Ch 17 Statistica

CHAPTER 17

Analysis of Variance

Multiple-Choice Questions

1. When doing an analysis of variance, where each population is assumed to have the same variance, we are more likely to reject the null hypothesis of equality of population means when there is:

A) high variability among (between) the sample means and high variability around within the sample means.

B) low variability among (between) the sample means and low variability around within the sample means.

C) high variability among (between) the sample means and low variability around within the sample means.

D) low variability among (between) the sample means and high variability around within the sample means.

ANSWER: C

2. In one-way ANOVA with n observations and k independent samples, the within-groups sum of squares is

A) referred to as the between-groups sum of squares, denoted by SSG.B) referred to as the error sum of squares, denoted by SSW.C) calculated by dividing SSG by (k-1).D) calculated by dividing SSW by (n-k).ANSWER: B

3. In a two-way ANOVA, with one observation per cell, if there are 5 groups and 3 blocks, then the total number of observations is

A) 15.B) 13.C) 11.D) 8.ANSWER: A

QUESTIONS 4 THROUGH 9 ARE BASED ON THE FOLLOWING INFORMATION:

473

Chapter 17

An analysis of variance table is displayed below with missing values, denoted by asterisks.

ANOVA: Single Factor

SUMMARY

Groups Count Sum Average Variance

Column 1 7 82 “ 14.238

Column 2 6 81 13.5 4.7

Column 3 7 112 16 10.667

Column 4 6 80 13.333 11.467

ANOVA

Source of Variation SS df MS F P-value F critical

Between Groups “ 3 21.874 “ “ 3.0491

Within Groups 230.262 22 “

Total 295.884 “

4. The average for group 1 is equal to:

A) 13.708.B) 12.433.C) 13.256.D) 11.714.ANSWER: D

5. The between groups sum of squares is equal to:

A) 76.801.B) 65.622.C) 58.718.D) 72.329.ANSWER: B

6. The total number of degrees of freedom is equal to:

A) 25.B) 23.C) 24.D) 26.ANSWER: A

7. The mean squares within groups is equal to

A) 10.937.B) 10.466.C) 11.408.D) 11.173.ANSWER: B

8. The value of the test statistic F is equal to:

474


A) 1.95.B) 1.89.C) 2.09.D) 2.04.ANSWER: C

9. The p-value is approximately equal to:

A) 0.01.B) 0.05.C) 0.005.D) 0.13.ANSWER: D

QUESTIONS 10 THROUGH 16 ARE BASED ON THE FOLLOWING INFORMATION:The results of a two-factor ANOVA without replication are displayed below with missing values, identified by “A” through “G “.

East South Midwest West

Low 87.3 73.2 82.3 76.0

Medium 84.3 69.4 83.2 75.2

High 74.5 72.3 73.4 68.4

ANOVA: Two-Factor Without Replication

SUMMARY

Count Sum Average Variance

Low 4 *A* 79.7 40.15333

Medium 4 78.025 49.50917

High 4 72.15 7.056667

East 3 82.03333 44.81333

South 3 71.63333 3.943333

Midwest 3 79.63333 29.34333

West 3 73.2 *B*

ANOVASource of Variation SS Df MS F P-value

Rows 125.765 2 62.8825 “F” 0.039939

Columns 224.843 3 74.946 “G” 0.02274

Error 65.315 “C” “E”

Total 415.923 “D”

10. What value should replace “A”?

475

Chapter 17

A) 19.92B) 318.8C) 225.4D) 64.46ANSWER: B

11. What value should replace “B”?

A) 21.56B) 9.29C) 23.64D) 17.44ANSWER: D

12. What value should replace “C”?

A) 12B) 9C) 6D) 3ANSWER: C

13. What value should replace “D”?

A) 11B) 12C) 13D) 10ANSWER: A

14. What value should replace “E”?

A) 11.344B) 10.886C) 9.568D) 8.822ANSWER: B

15. What value should replace “F”?

A) 5.777B) 6.885C) 6.331D) 4.695ANSWER: A

16. What value should “G” have?

A) 4.695B) 5.777C) 6.331D) 6.885ANSWER: D

UESTIONS 17 THROUGH 19 ARE BASED ON THE FOLLOWING INFORMATION:

476


The results of a two-factor ANOVA without replication are displayed below:


A 739 791 733 779

B 726 748 743 754

C 730 780 755 803


SUMMARYCount Sum Average Variance

A 4 3042 760.5 830.333

B 4 2971 742.75 144.917

C 4 3068 767 992.667

East 3 2195 731.667 44.333

South 3 2319 773 499

Midwest 3 2231 743.667 121.333

West 3 2336 778.667 600.333

ANOVA

Source of Variation SS df MS F P-value F critical

Rows 1260.5 2 630.25 2.9787 0.1263 5.143249

Columns 4634.25 3 1544.75 7.3009 0.0199 4.757055

Error 1269.5 6 211.583

Total 7164.25 11

17. What is the average over all rows and columns?

A) 753.25B) 756.75C) 755.55D) 751.25ANSWER: B

18. Is there sufficient evidence to reject that the average between different regions is the same?

A) Reject for = 0.01.

B) Reject for = 0.005.

C) Reject for = 0.05.

D) Unable to reject for < 0.10.ANSWER: C

477

Chapter 17

19. Is there sufficient evidence to reject that the average between programs A, B, and C regions is the same?

A) Unable to reject for < 0.10.



