Analysis Of Variance Sample Assignment
1. In a oneway ANOVA, if the computed F statistic exceeds the critical F value we may
 reject H_{0} since there is evidence all the means differ.
 reject H_{0} since there is evidence of a treatment effect.
 not reject H_{0} since there is no evidence of a difference.
 not reject H_{0} because a mistake has been made.
ANSWER:
b
TYPE: MC DIFFICULTY: Easy
KEYWORDS: oneway 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:
c
TYPE: MC DIFFICULTY: Easy
KEYWORDS: oneway analysis of variance, sum of squares, properties
3. Why would you use the TukeyKramer procedure?
 To test for normality.
 To test for homogeneity of variance.
 To test independence of errors.
 To test for differences in pairwise means.
ANSWER:
d
TYPE: MC DIFFICULTY: Easy
KEYWORDS: TukeyKramer 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:
a
TYPE: MC DIFFICULTY: Moderate
KEYWORDS: completely randomized design
5. The F test statistic in a oneway ANOVA is
 MSW/MSA.
 SSW/SSA.
 MSA/MSW.
 SSA/SSW.
ANSWER:
c
TYPE: MC DIFFICULTY: Easy
KEYWORDS: oneway analysis of variance, F test for factor
6. The degrees of freedom for the F test in a oneway ANOVA are
 (n – c) and (c – 1).
 (c – 1) and (n – c).
 (c – n) and (n – 1).
 (n – 1) and (c – n).
ANSWER:
b
TYPE: MC DIFFICULTY: Easy
KEYWORDS: oneway analysis of variance, degrees of freedom
7. In a oneway 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:
a
TYPE: MC DIFFICULTY: Easy
KEYWORDS: oneway analysis of variance, form of hypothesis
8. In a oneway 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:
c
TYPE: MC DIFFICULTY: Moderate
KEYWORDS: oneway analysis of variance, properties, interaction
9. Interaction in an experimental design can be tested in
 a completely randomized model.
 a randomized block model.
 a twofactor model.
 all ANOVA models.
ANSWER:
c
TYPE: MC DIFFICULTY: Easy
KEYWORDS: twofactor analysis of variance, interaction, properties
10. In a twoway ANOVA the degrees of freedom for the interaction term is
 (r – 1)(c – 1).
 rc(n – 1).
 (r – 1).
 rcn + 1.
ANSWER:
a
TYPE: MC DIFFICULTY: Easy
KEYWORDS: twofactor analysis of variance, interaction, degrees of freedom
11. In a twoway ANOVA the degrees of freedom for the "error" term is
 (r – 1)(c – 1).
 rc(n’ – 1).
 (r – 1).
 rcn' + 1.
ANSWER:
b
TYPE: MC DIFFICULTY: Easy
KEYWORDS: twofactor analysis of variance, degrees of freedom
TABLE 111
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 bottom 20%. 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 111, 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:
c
TYPE: MC DIFFICULTY: Moderate
KEYWORDS: twofactor analysis of variance, twofactor factorial design
13. Referring to Table 111, 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:
c
TYPE: MC DIFFICULTY: Moderate
KEYWORDS: twofactor analysis of variance, F test for interaction, decision, conclusion, interaction
14. Referring to Table 111, 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:
d
TYPE: MC DIFFICULTY: Difficult
KEYWORDS: twofactor 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.
 Oneway ANOVA F
 t test for the differences in means.
 t test for the mean difference.
