14: analysis of variance -...

14: ANALYSIS OF VARIANCE

MULTIPLE CHOICE QUESTIONS

In the following multiplechoice questions, please circle the correct answer.

1. In oneway ANOVA, the amount of total variation that is unexplained is measured by the:a. sum of squares for treatmentsb. sum of squares for errorc. total sum of squaresd. degrees of freedomANSWER: b

2. The test statistic of the singlefactor ANOVA equalsa. Sum of squares for treatments / Sum of squares for errorb. Sum of squares for error / Sum of squares for treatmentsc. Mean square for treatments / Mean square for errord. Mean square for error / Mean square for treatmentsANSWER: c

3. Which of the following statements is false?a. The sum of squares for treatments (SST) explains some of the variation.b. The sum of squares for error (SSE) measures the amount of variation that is unexplained.c. SS(Total) = SST + SSEd. SS(Total) measures the amount of variation within the samples.ANSWER: d

4. In oneway ANOVA, suppose that there are four treatments with 51 =n , 62 =n , 53 =n , and 44 =n . Then the rejection region for this test at the 5% level of significance is

a. F > 0.025,4,20F

b. F > 0.05,4,20F

c. F > 0.025,3,16F

d. F > 0.05,3,16F

ANSWER: d

5. In an ANOVA test, the test statistic is F = 6.75. The rejection region is F > 3.97 for the 5% level of significance, F > 5.29 for the 2.5% level, and F > 7.46 for the 1% level. For this test, the pvalue is a. greater than 0.05b. between 0.025 and 0.05

Analysis of Variance

c. between 0.01 and 0.025d. approximately 0.05ANSWER: c

6. In a twotail pooledvariance ttest (equalvariances ttest), the null and alternative hypotheses are exactly the same as in oneway ANOVA witha. exactly one treatmentb. exactly two treatmentsc. exactly three treatmentsd. any number of treatmentsANSWER: b

7. Which of the following is not a required condition for oneway ANOVA?a. The sample sizes must be equalb. The populations must all be normally distributedc. The population variances must be equal.d. The samples for each treatment must be selected randomly and independentlyANSWER: a

8. The following equation applies to which ANOVA model?

SS(Total) = SS(A) + SS(B) + SS(AB) + SSE

a. Oneway ANOVAb. Twoway ANOVAc. Completely randomized designd. Randomized block designANSWER: b

9. The following equation applies to which ANOVA model?

SS(Total) = SST + SSB + SSE

a. Oneway ANOVAb. Twoway ANOVAc. Completely randomized designd. Randomized block designANSWER: d


14. The analysis of variance is a procedure that allows statisticians to compare two or more population

a. meansb. proportionsc. variancesd. standard deviationsANSWER: a

15. The distribution of the test statistic for analysis of variance is the:a. normal distributionb. Student t distributionc. F distributiond. chisquared distributionANSWER: c

17. Which of the following is not true of the Fdistribution?a. Mean and median are equalb. It is skewed to the rightc. Its values are always positived. It is used in ANOVA testANSWER: a

18. In a singlefactor analysis of variance, MST is the mean square for treatments and MSE is the mean square for error. The null hypothesis of equal population means is rejected if:a. MST is much smaller than MSEb. MST is much larger than MSEc. MST is equal to MSEd. none of the above is correctANSWER: b

19. If we want to conduct a test to determine whether a population mean is greater than another population mean, wea. can use the analysis of varianceb. must use the independent samples ttest for difference between two meansc. must use the chisquare testd. both a and b are correctANSWER: b

20. In ANOVA, error variability is computed as the sum of the squared errors, SSE, for all values of the response variable. This variability is the:a. the total variationb. withingroup variationc. betweengroups variationd. none of the aboveANSWER: b

21. In a oneway ANOVA where there are k treatments and n observations, the degrees of freedom for the Fstatistic are equal to:


a. n and kb. k and nc. nk and k1d. k1 and nkANSWER: d

22. Oneway ANOVA is applied to three independent samples having means 10, 13, and 18, respectively. If each observation in the third sample were increased by 30, the value of the Fstatistics would:a. increaseb. decreasec. remain unchangedd. increase by 30ANSWER: a

23. The Fstatistic in a oneway ANOVA represents the variation:a. between the treatments plus the variation within the treatmentsb. within the treatments minus the variation between the treatmentsc. between the treatments divided by the variation within the treatmentsd. variation within the treatments divided by the variation between the treatmentsANSWER: c

26. In the oneway ANOVA where k is the number of treatments and n is the number of observations in all samples , the degrees of freedom for error is given by:a. k1b. nkc. n1d. nk+1ANSWER: b

27. In the oneway ANOVA where k is the number of treatments and n is the number of observations in all samples , the degrees of freedom for treatments is given by:a. k1b. nkc. n1d. nk+1ANSWER: a