D) Reject for = 0.05.ANSWER: A

QUESTIONS 20 THROUGH 29 ARE BASED ON THE FOLLOWING INFORMATION:Four stock analysts were asked to predict earnings for three different stocks for the next year. The following ANOVA table with missing values identified by “A through “H” summarizes the responses,

20. What is the value of “A”?


21. What is the value of “B”?

A) 6B) 5C) 3D) 2ANSWER: D

22. What is the value of “C”?

A) 6B) 5C) 3D) 2ANSWER: A

23. What is the value of “D”?

A) 15.20B) 23.24C) 12.77D) 38.44ANSWER: B

24. What is the value of “E”?

478

Source of Variation SS df MS FBetween Analysts 69.72 “A” “D” “G”Between Stocks 30.40 “B” “E” “H”Error 76.64 “C” “F”


A) 15.20B) 23.24C) 12.77D) 38.44ANSWER: A

25. What is the value of “F”?

A) 15.20B) 23.24C) 12.77D) 38.44ANSWER: C

26. What is the value of “G”?

A) 1.350B) 1.192C) 1.421D) 1.820ANSWER: D

27. What is the value of “H”?

A) 1.350B) 1.190C) 1.421D) 1.823ANSWER: B

28. Is there sufficient evidence to reject that there is no difference between the analysts’ forecasts?

A) Reject for = 0.05.



D) Unable to reject for < 0.10.ANSWER: D

29. Is there sufficient evidence to reject that there is no difference between the stocks?

A) Unable to reject for < 0.10.



D) Reject for = 0.01.ANSWER: A

479

Chapter 17

QUESTIONS 30 THROUGH 39 ARE BASED ON THE FOLLOWING INFORMATION:A consumer group wanted to examine how long different life bulbs last. They selected three different light bulbs (60 watt, 75 watt and 100 watt) from four different manufacturers. They generated the following statistics from the data.

Source of Variation SS df MS FBetween types of bulbs 1454 “A” “D” “G”Between manufacturers 739 “B” “E” “H”Error 1193 “C” “F”

30. What is the value of A?

A) 7B) 5C) 3D) 2ANSWER: D

31. What is the value of B?


32. What is the value of C?

A) 6B) 4C) 2D) 1ANSWER: A

33. What is the value of D?

A) 484.67B) 727.00C) 246.99D) 198.83ANSWER: B

34. What is the value of E?

A) 484.67B) 727.00C) 246.33D) 198.83ANSWER: C

35. What is the value of F?

480


A) 484.67B) 727.00C) 246.33D) 198.83ANSWER: D

36. What is the value of G?

A) 1.24B) 3.66C) 2.95D) 0.81ANSWER: B

37. What is the value of H?

A) 3.66B) 2.95C) 1.24D) 0.81ANSWER: C

38. Is there sufficient evidence to reject the null hypothesis that there is no difference between the different types of light bulbs?

A) Unable to reject for < 0.10

B) Reject for = 0.01

C) Reject for = 0.025

D) Reject for = 0.05ANSWER: A

39. Is there sufficient evidence to reject the null hypothesis that there is no difference between the different manufacturers of light bulbs?

A) Reject for = 0.01

B) Reject for = 0.25

C) Reject for = 0.05

D) Unable to reject for < 0.10ANSWER: D

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

A) reject H0 since there is sufficient evidence that all the means differ. B) reject H0 since there is sufficient evidence of a treatment effect.C) fail to reject H0 since there is no evidence of a difference between all the means.D) fail to reject H0 because a mistake has been made in the calculations.ANSWER: B

481

Chapter 17

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

A) Sum of squaresB) Degrees of freedomC) Mean squaresD) It is impossible to tell.ANSWER: C

42. A one-way analysis of variance design

A) has only one factor with several treatment groups.B) can have more than one factor, each with several treatment groups.C) has one factor and one block.D) has one factor and one block and multiple values.ANSWER: A

43. The F test statistic in a one-way ANOVA is given by

A) MSW / MSGB) SSW / SSGC) SSG / SSWD) MSG / MSWANSWER: D

44. The numerator and denominator degrees of freedom for the F test in a one-way ANOVA with n observations and K groups are, respectively, given by

A) (n – K) and (K – 1).B) (K – 1) and (n – K).C) (K – n) and (n – 1).D) (n – 1) and (K – n).ANSWER: B

45. In a one-way ANOVA, the null hypothesis is always written in a way to reflect that

A) there is no treatment effect.B) there is some treatment effect.C) all the population means are different.D) some of the population means are different.ANSWER: A

46. A nonparametric alternative to the one-way analysis of variance test is known as the:

A) Wilcoxon signed rank test for paired samples.B) Wilcoxon rank sum test.C) Wilcoxon signed rank test for one sample.D) Kruskal-Wallis test.

ANSWER: D

482


47. 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 five 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, 1Package 3: 10, 9, 6, 10, 12, Package 4: 7, 6, 6, 15, 12

A) F test for differences in variances.B) One-way ANOVA F test.C) t test for the differences in means.D) t test for the mean difference between matched pairs.ANSWER: B

48. 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: 10, 1.5, 1.1, 1.3B: 2.5, 2.1, 1.9, 1.6 D: 0.8, 1.3, 1.1, 0.7Interpret the results of the analysis summarized in the following table:

Source of Variation SS df MS F P-value

Neighborhoods 2.97 3 0.990 8.31 0.0260

Error 1.43 12 0.119

Total 4.40 15

A) At the 0.05 level of significance, the mean ratios for the four neighborhoods are not all the same.

