Deck 11: Analysis of Variance

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سؤال
ANOVA is a procedure intended to compare the means of several groups (treatments).
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سؤال
Three-factor ANOVA is required if we have three treatment groups (i.e., three data columns).
سؤال
In a 3×4 randomized block (two-factor unreplicated) ANOVA, we have 12 treatment groups.
سؤال
Sample sizes must be equal in one-factor ANOVA.
سؤال
ANOVA is a procedure intended to compare the variances of several groups (treatments).
سؤال
Tukey's test is similar to a two-sample t-test except that it pools the variances for all c samples.
سؤال
Hartley's test is the largest sample mean divided by the smallest sample mean.
سؤال
Interaction plots that show parallel lines would suggest interaction effects.
سؤال
Comparison of c means in one-factor ANOVA can equivalently be done by using c individual t-tests on c pairs of means at the same α.
سؤال
ANOVA assumes normal populations.
سؤال
Hartley's test measures the equality of the means for several groups.
سؤال
Tukey's test compares pairs of treatment means in an ANOVA.
سؤال
One-factor ANOVA with two groups is equivalent to a two-tailed t-test.
سؤال
ANOVA assumes equal variances within each treatment group.
سؤال
Interaction plots that show crossing lines indicate likely interactions.
سؤال
If you have four factors (call them A, B, C, and D) in an ANOVA experiment with replication, you could have a maximum of four different two-factor interactions.
سؤال
Tukey's test is not needed if we have the overall F statistic for the ANOVA.
سؤال
One-factor ANOVA stacked data for five groups will be arranged in five separate columns.
سؤال
Hartley's test is to check for unequal variances for c groups.
سؤال
In a two-factor ANOVA with three columns and four rows, there can be more than two interaction effects.
سؤال
Which of the following is not a characteristic of the F distribution?

A)It is always right-skewed.
B)It describes the ratio of two variances.
C)It is a family based on two sets of degrees of freedom.
D)It is negative when s12 is smaller than s22.
سؤال
Levene's test for homogeneity of variance is attractive because it does not depend on the assumption of normality.
سؤال
In a one-factor ANOVA, the computed value of F will be negative:

A)when there is no difference in the treatment means.
B)when there is no difference within the treatments.
C)when the SST (total) is larger than SSE (error).
D)under no circumstances.
سؤال
Tukey's test with seven groups would entail 21 comparisons of means.
سؤال
Which Excel function gives the right-tail p-value for an ANOVA test with a test statistic Fcalc = 4.52, n = 29 observations, and c = 4 groups?

A)=F.DIST.RT(4.52, 3, 25)
B)=F.INV(4.52, 4, 28)
C)=F.DIST(4.52, 4, 28)
D)=F.INV(4.52, 3, 25)
سؤال
ANOVA is robust to violations of the equal-variance assumption as long as group sizes are equal.
سؤال
Which is not an assumption of ANOVA?

A)Normality of the treatment populations.
B)Homogeneous treatment variances.
C)Independent sample observations.
D)Equal population sizes for groups.
سؤال
In an ANOVA, the SSE (error) sum of squares reflects:

A)the effect of the combined factor(s).
B)the overall variation in Y that is to be explained.
C)the variation that is not explained by the factors.
D)the combined effect of treatments and sample size.
سؤال
Which is not assumed in ANOVA?

A)Observations are independent.
B)Populations are normally distributed.
C)Variances of all treatment groups are the same.
D)Population variances are known.
سؤال
In an ANOVA, when would the F-test statistic be zero?

A)When there is no difference in the variances
B)When the treatment means are the same
C)When the observations are normally distributed
D)The F-test statistic cannot ever be zero
سؤال
It is desirable, but not necessary, that sample sizes be equal in a one-factor ANOVA.
سؤال
Tukey's test for five groups would require 10 comparisons of means.
سؤال
Tukey's test pools all the sample variances.
سؤال
Degrees of freedom for the between-group variation in a one-factor ANOVA with n1 = 8, n2 = 5, n3 = 7, n4 = 9 would be:

A)28.
B)3.
C)29.
D)4.
سؤال
To test the null hypothesis H0: μ1 = μ2 = μ3 using samples from normal populations with unknown but equal variances, we:

A)cannot safely use ANOVA.
B)can safely employ ANOVA.
C)would prefer three separate t-tests.
D)would need three-factor ANOVA.
سؤال
Which is the Excel function to find the critical value of F for α = .05, df1 = 3, df2 = 25?