37. Oneway ANOVA is performed on independent samples taken from three normally distributed populations with equal variances. The following summary statistics were calculated:

=1n 7 =1x 65 =1s 4.2=2n 8 =2x 65 =2s 4.9 =3n 9 =3x 65 =3s 4.6

The value of the test statistics, F, equals:a. 65b. 24c. 13.7d. 0ANSWER: d

40. The degrees of freedom for the denominator of a oneway ANOVA test for 4 population means with 15 observations sampled from each population is:a. 60b. 19c. 56d. 45ANSWER: c

41. In oneway ANOVA, the term x refers to:a. sum of the sample meansb. sum of the sample means divided by the total number of observationsc. sum of the population meansd. weighted mean of the sample meansANSWER: d

43. Oneway ANOVA is performed on three independent samples with 61 =n , 72 =n , and 83 =n . The critical value obtained from the Ftable for this test at the 2.5% level of significance

equals:a. 3.55b. 39.45c. 4.56d. 29.45ANSWER: c

44. Which of the following is a correct formulation for the null hypothesis in oneway ANOVA?a. 0321 =++ µµµb. 0321 ≠++ µµµc. 321 µµµ ==d. 321 µµµ ≠≠ANSWER: c

45. Oneway ANOVA is performed on independent samples taken from three normally distributed populations with equal variances. The following summary statistics were calculated:


=1n 6 =1x 50 =1s 5.2=2n 8 =2x 55 =2s 4.9 =3n 6 =3x 51 =3s 5.4

The grand mean equalsa. 50.0b. 52.0c. 52.3d. 53.0ANSWER: c

46. Oneway ANOVA is applied to independent samples taken from three normally distributed populations with equal variances. The following summary statistics were calculated:

=1n 8 =1x 15 =1s 2=2n 10 =2x 18 =2s 3 =3n 8 =3x 20 =3s 2

The withintreatments variation equalsa. 137b. 460c. 154d. 60ANSWER: a

50. Oneway ANOVA is applied to independent samples taken from three normally distributed populations with equal variances. The following summary statistics were calculated:

=1n 10 =1x 40 =1s 5=2n 10 =2x 48 =2s 6 =3n 10 =3x 50 =3s 4

The betweentreatments variation equalsa. 460b. 688c. 560d. 183ANSWER: c

51. Oneway ANOVA is applied to independent samples taken from four normally distributed populations with equal variances. If the null hypothesis is rejected, then we can infer thata. all population means are equalb. all population means differc. at least two population means are equald. at least two population means differANSWER: d

52. Consider the following partial ANOVA table:


Source of Variation SS df MS FTreatments 75 * 25 6.67Error 60 * 3.75Total 135 19

The numerator and denominator degrees of freedom (identified by asterisks) area. 4 and 15b. 3 and 16c. 15 and 4d. 16 and 3ANSWER: b

53. In singlefactor analysis of variance, betweentreatments variation stands for:a. sum of squares for errorb. sum of squares for treatmentsc. total sum of squaresd. both a and b ANSWER: b

54. Consider the following ANOVA table:

Source of Variation SS df MS FTreatments 4 2 2.0 0.80Error 30 12 2.5Total 34 14

The number of treatments isa. 13b. 5c. 3d. 33ANSWER: c


55. In oneway analysis of variance, withintreatments variation stands for:a. sum of squares for errorb. sum of squares for treatmentsc. total sum of squaresd. none of the above is correctANSWER: a

56. Consider the following ANOVA table:

Source of Variation SS df MS FTreatments 128 4 32 2.963Error 270 25 10.8Total 398 29

The number of observations in all samples is:a. 25b. 29c. 30d. 32ANSWER: c

57. In oneway analysis of variance, if all the sample means are equal, thena. total sum of squares is zerob. sum of squares for error is zeroc. sum of squares for treatments is zerod. sum of squares for error equals sum of squares for treatmentsANSWER: c

58. In singlefactor analysis of variance, if large differences exist among the sample means, it is then reasonable toa. reject the null hypothesisb. reject the alternative hypothesisc. fail to reject the null hypothesisd. none of the above is correctANSWER: a

59. Which of the following is not a required condition for oneway ANOVA?a. The populations are normally distributedb. The population variances are equalc. The samples are selected independently of each otherd. The population means are equalANSWER: d


60. In oneway ANOVA, suppose that there are five treatments with 5321 === nnn , and 754 == nn . Then the mean square for error, MSE, equals

a. SSE / 4b. SSE / 29c. SSE / 24d. SSE / 5ANSWER: c


TEST QUESTIONS

118. Given the following data drawn from three normal populations:

Treatment

1 2 38 11 16

16 13 2013 12 1514 15 139 10

11

Set up the ANOVA table and test at the 5% level of significance to determine whether differences exist among the population means.