B) At the 0.01 level of significance, the mean ratios for the four neighborhoods are not all the same.

C) At the 0.10 level of significance, the mean ratios for the four neighborhoods are not significantly different.

D) At the 0.05 level of significance, the mean ratios for the four neighborhoods are not significantly different from 0.

ANSWER: A

49. In a Kruskal-Wallis test at the 5% significance level, there are four samples and the value of the test statistic is calculated as W = 8.79. The most accurate statement that can be made about the p-value is that it is:

A) greater than 0.10B) greater than 0.05 but smaller than 0.10C) greater than 0.05D) greater than 0.025 but smaller than 0.05ANSWER: D

483

Chapter 17

50. A randomized block design ANOVA has five treatments and four blocks. The computed test statistic (value of F) is 4.35. With a 0.05 significance level, the appropriate table value and conclusion will, respectively, be:

A) 14.37; fail to reject the null hypothesis.B) 3.11; fail to reject the null hypothesis.C) 3.26; fail to reject the null hypothesis.D) 3.26; reject the null hypothesis.ANSWER: D

51. A randomized block experiment having five treatments and six blocks produced the following values: SST = 1446, SSG = 287 and SSE = 180. The value of SSB must be:

A) 30B) 979C) 467D) 20ANSWER: B

52. Three tennis players, a beginner, an intermediate, and advanced, have been randomly selected from the membership of a racquet facility club in a large city. Using the same tennis ball, each player hits ten serves, one with each of three racquet models, with the three racquet models selected randomly. The speed of each serve is measured with a machine and the result recorded. Among the ANOVA models listed below, the most likely model to fit this situation is the:

A) one-way ANOVAB) two-way ANOVAC) randomized block designD) None of the aboveANSWER: C

53. A randomized block design with 4 treatments and 5 blocks produced the following sum of

squares values: SST = 1951, SSG = 349, SSB = 1414. The value of SSE must be:

A) 188B) 537C) 1602D) 1763ANSWER: A

484


True-False Questions54. When doing an analysis of variance, we are more likely to reject the null hypothesis of

equality of population means when there is low variability between the sample means. ANSWER: F

55. When doing an analysis of variance, we are more likely to reject the null hypothesis of equality of population means when there is low variability around the sample means. ANSWER: T

56. The Kruskal - Wallis test is used in two-way ANOVA when we have reason to believe the

parent distribution is not normal. ANSWER: F

57. Consider a two-way ANOVA with multiple observations per cell. It is possible to reject the null hypothesis of no interaction, but be unable to reject the null hypothesis of no effect arising from either the block variable or the group variable. ANSWER: T

58. Three racquetball players, a beginner, an experienced, a professional, have been randomly selected from the membership of a large city racquetball club. Using the same ball, each person hits three serves, one with each of five racquet models, with the five racquet models selected randomly. Each serve is clocked with a radar gun and the result recorded. Among ANOVA models, this setup is most like the completely randomized design.ANSWER: F

59. The Kruskal-Wallis test is employed to test the equality of population means when an investigator has strong grounds for suspecting that the parent population distributions may be markedly different from the normal.ANSWER: T

60. In order to calculate the Kruskal-Wallis test, the values within each sample are ranked in ascending order, using the average of adjacent ranks in the case of ties.ANSWER: F

61. In one-way ANOVA, if the ratio F =MSG / MSW is quite close to 1.0, there is little cause to doubt the null hypothesis of equality of population mean.ANSWER: T

62. In one-way ANOVA, if the variability between groups is large compared to the variability within groups, we suspect the null hypothesis of equality of population means to be false.ANSWER: T

63. The Kruskal-Wallis test procedure is approximately valid, provided that the sample contains at least ten observations from each population.ANSWER: F

64. In a two-way ANOVA, with one observation per cell, the total sum of squares, SST, can be decomposed into three components: between – blocks sum of squares, SSB, between – groups sum of squares, SSG, and error sum of squares, SSE.ANSWER: T

65. In a two-way ANOVA, with one observation per cell, if there are 4 groups and 3 blocks, then there is a total of 7 observations.ANSWER: F

485

Chapter 17

66. A study will be undertaken to examine the effect of two kinds of background music and of two assembly methods on the output of workers at a fitness shoe factory. Two workers will be randomly assigned to each of four groups, for a total of eight in the study. Each worker will be given a headphone set so that the music type can be controlled. The number of shoes completed by each worker will be recorded. Does the kind of music or the assembly method or a combination of music and method affect output? The ANOVA model most likely to fit this situation is the Spearman rank correlation model.ANSWER: F

67. In a two-way ANOVA, a design with only one observation per cell allows the isolation of a further source of variability – the interaction between groups and bocks.ANSWER: F

68. In a two-way ANOVA, interaction between groups and blocks occur when differences in group efforts are not distributed uniformly across blocks.ANSWER: T

69. Three racquetball players, a beginner, an experienced, a professional, have been randomly selected from the membership of a large city racquetball club. Using the same ball, each person hits three serves, one with each of five racquet models, with the five racquet models selected randomly. Each serve is clocked with a radar gun and the result recorded. Among ANOVA models, this setup is most like the two-way ANOVA.ANSWER: F