ANSWER:
b
TYPE: MC DIFFICULTY: Easy
KEYWORDS: oneway analysis of variance, F test for factor
TABLE 112
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 
Pvalue 
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 112, the within groups degrees of freedom is
 3
 4
 16
 19
ANSWER:
c
TYPE: MC DIFFICULTY: Easy
KEYWORDS: oneway analysis of variance, degrees of freedom
17. Referring to Table 112, the total degrees of freedom is
 3
 4
 16
 19
ANSWER:
d
TYPE: MC DIFFICULTY: Easy
KEYWORDS: oneway analysis of variance, degrees of freedom
18. Referring to Table 112, the amonggroup (betweengroup) mean squares is
 525
 8
 4
 2
ANSWER:
b
TYPE: MC DIFFICULTY: Easy
KEYWORDS: oneway analysis of variance, mean squares
19. Referring to Table 112, 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:
c
TYPE: MC DIFFICULTY: Easy
KEYWORDS: oneway analysis of variance, F test for factor, decision, conclusion
TABLE 113
A realtor wants to compare the average salestoappraisal 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 113, the among group degrees of freedom is
 3
 4
 12
 16
ANSWER:
a
TYPE: MC DIFFICULTY: Easy
KEYWORDS: oneway analysis of variance, degrees of freedom
21. Referring to Table 113, the within group sum of squares is
 0606
 1825
 1819
 3644
ANSWER:
b
TYPE: MC DIFFICULTY: Easy
KEYWORDS: oneway analysis of variance, sum of squares
22. Referring to Table 113, the within group mean squares is
 10
 29
 06
 18
ANSWER:
a
TYPE: MC DIFFICULTY: Easy
KEYWORDS: oneway analysis of variance, mean squares
23. Referring to Table 113,
 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:
a
TYPE: MC DIFFICULTY: Easy
KEYWORDS: oneway analysis of variance, F test for factor, decision, conclusion
24. Referring to Table 113, the null hypothesis for Levene’s test for homogeneity of variances is
ANSWER:
c
TYPE: MC DIFFICULTY: Easy
KEYWORDS: Levene’s test, form of hypothesis
25. Referring to Table 113, the value of the test statistic for Levene’s test for homogeneity of variances is
 25
 37
 36
 76
ANSWER:
a
TYPE: MC DIFFICULTY: Moderate
KEYWORDS: Levene’s test, test statistic
26. Referring to Table 113, 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:
a
TYPE: MC DIFFICULTY: Moderate
KEYWORDS: Levene’s test, degrees of freedom
27. Referring to Table 113, the critical value of Levene’s test for homogeneity of variances at a 5% level of significance is
 64
 48
 29
 49
ANSWER:
d
TYPE: MC DIFFICULTY: Moderate
KEYWORDS: Levene’s test, critical value
28. Referring to Table 113, the pvalue of the test statistic for Levene’s test for homogeneity of variances is
 25
 64
 86
 49
ANSWER:
c
TYPE: MC DIFFICULTY: Moderate
KEYWORDS: Levene’s test, pvalue
29. Referring to Table 113, 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 pvalue is smaller than the level of significance.
 Reject the null hypothesis because the pvalue is larger than the level of significance.
 Do not eject the null hypothesis because the pvalue is smaller than the level of significance.
 Do not reject the null hypothesis because the pvalue is larger than the level of significance.
ANSWER:
d
TYPE: MC DIFFICULTY: Easy
KEYWORDS: Levene’s test, decision
30. Referring to Table 113, 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:
c
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:
c
TYPE: MC DIFFICULTY: Moderate
KEYWORDS: twofactor factorial design
TABLE 114
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 
Pvalue 
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 114, the degrees of freedom for the different building signs (factor A) is
 1
 2
 3
 8
ANSWER:
a
TYPE: MC DIFFICULTY: Easy
KEYWORDS: twofactor analysis of variance, degrees of freedom
33. Referring to Table 114, the within (error) degrees of freedom is
 1
 4
 8
 11
ANSWER:
c
TYPE: MC DIFFICULTY: Easy
KEYWORDS: twofactor analysis of variance, degrees of freedom
34. Referring to Table 114, the mean squares for starting location (factor B) is
 48
 4,413.17
 12,288
 14,008.3
ANSWER:
c
TYPE: MC DIFFICULTY: Easy
KEYWORDS: twofactor analysis of variance, mean squares
35. Referring to Table 114, the F test statistic for testing the main effect of types of signs is
 0109
 7844
 1742
 3176
ANSWER:
c
TYPE: MC DIFFICULTY: Easy
KEYWORDS: twofactor analysis of variance, F test for factor
36. Referring to Table 114, the F test statistic for testing the interaction effect between the types of signs and the starting location is
 0109
 7844
 1742
 3176
ANSWER:
a
TYPE: MC DIFFICULTY: Easy
KEYWORDS: twofactor analysis of variance, F test for interaction, interaction
37. Referring to Table 114, 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:
d
TYPE: MC DIFFICULTY: Easy
KEYWORDS: twofactor analysis of variance, F test for factor, F test for interaction, decision, conclusion
38. Referring to Table 114, 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:
b
TYPE: MC DIFFICULTY: Easy
KEYWORDS: twofactor analysis of variance, F test for factor, decision, conclusion
TABLE 115
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 costeffective than others. One theory is that Primary Specialty is an important factor in measuring the costeffectiveness 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 115, what was the total number of doctors included in the study?