ANSWER:

Source of Variation SS df MS F Pvalue F criticalTreatments 46.933 2 23.467 3.200 0.0749 3.885Error 88.000 12 7.333Total 134.933 14

3210 : µµµ ==H

:1H At least two means differConclusion: Don’t reject the null hypothesis. No

121. The following statistics were calculated based on samples drawn from four normal populations:

Treatment

Statistic 1 2 3 4jn 4 7 5 5

∑ jx 52 69 71 61

∑ 2jx 753 798 1248 912

Test at the 5% level of significance to determine whether differences exist among the population means.

ANSWER:


Source of Variation SS df MS F F criticalTreatments 60.500 3 20.167 0.569 3.197Error 602.453 17 35.438Total 662.953 20

43210 : µµµµ ===H


125. Fill in the blanks (identified by asterisks) in the following partial ANOVA table:

Source of Variation SS df MS FTreatments * * 195 *Error 625 * *Total 1600 25

ANSWER:

Source of Variation SS df MS FTreatments 975 5 195 6.24Error 625 20 31.25Total 1600 24

126. A statistician employed by a television rating service wanted to determine if there were differences in television viewing habits among three different cities in California. She took a random sample of five adults in each of the cities and asked each to report the number of hours spent watching television in the previous week. From the data shown below, can she infer at the 5% significance level that differences in hours of television watching exist among the three cities?

Hours Spent Watching Television

San Diego Los Angeles San Francisco25 28 2331 33 1818 35 2123 29 1727 36 15

ANSWER:

Source of Variation SS df MS F Pvalue F critical


Treatments 450.533 2 225.267 14.659 0.0006 3.885Error 184.400 12 15.367Total 634.933 14

3210 : µµµ ==H

:1H At least two means differ Conclusion: Reject the null hypothesis. Yes

127. A pharmaceutical manufacturer has been researching new formulas to provide quicker relief of minor pains. His laboratories have produced three different formulas, which he wanted to test. Fifteen people who complained of minor pains were recruited for an experiment. Five were given formula 1, five were given formula 2, and the last five were given formula 3. Each was asked to take the medicine and report the length of time until some relief was felt. The results are shown below. Do these data provide sufficient evidence to indicate that differences in the time of relief exist among the three formulas? Use α = 0.05.

Time in Minutes Until Relief is Felt

Formula 1 Formula 2 Formula 34 2 68 5 76 3 79 7 88 1 6

ANSWER:


3210 : µµµ ==H

:1H At least two means differConclusion: Reject the null hypothesis. Yes


129. The strength of a weld depends to some extent on the metal alloy used in the welding process. A scientist working in the research laboratory of a major automobile manufacturer has developed three new alloys. In order to test their strengths each alloy is used in several welds. The strengths of the welds are then measured with the results shown below. Can the scientist conclude at the 5% significance level that differences exist among the strengths of the welds with the different alloys?

Strength of Welds

Alloy 1 Alloy 2 Alloy 315 17 2523 21 2716 19 2429 25 31

28 2319

ANSWER:


3210 : µµµ ==H



32. The marketing manager of a pizza chain is in the process of examining some of the demographic characteristics of her customers. In particular, she would like to investigate the belief that the ages of the customers of pizza parlors, hamburger emporiums, and fastfood chicken restaurants are different. As an experiment, the ages of eight customers of each of the restaurants are recorded and listed below. Do these data provide enough evidence at the 5% significance level to infer that there are differences in ages among the customers of the three restaurants? From previous analyses we know that the ages are normally distributed.

Customers’ Ages

Pizza Hamburger Chicken23 26 2519 20 2825 18 3617 35 2336 33 3925 25 2728 19 3831 17 31

ANSWERS:


3210 : µµµ ==H



QUESTIONS 133 THROUGH 135 ARE BASED ON THE FOLLOWING INFORMATION:

In order to examine the differences in ages of teachers among five school districts, an educational statistician took random samples of six teachers’ ages in each district. The data are listed below.

Ages of Teachers among Five School District

1 2 3 4 541 39 36 45 5353 48 28 37 5528 41 29 46 4945 51 33 48 5640 49 27 51 4859 50 26 49 61

133. Test at the 5% significance level to determine if differences in teachers’ ages exist among the five districts.

ANSWER:


543210 : µµµµµ ====H

:1H At least two means differConclusion: Reject the null hypothesis. Yes

143. In a completely randomized design, 15 experimental units were assigned to each of four treatments. Fill in the blanks (identified by asterisks) in the partial ANOVA table shown below.