70. In a two-way ANOVA, with more than one observation per cell, the total sum of squares, SST, can be decomposed into four components: between – groups sum of squares, SSG, between – blocks sum of squares, SSB, interaction sum of squares, SSI, and error sum of squares, SSE.ANSWER: T

71. In a two-way ANOVA, with one or more than one observation per cell, there are three tests of hypothesis that can be carried out with reference to the F-distribution with the corresponding numerator and denominator degrees of freedom.ANSWER: F

72. The analysis of variance (ANOVA) tests hypotheses about the population variance.ANSWER: F

73. The F test in one-way ANOVA model is just an expansion of the t test for independent samples.ANSWER: T

74. Three racquetball players, a beginner, an experienced, a professional, have been randomly selected from the membership of a large city racquetball club. Using the same ball, each person hits three serves, one with each of five racquet models, with the five racquet models selected randomly. Each serve is clocked with a radar gun and the result recorded. Among ANOVA models, this setup is most like the randomized block design.ANSWER: T

75. Subject to the value of the sample sizes, the MSE can be negative (especially for very small sample sizes) or positive.ANSWER: F

486


76. A study will be undertaken to examine the effect of two kinds of background music and of two assembly methods on the output of workers at a fitness shoe factory. Two workers will be randomly assigned to each of four groups, for a total of eight in the study. Each worker will be given a headphone set so that the music type can be controlled. The number of shoes completed by each worker will be recorded. Does the kind of music or the assembly method or a combination of music and method affect output? The ANOVA model most likely to fit this situation is the one-way analysis of variance.ANSWER: F

77. When the F test is used for ANOVA, the rejection region is always in the right tail.ANSWER: T

78. The critical value of the Kruskal-Wallis test for comparing more than two independent samples is obtained from the chi-square distribution whenever each sample size is at least 5.ANSWER: T

79. A one-sample t-test is the parametric counterpart to the Kruskal-Wallis testANSWER: F

80. Analysis of variance (ANOVA) is a set of techniques that allow us to compare two or more sample means at the same time.ANSWER: T

81. The Mann-Whitney test is a nonparametric alternative to the one-way analysis of variance.ANSWER: F

82. Computer assistance is especially useful in ANOVA because the calculations can be quite extensive even for small amounts of data.ANSWER: T

83. In the one-way ANOVA test, if the sample sizes are large, then the assumption that the underlying population distributions are normal is not necessary.ANSWER: F

84. When there are two treatment groups, one-way ANOVA is equivalent to the pooled-variances t-test for difference between means.ANSWER: T

85. A study will be undertaken to examine the effect of two kinds of background music and of two assembly methods on the output of workers at a fitness shoe factory. Two workers will be randomly assigned to each of four groups, for a total of eight in the study. Each worker will be given a headphone set so that the music type can be controlled. The number of shoes completed by each worker will be recorded. Does the kind of music or the assembly method or a combination of music and method affect output? The ANOVA model most likely to fit this situation is the simple regression model.ANSWER: F

86. In ANOVA, if the calculated F-statistic exceeds the critical F for a given test, the null hypothesis of equal population means is rejected.ANSWER: T

487

Chapter 17

87. If we simultaneously examine the effects of two factors on the dependent variable, along with the effects of interactions between the different levels of those factors, we are performing two-way ANOVA.ANSWER: T

88. Basic to ANOVA is the comparison of variation between samples versus the amount of variation within the samples. The test statistic is an F-ratio in which the numerator reflects variation within the samples and the denominator reflects the variation between them.ANSWER: F

89. Like the majority of the nonparametric tests, the Kruskal-Wallis test is based the ranks of the sample observations.ANSWER: T

90. A study will be undertaken to examine the effect of two kinds of background music and of two assembly methods on the output of workers at a fitness shoe factory. Two workers will be randomly assigned to each of four groups, for a total of eight in the study. Each worker will be given a headphone set so that the music type can be controlled. The number of shoes completed by each worker will be recorded. Does the kind of music or the assembly method or a combination of music and method affect output? The ANOVA model most likely to fit this situation is the two-way analysis of variance.ANSWER: T

488


Basic and Applied Questions

QQUESTIONS 91 THROUGH 98 ARE BASED ON THE FOLLOWING INFOEMATION:A hotel chain has identically sized resorts in five locations. The data that follow resulted from analyzing the hotel occupancies on randomly selected days in the five locations.

Caymen Pennkamp California Mayaguez Maui28 40 21 37 2233 35 21 47 1941 33 27 45 25


91. What are the appropriate null and alternative hypotheses?

ANSWER:

for at least one pair .

92. What is the value of the element in the ANOVA table that always provides an estimate of the population variance?

ANSWER:MSE = 21.0

93. The test statistic F is the ratio of two elements of the ANOVA table. Of these elements, what is the value of the one that provides an estimate of the population variance only when the null hypothesis is true?

ANSWER:MSG = 240.9

94. What is the value that represents the between group variation?

ANSWER:SSG = 963.6

95. What is the value that represents the within group variation?

ANSWER:SSW = 210.0

489


Location 963.6 4 240.9 11.47 0.001

Error 210.0 10 21.0

Total

Chapter 17

96. If a level of significance of 0.05 is chosen, what is your conclusion?

ANSWER:Since p-value = 0.001 < = 0.05, the null hypothesis is rejected. We conclude that at least two of the population means are not equal.

97. What is the value of the total variation?

ANSWER:SST = 963.6 + 210.0 = 1173.6

98. What are the numerator and denominator degrees of freedom of the test ratio in testing the hypotheses in Question 50?

