mean squares is 261.90.

ANSWER:

False

TYPE: TF DIFFICULTY: Easy

KEYWORDS: one-way analysis of variance, sum of squares

109. Referring to Table 11-8, the null hypothesis for Levene’s test for homogeneity of variances is

ANSWER:

TYPE: MC DIFFICULTY: Easy

KEYWORDS: Levene’s test, form of hypothesis

110. Referring to Table 11-8, what is the value of the test statistic for Levene’s test for homogeneity of variances?

ANSWER:

0.28

TYPE: PR DIFFICULTY: Moderate

KEYWORDS: Levene’s test, test statistic

111. Referring to Table 11-8, what are the numerator and denominator degrees of freedom for Levene’s test for homogeneity of variances respectively?

ANSWER:

4, 10

TYPE: PR DIFFICULTY: Moderate

KEYWORDS: Levene’s test, degrees of freedom

112. Referring to Table 11-8, what is the critical value of Levene’s test for homogeneity of variances at a 5% level of significance?

ANSWER:

3.48

TYPE: PR DIFFICULTY: Moderate

KEYWORDS: Levene’s test, critical value

113. Referring to Table 11-8, what is the p-value of the test statistic for Levene’s test for homogeneity of variances?

ANSWER:

0.88

TYPE: PR DIFFICULTY: Moderate

KEYWORDS: Levene’s test, p-value

114. Referring to Table 11-8, what should be the decision for the Levene’s test for homogeneity of variances at a 5% level of significance?

Reject the null hypothesis because the p-value is smaller than the level of significance.
Reject the null hypothesis because the p-value is larger than the level of significance.
Do not eject the null hypothesis because the p-value is smaller than the level of significance.
Do not reject the null hypothesis because the p-value is larger than the level of significance.

ANSWER:

TYPE: MC DIFFICULTY: Easy

KEYWORDS: Levene’s test, decision

115. Referring to Table 11-8, what should be the conclusion for the Levene’s test for homogeneity of variances at a 5% level of significance?

There is not sufficient evidence that the variances are all the same.
There is sufficient evidence that the variances are all the same.
There is not sufficient evidence that the variances are not all the same.
There is sufficient evidence that the variances are not all the same.

ANSWER:

TYPE: MC DIFFICULTY: Easy

KEYWORDS: Levene’s test, conclusion

TABLE 11-9

The marketing manager of a company producing a new cereal aimed for children wants to examine the effect of the color and shape of the box's logo on the approval rating of the cereal. He combined 4 colors and 3 shapes to produce a total of 12 designs. Each logo was presented to 2 different groups (a total of 24 groups) and the approval rating for each was recorded and is shown below. The manager analyzed these data using the = 0.05 level of significance for all inferences.

COLORS
SHAPES	Red	Green	Blue	Yellow
Circle	54	67	36	45
44	61	44	41
Square	34	56	36	21
36	58	30	25
Diamond	46	60	34	31
48	60	38	33

Analysis of Variance

Source	df	SS	MS	F	p
Colors	3	2711.17	903.72	72.30	0.000
Shapes	2	579.00	289.50	23.16	0.000
Interaction	6	150.33	25.06	2.00	0.144
Error	12	150.00	12.50
Total	23	3590.50

116. Referring to Table 11-9, the mean square for the factor color is ________.

ANSWER:

903.72

TYPE: FI DIFFICULTY: Easy

KEYWORDS: two-factor analysis of variance, mean squares

117. Referring to Table 11-9, the mean square for the factor shape is ________.

ANSWER:

289.50

TYPE: FI DIFFICULTY: Easy

KEYWORDS: two-factor analysis of variance, mean squares

118. Referring to Table 11-9, the mean square for the interaction of color and shape is ________.

ANSWER:

25.06

TYPE: FI DIFFICULTY: Easy

KEYWORDS: two-factor analysis of variance, mean squares

119. Referring to Table 11-9, the mean square for error is ________.

ANSWER:

12.50

TYPE: FI DIFFICULTY: Easy

KEYWORDS: two-factor analysis of variance, mean squares

120. Referring to Table 11-9, the critical value of the test for significant differences between colors is ________.

ANSWER:

3.49

TYPE: FI DIFFICULTY: Easy

KEYWORDS: two-factor analysis of variance, F test for factor, critical value

121. Referring to Table 11-9, the value of the statistic used to test for significant differences between colors is ________.

ANSWER:

72.30

TYPE: FI DIFFICULTY: Easy

KEYWORDS: two-factor analysis of variance, F test for factor, test statistic

122. True or False: Referring to Table 11-9, based on the results of the hypothesis test, it appears that there is a significant effect on the approval rating associated with the color of the logo.

ANSWER:

True

TYPE: TF DIFFICULTY: Easy

KEYWORDS: two-factor analysis of variance, F test for factor, decision, conclusion

123. Referring to Table 11-9, the critical value in the test for significant differences between shapes is ________.

ANSWER:

3.89

TYPE: FI DIFFICULTY: Easy

KEYWORDS: two-factor analysis of variance, F test for factor, critical value

124. Referring to Table 11-9, the value of the statistic used to test for significant differences between shapes is ________.