A)=F.DIST(.05, 2, 24)
B)=F.INV.RT(.05, 3, 25)
C)=F.DIST(.05, 3, 25)
D)=F.INV(.05, 2, 24)
سؤال
Degrees of freedom for the between-group variation in a one-factor ANOVA with n1 = 5, n2 = 6, n3 = 7 would be:

A)18.
B)17.
C)6.
D)2.
سؤال
Variation "within" the ANOVA treatments represents:

A)random variation.
B)differences between group means.
C)differences between group variances.
D)the effect of sample size.
سؤال
Analysis of variance is a technique used to test for:

A)equality of two or more variances.
B)equality of two or more means.
C)equality of a population mean and a given value.
D)equality of more than two variances.
سؤال
ANOVA is used to compare:

A)proportions of several groups.
B)variances of several groups.
C)means of several groups.
D)both means and variances.
سؤال
Refer to the following partial ANOVA results from Excel (some information is missing).  Source of Variation SSdfMSF Between groups 210.2778 Within groups 148374.15 Total 2113.833\begin{array} { | l | c | c | c | c | } \hline \text { Source of Variation } & S S & d f & M S & F \\\hline \text { Between groups } & & & 210.2778 & \\\hline \text { Within groups } & 1483 & & 74.15 & \\\hline \text { Total } & 2113.833 & & & \\\hline\end{array} The number of treatment groups is:

A)4.
B)3.
C)2.
D)1.
سؤال
Refer to the following partial ANOVA results from Excel (some information is missing).  Source of Variation SS df MSF Between groups 210.2778 Within groups 148374.15 Total 2113.833\begin{array} { | l | c | c | c | c | } \hline \text { Source of Variation } & S S & \text { df } & M S & F \\\hline \text { Between groups } & & & 210.2778 & \\\hline \text { Within groups } & 1483 & & 74.15 & \\\hline \text { Total } & 2113.833 & & & \\\hline\end{array} Degrees of freedom for the F-test are:

A)5, 22.
B)4, 21.
C)3, 20.
D)impossible to determine.
سؤال
The Internal Revenue Service wishes to study the time required to process tax returns in three regional centers. A random sample of three tax returns is chosen from each of three centers. The time (in days) required to process each return is recorded as shown below.  East  West  Midwest 494754395249455156\begin{array} { | c | c | c | } \hline \text { East } & \text { West } & \text { Midwest } \\\hline 49 & 47 & 54 \\39 & 52 & 49 \\45 & 51 & 56 \\\hline\end{array} The test to use to compare the means for all three groups would require:

A)three-factor ANOVA.
B)one-factor ANOVA.
C)repeated two-sample test of means.
D)two-factor ANOVA with replication.
سؤال
Refer to the following partial ANOVA results from Excel (some information is missing).  Source of Variation SSdfMSF Between groups 210.2788 Within groups 148374.15 Total 2113.833\begin{array} { | l | c | c | c | c | } \hline \text { Source of Variation } & S S & d f & M S & F \\\hline \text { Between groups } & & & 210.2788 & \\\hline \text { Within groups } & 1483 & & 74.15 & \\\hline \text { Total } & 2113.833 & & & \\\hline\end{array} The F-test statistic is:

A)2.84.
B)3.56.
C)2.80.
D)2.79.
سؤال
For this one-factor ANOVA (some information is missing), how many treatment groups were there?  Source  Sum of Squares df Mean Square F Treatment 654218 Error 3,456128 Total 4,110\begin{array} { l c c c c } \hline \text { Source } & \text { Sum of Squares } & d f & \text { Mean Square } & F \\\hline \text { Treatment } & 654 && 218 & \\\text { Error } & 3,456 && 128 \\\hline \text { Total } & 4,110 & \\\hline\end{array}