 20
 40
 159
 160
ANSWER:
d
TYPE: MC DIFFICULTY: Easy
KEYWORDS: twofactor analysis of variance, properties
40. Referring to Table 115, 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:
d
TYPE: MC DIFFICULTY: Easy
KEYWORDS: twofactor analysis of variance, F test for interaction, degrees of freedom
41. Referring to Table 115, 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:
a
TYPE: MC DIFFICULTY: Difficult
KEYWORDS: twofactor analysis of variance, interaction, interpretation
42. Referring to Table 115, 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:
d
TYPE: MC DIFFICULTY: Easy
KEYWORDS: twofactor analysis of variance, F test for factor, degrees of freedom
43. Referring to Table 115, 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:
c
TYPE: MC DIFFICULTY: Easy
KEYWORDS: twofactor analysis of variance, F test for factor, degrees of freedom
44. Referring to Table 115, 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:
b
TYPE: MC DIFFICULTY: Easy
KEYWORDS: twofactor analysis of variance, F test for factor, decision, conclusion
45. Referring to Table 115, 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:
d
TYPE: MC DIFFICULTY: Moderate
KEYWORDS: twofactor 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: oneway 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 threeway ANOVA design.
ANSWER: False
TYPE: TF DIFFICULTY: Easy
KEYWORDS: oneway 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 twoway ANOVA, it is easier to interpret main effects when the interaction component is not significant.
ANSWER: True
TYPE: TF DIFFICULTY: Moderate
KEYWORDS: twofactor analysis of variance, interpretation
53. True or False: In a onefactor 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: oneway analysis of variance, sum of squares, properties
54. True or False: In a twofactor 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: twofactor analysis of variance, sum of squares, properties
TABLE 116
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 116, 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: oneway analysis of variance, sum of squares, interpretation
56. Referring to Table 116, the among group variation or SSA is __________.
ANSWER: 81.33
TYPE: FI DIFFICULTY: Moderate
KEYWORDS: oneway analysis of variance, sum of squares
57. Referring to Table 116, the within group variation or SSW is __________.
ANSWER: 21.33
TYPE: FI DIFFICULTY: Moderate
KEYWORDS: oneway analysis of variance, sum of squares
58. Referring to Table 116, the value of MSA is __________, while MSW is __________.
ANSWER: 27.11; 2.67
TYPE: FI DIFFICULTY: Moderate
KEYWORDS: oneway analysis of variance, mean squares
59. Referring to Table 116, 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: oneway analysis of variance, F test for factor, decision
60. Referring to Table 116, 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: oneway analysis of variance, F test for factor, test statistic
61. Referring to Table 116, 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: oneway analysis of variance, properties
62. True or False: Referring to Table 116, the null hypothesis should be rejected at a 5% level of significance.
ANSWER:
True
TYPE: TF DIFFICULTY: Easy
KEYWORDS: oneway analysis of variance, decision
63. True or False: Referring to Table 116, the decision made implies that all 4 means are significantly different.
ANSWER:
False
TYPE: TF DIFFICULTY: Moderate
KEYWORDS: oneway analysis of variance, conclusion, interpretation
64. True or False: Referring to Table 116, 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: oneway analysis of variance, assumption
65. True or False: Referring to Table 116, 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: oneway analysis of variance, assumption
66. True or False: Referring to Table 116, the test is valid only if the population of speeds is normally distributed.
ANSWER:
True
TYPE: TF DIFFICULTY: Easy
KEYWORDS: oneway analysis of variance, assumption
67. True or False: Referring to Table 116, 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: oneway analysis of variance, assumption
68. Referring to Table 116, the sporting goods retailer decided to compare the 4 treatment means by using the TukeyKramer procedure with an overall level of significance of 0.05. There are ________ pairwise comparisons that can be made.