ANSWER:The numerator and denominator degrees of freedom are 4 and 10, respectively.

QUESTIONS 99 THROUGH 103 ARE BASED ON THE FOLLOWING INFORMATION:Consider the following analysis of variance table:

Source of Variation Sum of Squares Degrees of FreedomBetween groups 1280 4Within groups 780 15Total 2060 19


ANSWER:


100. Compute mean squares for between groups

ANSWER:MSG = SSG / (K– 1) = 1280 / 4 = 320

101. Compute mean squares for within groups.

ANSWER:MSW = SSW / (n – K) = 780 / 15 = 52

102. Compute the F- ratio

ANSWER:F = MSG / MSW = 320 / 52 = 6.154

103. Test the null hypothesis in Question 102 at the 5% level of significance.

ANSWER:

Reject if 3.06. Therefore, is rejected. We conclude that at least two

of the population means are not equal.

490


QUESTIONS 104 THROUGH 106 ARE BASED ON THE FOLLOWING INFORMATION:A machine press manufacturer is comparing orders coming from different regions in the country and has collected the following data from 20 orders.


7 5 2 5

9 6 4 7

9 5 6 9

8 7 6 7

10 7 6

4

104. Develop the analysis of variance table.

ANSWER:


SUMMARY


Column 1 4 33 8.25 0.9167

Column 2 6 37 6.1667 4.5667

Column 3 5 25 5 4

Column 4 5 34 6.8 2.2

ANOVA


Between Groups 24.5667 3 8.1889 2.6005 0.088

Within Groups 50.3833 16 3.1490

Total 74.95 19

105. Is there sufficient evidence to reject the null hypothesis that the average number of machines per order is the same for different regions of the country?

ANSWER:


Since p-value = 0.088, we fail to reject for all levels of significance greater than 8.8%, and conclude that there is sufficient statistical evidence that the average number of machines per order is the same for the four different regions of the country.

491

Chapter 17

106. Carry out the Kruskal-Wallis test for this data using statistical software and write your conclusion?

ANSWER:

Since p-value = 0.0949, we fail to reject for all levels of significance greater than 9.49%, and conclude that there is sufficient statistical evidence that the average number of machines per order is the same for the four different regions of the country.

QUESTIONS 107 THROUGH 109 ARE BASED ON THE FOLLOWING INFORMATION:A marketing manager is considering adopting one of three new marketing schemes. He chooses three sales people in each region, and has them try the new schemes. He then records the resulting sales as shown below.

MarketingSchemes


A 4 5 7 4B 7 6 8 9C 6 7 9 5

492

Level of Significance 0.05

Group 1Sum of Ranks 65Sample Size 4


Group 3Sum of Ranks 33.5Sample Size 5


Sum of Squared Ranks/Sample Size 2427.95Sum of Sample Sizes 20Number of groups 4W Test Statistic 6.37

Critical Value 7.8147p -Value 0.0949

Intermediate Calculations

Orders Data

Test Result


107. Perform a two-way ANOVA with this data.

ANSWER:

108. Test at the 10% level of significance the null hypothesis that the population mean sales are the same for the four regions.

ANSWER:

for at least one pair

Since p-value = 0.1942, we fail to reject at the 10% level of significance, and conclude that the population mean sales are the same for the four regions.

109. Test at the 10% level of significance the null hypothesis that the population mean sales are the same for the three marketing schemes.

ANSWER:


Since p-value = 0.0736, we reject at the 10% level of significance, and conclude that there is difference in sales for at lease two of the three marketing schemes.

493


SUMMARY

Count Sum Average Variance

A 4 20 5 2

B 4 30 7.5 1.6667

C 4 27 6.75 2.9167

East 3 17 5.6667 2.3333

South 3 18 6 1

Midwest 3 24 8 1

West 3 18 6 7

ANOVA


Rows 13.1667 2 6.5834 4.1580 0.0736

Columns 10.25 3 3.4167 2.1580 0.1942

Error 9.5 6 1.5833

Total 32.9167 11

Chapter 17

QUESTIONS 110 THROUGH 112 ARE BASED ON THE FOLLOWING INFORMATION:Five stock analysts were asked to predict earnings for four different stocks for the next year. The following statistics summarize their responses:

Source of Variation Sum of SquaresBetween Analysts 74.6Between Stocks 35.7Error 82.2Total 192.5

110. Complete the ANOVA table.

ANSWER:

Source of Variation SS df MS F-ratioBetween Analysts 74.6 4 18.65 2.723Between Stocks 35.7 3 11.90 1.737Error 82.2 12 6.85Total 192.5 19

111. Is there sufficient evidence to reject the null hypothesis that there is no difference between the analysts forecast?

ANSWER:

Since = 3.26 and F =2.723, we fail to reject at . There is no

sufficient statistical evidence to conclude that there is a difference between the analysts’ forecast.

112. Is there sufficient evidence to reject the null hypothesis that there is no difference between the stocks?

ANSWER:

Since =3.49 and F =1.737 we fail to reject . There is no sufficient

statistical evidence to conclude that there is a difference between the stocks.

QUESTIONS 113 THROUGH 115 ARE BASED ON THE FOLLOWING INFORMATION:A firm is considering adopting an employee screening exam and has three to choose from. The exams are administered to 10 different people, and their responses are recorded.