ANSWER:

23.16

TYPE: FI DIFFICULTY: Easy

KEYWORDS: two-factor analysis of variance, F test for factor, test statistic

125. True or False: Referring to Table 11-9, based on the results of the hypothesis test, it appears that there is a significant effect associated with the shape of the logo.

ANSWER:

True

TYPE: TF DIFFICULTY: Easy

KEYWORDS: two-factor analysis of variance, F test for factor, decision, conclusion

126. Referring to Table 11-9, the critical value in the test for a significant interaction is ________.

ANSWER:

3.00

TYPE: FI DIFFICULTY: Easy

KEYWORDS: two-factor analysis of variance, F test for interaction, critical value

127. Referring to Table 11-9, the value of the statistic used to test for an interaction is ________.

ANSWER:

2.00

TYPE: FI DIFFICULTY: Easy

KEYWORDS: two-factor analysis of variance, F test for interaction, test statistic

128. True or False: Referring to Table 11-9, based on the results of the hypothesis test, it appears that there is a significant interaction.

ANSWER:

False

TYPE: TF DIFFICULTY: Easy

KEYWORDS: two-factor analysis of variance, F test for interaction, decision, conclusion

TABLE 11-10

An agronomist wants to compare the crop yield of 3 varieties of chickpea seeds. She plants all 3 varieties of the seeds on each of 5 different patches of fields. She then measures the crop yield in bushels per acre. Treating this as a randomized block design, the results are presented in the table that follows.

Fields	Smith	Walsh	Trevor
1	11.1	19.0	14.6
2	13.5	18.0	15.7
3	15.3	19.8	16.8
4	14.6	19.6	16.7
5	9.8	16.6	15.2

129. Referring to Table 11-10, the agronomist decided to perform a randomized block F test for the difference in the means. The amount of total variation or SST is __________.

ANSWER:

114.82

TYPE: FI DIFFICULTY: Moderate

KEYWORDS: randomized block design, sum of squares

130. Referring to Table 11-10, the among-group variation or SSA is __________.

ANSWER:

82.39

TYPE: FI DIFFICULTY: Moderate

KEYWORDS: randomized block design, sum of squares

131. Referring to Table 11-10, the among-block variation or SSBL is __________.

ANSWER:

24.46

TYPE: FI DIFFICULTY: Moderate

KEYWORDS: randomized block design, sum of squares

132. Referring to Table 11-10, the value of MSA is __________, while MSBL is __________.

ANSWER:

41.19; 6.11

TYPE: FI DIFFICULTY: Moderate

KEYWORDS: randomized block design, mean squares

133. Referring to Table 11-10, the null hypothesis for the randomized block F test for the difference in the means is

ANSWER:

TYPE: MC DIFFICULTY: Easy

KEYWORDS: randomized block design, F test for factor, form of hypothesis

134. Referring to Table 11-10, what are the degrees of freedom of the randomized block F test for the difference in the means at a level of significance of 0.01?

ANSWER:

2 numerator and 8 denominator degrees of freedom

TYPE: PR DIFFICULTY: Easy

KEYWORDS: randomized block design, F test for factor, degrees of freedom

135. Referring to Table 11-10, what is the critical value of the randomized block F test for the difference in the means at a level of significance of 0.01?

ANSWER:

8.65

TYPE: PR DIFFICULTY: Easy

KEYWORDS: randomized block design, F test for factor, critical value

136. Referring to Table 11-10, what is the value of the test statistic for the randomized block F test for the difference in the means?

ANSWER:

41.32

TYPE: PR DIFFICULTY: Moderate

KEYWORDS: randomized block design, F test for factor, test statistic

137. Referring to Table 11-10, what is the p-value of the test statistic for the randomized block F test for the difference in the means?

ANSWER:

6.07E-05

TYPE: PR DIFFICULTY: Moderate

KEYWORDS: randomized block design, F test for factor, p-value

138. True or False: Referring to Table 11-10, the null hypothesis for the randomized block F test for the difference in the means should be rejected at a 0.01 level of significance.

ANSWER:

True

TYPE: TF DIFFICULTY: Easy

KEYWORDS: randomized block design, F test for factor, decision

139. True or False: Referring to Table 11-10, the decision made at a 0.01 level of significance on the randomized block F test for the difference in means implies that all 3 means are significantly different.

ANSWER:

False

TYPE: TF DIFFICULTY: Moderate

KEYWORDS: randomized block design, F test for factor, conclusion

140. True or False: Referring to Table 11-10, the randomized block F test is valid only if the population of crop yields has the same variance for the 3 varieties.