ANSWER:
6
TYPE: FI DIFFICULTY: Moderate
KEYWORDS: oneway analysis of variance, TukeyKramer procedure
69. Referring to Table 116, 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 TukeyKramer procedure is ________.
ANSWER:
4.53
TYPE: FI DIFFICULTY: Moderate
KEYWORDS: oneway analysis of variance, TukeyKramer procedure, critical value
70. Referring to Table 116, using an overall level of significance of 0.05, the critical range for the TukeyKramer procedure is ________.
ANSWER:
4.27
TYPE: FI DIFFICULTY: Easy
KEYWORDS: oneway analysis of variance,, TukeyKramer procedure, critical value
71. True or False: Referring to Table 116, based on the TukeyKramer 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: oneway analysis of variance, TukeyKramer procedure, decision, conclusion
72. True or False: Referring to Table 116, based on the TukeyKramer 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: oneway analysis of variance, TukeyKramer procedure, decision, conclusion
73. True or False: Referring to Table 116, based on the TukeyKramer 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: oneway analysis of variance, TukeyKramer procedure, decision, conclusion
74. True or False: Referring to Table 116, based on the TukeyKramer 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: oneway analysis of variance, TukeyKramer procedure, decision, conclusion
75. Referring to Table 116, the null hypothesis for Levene’s test for homogeneity of variances is
ANSWER:
c
TYPE: MC DIFFICULTY: Easy
KEYWORDS: Levene’s test, form of hypothesis
76. Referring to Table 116, 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 116, 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 116, 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 116, what is the pvalue of the test statistic for Levene’s test for homogeneity of variances?
ANSWER:
0.94
TYPE: PR DIFFICULTY: Moderate
KEYWORDS: Levene’s test, pvalue
80. Referring to Table 116, 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 pvalue is smaller than the level of significance.
 Reject the null hypothesis because the pvalue is larger than the level of significance.
 Do not eject the null hypothesis because the pvalue is smaller than the level of significance.
 Do not reject the null hypothesis because the pvalue is larger than the level of significance.
ANSWER:
d
TYPE: MC DIFFICULTY: Easy
KEYWORDS: Levene’s test, decision
81. Referring to Table 116, 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:
c
TYPE: MC DIFFICULTY: Easy
KEYWORDS: Levene’s test, conclusion
TABLE 117
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 117, 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: oneway analysis of variance, sum of squares
83. Referring to Table 117, the amonggroup variation or SSA is __________.
ANSWER:
82.39
TYPE: FI DIFFICULTY: Moderate
KEYWORDS: oneway analysis of variance, sum of squares
84. Referring to Table 117, the withingroup variation or SSW is __________.
ANSWER:
32.43
TYPE: FI DIFFICULTY: Moderate
KEYWORDS: oneway analysis of variance, sum of squares
85. Referring to Table 117, the value of MSA is __________, while MSW is __________.
ANSWER:
41.19; 2.70
TYPE: FI DIFFICULTY: Moderate
KEYWORDS: oneway analysis of variance, mean squares
86. Referring to Table 117, 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: oneway analysis of variance, F test for factor, critical value
87. Referring to Table 117, the value of the test statistic is __________.
ANSWER:
15.24
TYPE: FI DIFFICULTY: Moderate
KEYWORDS: oneway analysis of variance, F test for factor, test statistic
88. Referring to Table 117, 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: oneway analysis of variance, F test for factor, properties
89. Referring to Table 117, state the null hypothesis that can be tested.
ANSWER:
H_{0}:
TYPE: PR DIFFICULTY: Easy
KEYWORDS: oneway analysis of variance, F test for factor, form of hypothesis
90. True or False: Referring to Table 117, the null hypothesis should be rejected at 0.005 level of significance.
ANSWER:
True
TYPE: TF DIFFICULTY: Moderate
KEYWORDS: oneway analysis of variance, F test for factor, decision
91. True or False: Referring to Table 117, the decision made at 0.005 level of significance implies that all 3 means are significantly different.
ANSWER:
False
TYPE: TF DIFFICULTY: Moderate
KEYWORDS: oneway analysis of variance, F test for factor, conclusion
92. True or False: Referring to Table 117, 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: oneway analysis of variance, F test for factor, assumption
93. True or False: Referring to Table 117, 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: oneway analysis of variance, F test for factor, assumption
94. Referring to Table 117, the agronomist decided to compare the 3 treatment means by using the TukeyKramer procedure with an overall level of significance of 0.01. There are ________ pairwise comparisons that can be made.