ANSWER:

Source of Variation SS df MS F-ratioBetween Exams 174.50 2 87.25 5.193Between Applicants 258.75 9 28.75 1.711Error 302.40 18 16.8Total 29

494

Source of Variation Sum of SquaresBetween Exams 174.50Between Applicants 258.75Error 302.40


114. Is there sufficient evidence to reject the null hypothesis that there is no difference between the three exam forms? Use = 0.05

ANSWER:

Since = 3.55 and F = 5.193, we reject at = 0.05 and conclude that there

is sufficient statistical evidence that there a difference between at least two of the three exam forms.

115. Is there sufficient evidence to reject the null hypothesis that there is no difference between the applicants? Use = 0.05

ANSWER:

Since = 2.46 and F =1.711, we fail to reject , and conclude that

there is no difference between the applicants.

QUESTIONS 116 THROUGH 119 ARE BASED ON THE FOLLOWING INFORMATION:Four different brokerage houses were asked for stocks earnings forecasts for the next year for five different corporations. The brokerage houses asked the analysts on staff who had experience with the corporations in question. Four analysts were surveyed at each brokerage house. Some descriptive statistics are listed below:

Source of Variation Sum of SquaresBetween brokerage houses 240.3Between stocks’ earnings forecasts 122.5Interaction 20.3Error 325.3


ANSWER:

117. Is there sufficient evidence to reject the null hypothesis that there is no difference between the brokerage houses? Use = 0.05

ANSWER:

Since =2.76 and F =14.773, we reject .There is sufficient statistical evidence

to conclude that there is a difference between at least two of the four brokerage houses.

495

Source of Variation SS df MS F-ratioBetween brokerage houses 240.3 3 80.1 14.773Between stocks’ earnings forecasts 122.5 4 30.625 5.648Interaction 20.3 12 1.692 0.312Error 325.3 60 5.422Total 708.4 79

Chapter 17

118. Is there sufficient evidence to reject the null hypothesis that there is no difference between the stocks’ earning forecasts? Use = 0.05

ANSWER:

Since =2.53 and F = 5.648, we reject at = 0.05. There is sufficient

statistical evidence to conclude that there is a difference between at least two of the five stocks’ earning forecasts.

119. Is there sufficient evidence to reject the null hypothesis that there is no interaction between brokerage houses and stocks’ earning forecasts? Use = 0.05

ANSWER:

Since =1.92 and F = 0.312, we fail to reject at , and conclude that

there is no interaction between the brokerage houses and the stocks’ earnings forecasts.

QUESTIONS 120 THROUGH 123 ARE BASED ON THE FOLLOWING INFORMATION:A sales manager is interested in evaluating the effectiveness of product literature. There are five different types of brochures currently in use. He asks four salespeople in each of the four regions of the country to evaluate the effectiveness of each of the brochures. The results of a two-way ANOVA are presented below.

Source of Variation Sum of SquaresRegion of country 36.40Brochure type 13.40Interaction 31.32Error 125.2


ANSWER:

Source of Variation SS df MS F-ratioRegion of country 36.40 3 12.133 5.814Brochure type 13.40 4 3.35 1.605Interaction 31.32 12 2.61 1.251Error 125.2 60 2.087Total 206.32 79

121. Is there sufficient evidence to reject the null hypothesis that there is no difference between the regions of the country? Use = 0.05

ANSWER:

Since =2.76 and F = 5.814, we reject . There is sufficient statistical evidence

to conclude that differences exist in at least two regions of the country.

122. Is there sufficient evidence to reject the null hypothesis that there is no difference between different types of brochures? Use = 0.05

ANSWER:

496


Since =2.53 and F =1.605, we fail to reject at , and conclude that

there is no difference between the different types of brochures.

123. Is there sufficient evidence to reject the null hypothesis that there is no interaction between region of the country and brochure type? Use = 0.05

ANSWER:

Since =1.92 and F =1.251, we fail to reject at , and conclude that

there is no interaction between regions of the country and brochure type.

QUESTIONS 124 THROUGH 130 ARE BASED ON THE FOLLOWING INFORMATION:Consider the following analysis of variance table with missing values identified by asterisks.

Source of Variation Sum of Squares Degrees of Freedom

Between groups 1,600 *Within groups * 15Total 2790 17

124. What are the degrees of freedom for between groups?

ANSWER:df for between groups = 17 – 15 = 2

125. What is the sum of squares for within groups?

ANSWER:SSW = SST – SSG = 2790 – 1600 = 1190


ANSWER:


127. Compute mean squares for between groups

ANSWER:MSG = SSG / (K– 1) = 1600 / 2 = 800

128. Compute mean squares for within groups.

ANSWER:MSW = SSW / (n – K) = 1190 / 15 = 79.333

129. Compute the F- ratio

ANSWER:F = MSG / MSW = 800 / 79.333 = 10.084

130. Test the null hypothesis in Question 124 at the 5% level of significance.

497

Chapter 17

ANSWER:

Reject if 3.68. Therefore, is rejected. We conclude that at least two

of the population means are not equal.

QUESTIONS 131 THROUGH 135 ARE BASED ON THE FOLLOWING INFORMATION:A production manager monitors the output per hour of four workers over the course of four hours. The results are reproduced below.

Worker A Worker B Worker C Worker D

7 5 5 6

3 3 8 2

6 4 8 3

6 2 7 5

Assume that the populations from which the sample drawn are normally distributed.