ANSWER:

True

TYPE: TF DIFFICULTY: Easy

KEYWORDS: randomized block design, F test for factor, assumption

141. True or False: Referring to Table 11-10, the randomized block F test is valid only if the population of crop yields is normally distributed for the 3 varieties.

ANSWER:

True

TYPE: TF DIFFICULTY: Easy

v KEYWORDS: randomized block design, F test for factor, assumption

142. True or False: Referring to Table 11-10, the randomized block F test is valid only if there is no interaction between the variety of seeds and the patches of fields.

ANSWER:

True

TYPE: TF DIFFICULTY: Easy

KEYWORDS: randomized block design, F test for factor, assumption

143. Referring to Table 11-10, the agronomist decided to compare the 3 treatment means by using the Tukey multiple comparison procedure with an overall level of significance of 0.01. How many pair-wise comparisons can be made?

ANSWER:

TYPE: PR DIFFICULTY: Moderate

KEYWORDS: randomized block design, Tukey procedure, properties

144. Referring to Table 11-10, using an overall level of significance of 0.01, what is the critical value of the Studentized range Q used in calculating the critical range for the Tukey multiple comparison procedure?

ANSWER:

5.63

TYPE: PR DIFFICULTY: Easy

KEYWORDS: randomized block design, Tukey procedure, critical value

145. Referring to Table 11-10, using an overall level of significance of 0.01, what is the critical range for the Tukey multiple comparison procedure?

ANSWER:

2.51

TYPE: PR DIFFICULTY: Moderate

KEYWORDS: randomized block design, Tukey procudure, critical value

146. True or False: Referring to Table 11-10, based on the Tukey multiple comparison procedure with an overall level of significance of 0.01, the agronomist would decide that there is a significant difference between the crop yield of Smith and Walsh seeds.

ANSWER:

True

TYPE: TF DIFFICULTY: Moderate

KEYWORDS: randomized block design, Tukey procedure, decision, conclusion

147. True or False: Referring to Table 11-10, based on the Tukey-Kramer procedure with an overall level of significance of 0.01, the agronomist would decide that there is a significant difference between the crop yield of Smith and Trevor seeds.

ANSWER:

True

TYPE: TF DIFFICULTY: Moderate

KEYWORDS: randomized block design, Tukey procedure, decision, conclusion

148. True or False: Referring to Table 11-10, based on the Tukey multiple comparison procedure with an overall level of significance of 0.01, the agronomist would decide that there is a significant difference between the crop yield of Walsh and Trevor seeds.

ANSWER:

True

TYPE: TF DIFFICULTY: Moderate

KEYWORDS: randomized block design, Tukey procedure, decision, conclusion

149. Referring to Table 11-10, what is the null hypothesis for testing the block effects?

ANSWER:

TYPE: MC DIFFICULTY: Easy

KEYWORDS: randomized block design, F test for block effects, form of hypothesis

150. Referring to Table 11-10, what are the degrees of freedom of the F test statistic for testing the block effects?

ANSWER:

4 numerator and 8 denominator degrees of freedom

TYPE: PR DIFFICULTY: Easy

KEYWORDS: randomized block design, F test for block effects, degrees of freedom

151. Referring to Table 11-10, what is the value of the F test statistic for testing the block effects?

ANSWER:

6.13

TYPE: PR DIFFICULTY: Moderate

KEYWORDS: randomized block design, F test for block effects, test statistic

152. Referring to Table 11-10, what is the critical value for testing the block effects at a 0.01 level of significance?

ANSWER:

7.01

TYPE: PR DIFFICULTY: Moderate

KEYWORDS: randomized block design, F test for block effects, critical value

153. Referring to Table 11-10, what is the p-value of the F test statistic for testing the block effects?

ANSWER:

0.015 or between 0.01 and 0.025

TYPE: PR DIFFICULTY: Moderate

KEYWORDS: randomized block design, F test for block effects, p-value

154. True or False: Referring to Table 11-10, the null hypothesis for the F test for the block effects should be rejected at a 0.01 level of significance.

ANSWER:

False

TYPE: TF DIFFICULTY: Easy

KEYWORDS: randomized block design, F test for block effects, decision

155. True or False: Referring to Table 11-10, the decision made at a 0.01 level of significance on the F test for the block effects implies that the blocking has been advantageous in reducing the experiment error.

ANSWER:

False

TYPE: TF DIFFICULTY: Moderate

KEYWORDS: randomized block design, F test for block effects, conclusion

156. Referring to Table 11-10, what is the estimated relative efficiency?

ANSWER:

2.47

TYPE: PR DIFFICULTY: Moderate

KEYWORDS: randomized block design, F test for block effects, relative efficiency

157. True or False: Referring to Table 11-10, the relative efficiency means that 2.47 times as many observations in each variety group would be needed in a one-way ANOVA design as compared to the randomized block design in order to obtain the same precision for comparison of the variety means.

ANSWER:

True

TYPE: TF DIFFICULTY: Moderate

KEYWORDS: randomized block design, F test for block effects, relative efficiency, interpretation

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