ANSWER:
3
TYPE: FI DIFFICULTY: Moderate
KEYWORDS: oneway analysis of variance, TukeyKramer procedure, properties
95. Referring to Table 117, 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 TukeyKramer procedure is ________.
ANSWER:
5.04
TYPE: FI DIFFICULTY: Easy
KEYWORDS: oneway analysis of variance, TukeyKramer procedure, critical value
96. Referring to Table 117, using an overall level of significance of 0.01, the critical range for the TukeyKramer procedure is ________.
ANSWER:
3.70
TYPE: FI DIFFICULTY: Moderate
KEYWORDS: oneway analysis of variance, TukeyKramer procedure, critical value
97. True or False: Referring to Table 117, based on the TukeyKramer 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: oneway analysis of variance, TukeyKramer procedure, decision, conclusion
98. True or False: Referring to Table 117, based on the TukeyKramer 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: oneway analysis of variance, TukeyKramer procedure, decision, conclusion
99. True or False: Referring to Table 117, based on the TukeyKramer 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: oneway analysis of variance, TukeyKramer procedure, decision, conclusion
TABLE 118
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 118, if a level of significance of 0.05 is chosen, the null hypothesis should be rejected.
ANSWER:
True
TYPE: TF DIFFICULTY: Easy
KEYWORDS: oneway analysis of variance, F test for factor, decision
101. True or False: Referring to Table 118, 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: oneway analysis of variance, F test for factor, conclusion
102. True or False: Referring to Table 118, 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: oneway analysis of variance, F test for factor, conclusion
103. Referring to Table 118, the amonggroup variation or SSA is _________.
ANSWER:
963.6
TYPE: FI DIFFICULTY: Easy
KEYWORDS: oneway analysis of variance, sum of squares
104. Referring to Table 118, the withingroup variation or SSW is _________.
ANSWER:
210.0
TYPE: FI DIFFICULTY: Easy
KEYWORDS: oneway analysis of variance, sum of squares
105. Referring to Table 118, the total variation or SST is ________.
ANSWER:
1,173.6
TYPE: FI DIFFICULTY: Easy
KEYWORDS: oneway analysis of variance, sum of squares
106. Referring to Table 118, the value of MSA is ______ while MSW is _______.
ANSWER:
240.9; 21.0
TYPE: FI DIFFICULTY: Easy
KEYWORDS: oneway analysis of variance, mean squares
107. Referring to Table 118, the numerator and denominator degrees of freedom of the test ratio are ________ and ________, respectively.
ANSWER:
4 and 10
TYPE: FI DIFFICULTY: Easy
KEYWORDS: oneway analysis of variance, degrees of freedom
108. True or False: Referring to Table 118, the total mean squares is 261.90.
ANSWER:
False
TYPE: TF DIFFICULTY: Easy
KEYWORDS: oneway analysis of variance, sum of squares
109. Referring to Table 118, the null hypothesis for Levene’s test for homogeneity of variances is
ANSWER:
c
TYPE: MC DIFFICULTY: Easy
KEYWORDS: Levene’s test, form of hypothesis
110. Referring to Table 118, 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 118, 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 118, 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 118, what is the pvalue of the test statistic for Levene’s test for homogeneity of variances?
ANSWER:
0.88
TYPE: PR DIFFICULTY: Moderate
KEYWORDS: Levene’s test, pvalue
114. Referring to Table 118, 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 pvalue is smaller than the level of significance.
 Reject the null hypothesis because the pvalue is larger than the level of significance.
 Do not eject the null hypothesis because the pvalue is smaller than the level of significance.
 Do not reject the null hypothesis because the pvalue is larger than the level of significance.
ANSWER:
d
TYPE: MC DIFFICULTY: Easy
KEYWORDS: Levene’s test, decision
115. Referring to Table 118, 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:
c
TYPE: MC DIFFICULTY: Easy
KEYWORDS: Levene’s test, conclusion
TABLE 119
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 119, the mean square for the factor color is ________.