131. Perform the appropriate analysis of variance and report the results.

ANSWER:


SUMMARY


Worker A 4 22 5.5 3

Worker B 4 14 3.5 1.6667

Worker C 4 28 7 2

Worker D 4 16 4 3.3333

ANOVA


Between Groups 30 3 10 4 0.0346

Within Groups 30 12 2.5

Total 60 15

132. What are the appropriate null and alternative hypothesis using this information?

ANSWER:


133. Assume that the populations from which the samples are drawn are normally distributed, is there sufficient evidence to reject the that there is no difference in the output per hour of the four workers?

498


ANSWER:Since p-value = 0.0346, we reject at =0.05 and conclude that there is sufficient statistical evidence that there is a difference in the output per hour for at least one pair of workers (possibly workers B and C or C and D).

134. Carry out the Kruskal-Wallis test for this data at =0.05 and write your conclusion.

ANSWER:

Since p-value = 0.0596, we fail to reject at = 0.05. We conclude that there is no difference in the output per hour of the four workers.

135. Compare your answers to Questions 133 and 134 and explain how to interpret these results.

ANSWER:In Question 133 we rejected at = 0.05, but failed to reject under same level of significance in Question 134. The reason is that under the assumption of normality in the populations, nonparametric tests such as Kruskal-Wallis are less powerful than parametric tests, such as ANOVA.

QUESTIONS 136 AND 137 ARE BASED ON THE FOLLOWING INFORMATION:Independent random samples of 100 college sophomores, 110 college juniors, and 95 college seniors were asked to rate, on a scale from 1 to 7, the importance attached to brand name when purchasing a computer. The value of Kruskal-Wallis statistic obtained was 0.75.

136. What null hypothesis can be tested using this information?

499






Sum of Squared Ranks/Sample Size 1324.25Sum of Sample Sizes 16Number of groups 4H Test Statistic 7.422794



Production Output Data

Test Result

Chapter 17

ANSWER:The null hypothesis tests the equality of the population mean ratings across the classes; that is,

137. Carry out the appropriate test for the null hypothesis in Question 136 at the 10% level.

ANSWER:W = 0.75

, therefore, do not reject the null hypothesis at the 10% level. We conclude

that the population mean ratings across the classes (sophomores, sophomores, and seniors) are the same.

QUESTIONS 138 THROUGH 140 ARE BASED ON THE FOLLOWING INFORMATION:Random samples of eight freshmen, eight sophomores, and eight juniors taking a business statistics class were drawn. The accompanying table shows scores on the final examination.

Freshmen Sophomores Juniors81 80 9073 74 8156 69 9077 97 9472 81 5996 88 7664 65 7885 74 67

138. Set out the analysis of variance table.

ANSWER:

139. Test the null hypothesis that the three population mean scores are equal. Use = 0.05

ANSWER:


500

Anova: Single Factor

SUMMARYGroups Count Sum Average VarianceFreshmen 8 604 75.5 153.4286Sophomores 8 628 78.5 107.7143Juniors 8 635 79.375 146.2679

ANOVASource of Variation SS df MS F P-value F crit

Between Groups 66.0833 2 33.04167 0.2433 0.7862 3.4668Within Groups 2851.8750 21 135.8036Total 2917.9583 23


Since p-value = 0.7862, we fail to reject at = 0.05, and conclude that the three population mean scores are equal

140. Carry out a nonparametric test of the null hypothesis of equality of population mean examination scores for freshman, sophomore, and juniors. Use = 0.05

ANSWER:We use the Kruskal-Wallis test as the nonparametric alternative to the one-way ANOVA test. The results are shown below





Sum of Squared Ranks/Sample Size 3781Sum of Sample Sizes 24Number of groups 3W Test Statistic 0.62



Exam Scores Data

Test Result

Again, since p-value = 0.7334, we fail to reject the null hypothesis at = 0.05, and conclude that the three population mean scores are equal

141. Consider a problem with three subgroups with the sum of ranks in each of the subgroups equal to 32, 101.5, and 76.5, with subgroup sizes equal to 7, 7, and 6, respectively. Complete the Kruskal-Wallis test and test the null hypothesis of equal subgroup ranks at the 1% level.

ANSWER:


Kruskal-Wallis test statistic is

= 11.097

501

Chapter 17

Reject if = 9.21. Therefore is rejected at the 1% kevel. We

conclude that at least two of the three population means are not the same.

QUESTIONS 142 THROUGH 144 ARE BASED ON THE FOLLOWING INFORMATION:Independent random samples of seven assistant professors, five associate professors, and six full professors were asked to estimate the amount of time outside the classroom spent on teaching responsibilities in the last week. Results, in hours, are shown in the accompanying table.

Assistant Associate Full12 14 915 9 810 16 1016 13 712 16 813 128


ANSWER:

143. Test the null hypothesis that the three population mean times are equal. Use = 0.05

ANSWER:


Since p-value = 0.0209, we reject at = 0.05, and conclude that at least two of the three population mean times are not the same.

144. Without assuming normal population distributions, test the null hypothesis that the population mean times spent outside the classroom on teaching responsibilities are the same for assistant, associate, and full professors.

ANSWER:

502

Anova: Single Factor

SUMMARYGroups Count Sum Average Variance

Assistant 7 86 12.28571 7.571429Associate 5 68 13.6 8.3Full 6 54 9 3.2

ANOVASource of Variation SS df MS F P-value F crit

Between Groups 63.8159 2 31.9079 5.0579 0.0209 3.6823Within Groups 94.6286 15 6.3086Total 158.4445 17


Without assuming normal population distributions, we will employ Kruskal-Wallis test as the nonparametric alternative to the one-way ANOVA test. The results are shown below.