Source of Variation SS df MS FTreatments * * 240 *Error * * *Total 2512 *

ANSWER:

Source of Variation SS df MS FTreatments 720 3 240 7.5Error 1792 56 32Total 2512 59

144. A statistics professor has carried out a study to compare different teaching methods used in three different sections of an elementary statistics course. A sample of students have been randomly


selected form each section, and their grades in the final test, as shown below, are used to determine whether the teaching methods made any difference.

Method 1 Method 2 Method 384 78 9770 85 8972 93 8167 66 73

99 77

Can we infer at the 5% significance level that the population means of the three methods differ?

ANSWER:

Source of Variation SS df MS FTreatments 319.607 2 159.804 1.459Error 1,204.75 11 109.523Total 1,524.357 13

3210 : µµµ ==H

:1H At least two means differ Rejection region: F > 98.311,2,05. =F

Conclusion: Don’t reject the null hypothesis. No


QUESTIONS 150 AND 151 ARE BASED ON THE FOLLOWING INFORMATION:

In a completely randomized design, 7 experimental units were assigned to the first treatment, 13 units to the second treatment, and 10 units to the third treatment. A partial ANOVA table for this experiment is shown below:

Source of Variation SS df MS FTreatments * * * 1.50Error * * 4Total * *

150. Fill in the blanks (identified by asterisks) in the above ANOVA. Table.

ANSWER:

Source of Variation SS df MS FTreatments 12 2 6 1.50Error 108 27 4Total 130 29

151. Test at the 5% significance level to determine if differences exist among the three treatment means.

ANSWER:3210 : µµµ ==H

:1H At least two means differ

Rejection region: F > 0.05,2,27 3.35F =Test statistics: F = 1.50Conclusion: Don’t reject the null hypothesis. No


QUESTIONS 158 THROUGH 161 ARE BASED ON THE FOLLOWING INFORMATION:

An insurance company is considering opening a new branch in Lansing. The company will choose the final location from two locations within the city. One of the factors in the decision is the annual family income (in thousands of dollars) of five families randomly sampled from a radius of five miles from the potential locations.

Area 1 Area 273 7448 5046 8153 4951 61

158. Perform equalvariances ttest at the 5% significance level to determine whether the population means differ.

ANSWER:210 : µµ =H

211 : µµ ≠H

Rejection region: | t | > 0.025,8t = 2.306 Test statistics: t = 1.098

Conclusion: Don’t reject the null hypothesis. No

159. Perform an Ftest for oneway ANOVA at the 5% significance level to determine whether the population means differ.

ANSWER:

Source of Variation SS df MS FTreatments 193.6 1 193.6 1.206Error 1,284.8 8 160.6Total 1,478.4 9

Rejection region: F > 0.05,1,8F = 5.32 Test statistic: F = 1.206 Conclusion: Don’t reject the null hypothesis. No

160. What is the relation between the observed t and observed F test statistics from Questions 158 and 159? Does the same relation hold true for the corresponding critical values?

ANSWER:Ft ==−= 206.1)098.1()( 22

Yes, since 2 20.025,8 0.05,1,8( ) (2.306) 5.32t F= = =


161. If we want to determine whether the population mean income for area 2 is higher than that for area 1, can we still use both tests applied in Questions 158 and 159? Explain.

ANSWER:No. We must use the equal variances ttest of 21 µµ − . We cannot use the analysis of variance Ftest since this technique only allows us to test for a difference.

QUESTIONS 162 AND 163 ARE BASED ON THE FOLLOWING INFORMATION:

An investor studied the percentage rates of return of three different types of mutual funds. Random samples of percentage rates of return for four periods were taken from each fund. The results appear in the table below:

Mutual Funds Percentage Rates

Fund 1 Fund 2 Fund 312 4 915 8 313 6 514 5 717 4 4

162. Test at the 5% significance level to determine whether the mean percentage rates for the three funds differ.

ANSWER:

Source of Variation SS df MS FTreatments 64.933 2 32.467 12.816Error 30.400 12 2.533Total 95.333 14

3210 : µµµ ==H

:1H At least two means differRejection region: F > 89.312,2,05. =F

Conclusion: Reject the null hypothesis. Yes


164. A random sample of 10 observations was selected from each of four normal populations. A partial one way ANOVA table is shown below:

Source of Variation SS df MS FTreatments * * 270 *Error * * *Total 1,350 *

a. Complete the missing entries (identified by asterisks) in the ANOVA table.b. How many groups were in this study?c. How many experimental units were in this study?d. At the 5% significance level, can we infer that the means of the populations differ?

ANSWERS:a.

Source of Variation SS df MS FTreatments 810 3 270 18Error 540 36 15Total 1,350 39

b. 4c. 40d. 43210 : µµµµ ===H

:1H At least two means differ

Rejection region: F > 0.05,3,36 2.88F = Test statistics: F = 18

Conclusion: Reject the null hypothesis. Yes

14: analysis of variance -...

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