ANSWER:
903.72
TYPE: FI DIFFICULTY: Easy
KEYWORDS: twofactor analysis of variance, mean squares
117. Referring to Table 119, the mean square for the factor shape is ________.
ANSWER:
289.50
TYPE: FI DIFFICULTY: Easy
KEYWORDS: twofactor analysis of variance, mean squares
118. Referring to Table 119, the mean square for the interaction of color and shape is ________.
ANSWER:
25.06
TYPE: FI DIFFICULTY: Easy
KEYWORDS: twofactor analysis of variance, mean squares
119. Referring to Table 119, the mean square for error is ________.
ANSWER:
12.50
TYPE: FI DIFFICULTY: Easy
KEYWORDS: twofactor analysis of variance, mean squares
120. Referring to Table 119, the critical value of the test for significant differences between colors is ________.
ANSWER:
3.49
TYPE: FI DIFFICULTY: Easy
KEYWORDS: twofactor analysis of variance, F test for factor, critical value
121. Referring to Table 119, the value of the statistic used to test for significant differences between colors is ________.
ANSWER:
72.30
TYPE: FI DIFFICULTY: Easy
KEYWORDS: twofactor analysis of variance, F test for factor, test statistic
122. True or False: Referring to Table 119, 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: twofactor analysis of variance, F test for factor, decision, conclusion
123. Referring to Table 119, the critical value in the test for significant differences between shapes is ________.
ANSWER:
3.89
TYPE: FI DIFFICULTY: Easy
KEYWORDS: twofactor analysis of variance, F test for factor, critical value
124. Referring to Table 119, the value of the statistic used to test for significant differences between shapes is ________.
ANSWER:
23.16
TYPE: FI DIFFICULTY: Easy
KEYWORDS: twofactor analysis of variance, F test for factor, test statistic
125. True or False: Referring to Table 119, 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: twofactor analysis of variance, F test for factor, decision, conclusion
126. Referring to Table 119, the critical value in the test for a significant interaction is ________.
ANSWER:
3.00
TYPE: FI DIFFICULTY: Easy
KEYWORDS: twofactor analysis of variance, F test for interaction, critical value
127. Referring to Table 119, the value of the statistic used to test for an interaction is ________.
ANSWER:
2.00
TYPE: FI DIFFICULTY: Easy
KEYWORDS: twofactor analysis of variance, F test for interaction, test statistic
128. True or False: Referring to Table 119, based on the results of the hypothesis test, it appears that there is a significant interaction.
ANSWER:
False
TYPE: TF DIFFICULTY: Easy
KEYWORDS: twofactor analysis of variance, F test for interaction, decision, conclusion
TABLE 1110
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 1110, 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 1110, the amonggroup variation or SSA is __________.
ANSWER:
82.39
TYPE: FI DIFFICULTY: Moderate
KEYWORDS: randomized block design, sum of squares
131. Referring to Table 1110, the amongblock variation or SSBL is __________.
ANSWER:
24.46
TYPE: FI DIFFICULTY: Moderate
KEYWORDS: randomized block design, sum of squares
132. Referring to Table 1110, 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 1110, the null hypothesis for the randomized block F test for the difference in the means is
ANSWER:
b
TYPE: MC DIFFICULTY: Easy
KEYWORDS: randomized block design, F test for factor, form of hypothesis
134. Referring to Table 1110, 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 1110, 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 1110, 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 1110, what is the pvalue of the test statistic for the randomized block F test for the difference in the means?
ANSWER:
6.07E05
TYPE: PR DIFFICULTY: Moderate
KEYWORDS: randomized block design, F test for factor, pvalue
138. True or False: Referring to Table 1110, 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 1110, 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 1110, 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 1110, 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 1110, 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 1110, 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 pairwise comparisons can be made?
ANSWER:
3
TYPE: PR DIFFICULTY: Moderate
KEYWORDS: randomized block design, Tukey procedure, properties
144. Referring to Table 1110, 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 1110, 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 1110, 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 1110, based on the TukeyKramer 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 1110, 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 1110, what is the null hypothesis for testing the block effects?
ANSWER:
a
TYPE: MC DIFFICULTY: Easy
KEYWORDS: randomized block design, F test for block effects, form of hypothesis
150. Referring to Table 1110, 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 1110, 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 1110, 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 1110, what is the pvalue 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, pvalue
154. True or False: Referring to Table 1110, 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 1110, 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 1110, 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 1110, the relative efficiency means that 2.47 times as many observations in each variety group would be needed in a oneway 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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