Again, Since p-value = 0.0298, we reject at = 0.05, and conclude that at least two of the three population mean times are not the same.

QUESTIONS 145 THROUGH 150 ARE BASED ON THE FOLLOWING INFORMATION:Consider a two-way analysis of variance with one observation per cell and randomized blocks with the following results:

Source of Variation Sum of Squares Degrees of FreedomBetween groups 456.0 5Between blocks 278.4 5Error 464.4 25Total 1198.8 35

145. Compute the mean squares between groups

ANSWER:MSG = SSG / (K – 1) = 456 / 5 = 91.2

146. Compute the mean squares between blocks

ANSWER:MSB = SSB / (H – 1) = 278.4 / 5 = 55.68

147. Compute the error mean squares

ANSWER:MSE = SSE / [(K – 1) (H – 1)] = 464.4 / 25 = 18.576

503





Sum of Squared Ranks/Sample Size 1824.771Sum of Sample Sizes 18Number of groups 3W Test Statistic 7.027068



Times Data

Test Result

Chapter 17

148. Set up the two-way ANOVA table

ANSWER:

149. Test the hypotheses that between-group means are equal. Use = 0.05

ANSWER:


Reject if = 2.60. Since F = 4.1. is rejected at = 0.05. We conclude

that at least two of the six between-group means are not the same.

150. Test the hypotheses that between-block means are equal

ANSWER:


Reject if = 2.60. Since F = 3.0, is rejected at = 0.05. We conclude

that at least two of the six between-block means are not the same.

QUESTIONS 151 THROUGH 153 ARE BASED ON THE FOLLOWING INFORMATION:The data shown in the table below are collected from three samples. An investigator has strong grounds for suspecting that the parent population distributions may be markedly different from the normal. Sample

1 2 323 25 2522 27 2225 17 1920 19 2118 20 26

151. What is the appropriate statistical technique that can be used to analyze such data?

ANSWER:The Kruskal - Wallis test

504

Source of Variation SS df MS F-ratioBetween groups 456.0 5 91.2 4.1Between blocks 278.4 5 55.68 3.0Error 464.4 25 18.576Total 1198.8 35


152. Apply the technique in Question 158 and test to determine if there is enough evidence at the 5% significance level to infer that at least one of the population means differs from the others.

ANSWER:

At least one mean differs from the others

Reject if

Test statistic: W = 0.38; therefore, we fail to reject the null hypothesis. We may infer that the population means are equal.

153. Use statistical software to verify your answers to Question 152

ANSWER:

QUESTIONS 154 THROUGH 156 ARE BASED ON THE FOLLOWING INFORMATION:An instructor in an economics class is considering three different texts. She is also considering three types of examinations – multiple choice, essays, and a mix of multiple choice and essays. During the year she teaches nine sections of the course and randomly assigns a text– examination type combination of each section. At the end of the course she obtained students’ evaluations for each section. These ratings are shown in the accompanying table.

TextExamination A B CMultiple choice 5.2 4.7 4.8Essays 4.9 4.5 4.2

505





Sum of Squared Ranks/Sample Size 967.6Sum of Sample Sizes 15Number of groups 3H Test Statistic 0.38



Sample Data

Test Result

Chapter 17

Mix 5.0 4.5 4.7


ANSWER

155. Test the null hypothesis of equality of population mean ratings for the three texts. Use = 0.05

ANSWER:


Since p-value = 0.023, is rejected at = 0.05. We conclude that at least two of the population mean ratings for the three texts are not the same.

156. Test the null hypothesis of equality of population mean ratings for the three examination types. Use = 0.05

ANSWER:


Since p-value = 0.0772, is not rejected at = 0.05. We conclude that the population mean ratings for the three examination types are the same.

506


QUESTIONS 157 THROUGH 160 ARE BASED ON THE FOLLOWING INFORMATION:Random samples of two freshman, two sophomores, two juniors, and two seniors each from four dormitories were asked to rate on a scale from 1 (poor) to 10 (excellent) the quality of the dormitory environment for studying. The results are shown in the table.

DormitoryYear A B C DFreshman 7 5 8 6 9 8 9 9Sophomore 6 8 5 5 7 8 8 9Junior 5 4 7 6 6 7 7 8Senior 7 4 6 8 7 5 6 7

157. Set up the analysis of variance table.

ANSWER:

158. Test the null hypothesis that the population mean ratings are the same for the four dormitories.

ANSWER:


Since p-value = 0.0067, is rejected at = 0.05. We conclude that the population mean ratings for at least two of the four dormitories are not the same.

159. Test the null hypothesis that the population mean ratings are the same for the four student years.

ANSWER:


Since p-value = 0.0587, is rejected at = 0.05. We conclude that the population mean ratings for at least two of the four student years (Freshman, Sophomore, Junior, and Senior) are not the same.

507

Chapter 17

160. Test the null hypothesis of no interaction between student year and dormitory rating.

ANSWER: No interaction exists between year and dormitory ratings

There is interaction between year and dormitory ratings

Since p-value = 0.216, is not rejected at = 0.05. We conclude that No interaction exists between year and dormitory ratings

508

Documents

Ch 17 Statistica