A)Cannot be determined
B)3
C)4
D)2
سؤال
Refer to the following partial ANOVA results from Excel (some information is missing).  Source of Variation  SS df MS FP-value  Between groups 210.27780.064139 Within groups 148374.15 Total 2113.833\begin{array} { | l | c | c | c | c | c | } \hline \text { Source of Variation } & \text { SS } & d f & \text { MS } & F & P \text {-value } \\\hline \text { Between groups } & & & 210.2778 & & 0.064139 \\\hline \text { Within groups } & 1483 & & 74.15 & & \\\hline \text { Total } & 2113.833 & & & & \\\hline\end{array} At ? = 0.05, the difference between group means is:

A)highly significant.
B)barely significant.
C)not quite significant.
D)clearly insignificant.
سؤال
Refer to the following partial ANOVA results from Excel (some information is missing).  Source of Variation SSdfMSF Between groups 210.2778 Within groups 148374.15 Total 2113.833\begin{array} { | l | c | c | c | c | } \hline \text { Source of Variation } & S S & d f & M S & F \\\hline \text { Between groups } & & & 210.2778 & \\\hline \text { Within groups } & 1483 & & 74.15 & \\\hline \text { Total } & 2113.833 & & & \\\hline\end{array} SS for between-groups variation will be:

A)129.99.
B)630.83.
C)1233.4.
D)Can't tell.
سؤال
Refer to the following partial ANOVA results from Excel (some information is missing).  Source of Variation  SS dfMSFP-value F crit  Between groups 210.27780.064139 Within groups 148374.15 Total 2113.833\begin{array} { | l | c | c | c | c | c | c | } \hline \text { Source of Variation } & \text { SS } & d f & M S & F & P \text {-value } & F \text { crit } \\\hline \text { Between groups } & & & 210.2778 & & 0.064139 & \\\hline \text { Within groups } & 1483 & & 74.15 & & & \\\hline \text { Total } & 2113.833 & & & & & \\\hline\end{array} The critical value of F at ? = 0.05 is:

A)1.645.
B)2.84.
C)3.10.
D)4.28.
سؤال
Refer to the following partial ANOVA results from Excel (some information is missing).  Source of Variation SSdfMSF Between groups 210.2778 Within groups 148374.15 Total 2113.833\begin{array} { | l | c | c | c | c | } \hline \text { Source of Variation } & S S & d f & M S & F \\\hline \text { Between groups } & & & 210.2778 & \\\hline \text { Within groups } & 1483 & & 74.15 & \\\hline \text { Total } & 2113.833 & & & \\\hline\end{array} Degrees of freedom for between-groups variation are:

A)3.
B)4.
C)5.
D)Can't tell from given information.
سؤال
For this one-factor ANOVA (some information is missing), what is the F-test statistic?  Source  Sum of Squares df Mean Square F Treatment 654218 Error 3,456128 Total 4,110\begin{array} { l c c c c } \hline \text { Source } & \text { Sum of Squares } & d f & \text { Mean Square } & F \\\hline \text { Treatment } & 654 && 218 & \\\text { Error } & 3,456 & & 128 & \\\hline \text { Total } & 4,110 & & \\\hline\end{array}

A)0.159
B)2.833
C)1.703
D)Cannot be determined
سؤال
Identify the degrees of freedom for the treatment and error in this one-factor ANOVA (blanks indicate missing information).  Source  Sum of Squares df Mean Square  Treatment 993331.0 Error 1,00250.1 Total 1,99523\begin{array} { l c c c } \hline \text { Source } & \text { Sum of Squares } & d f & \text { Mean Square } \\\hline \text { Treatment } & 993 & & 331.0 \\\text { Error } & 1,002 & & 50.1 \\\hline \text { Total } & 1,995 & 23 & \\\hline\end{array}

A)4, 24
B)3, 20
C)5, 23
سؤال
The within-treatment variation reflects:

A)variation among individuals of the same group.
B)variation between individuals in different groups.
C)variation explained by factors included in the ANOVA model.
D)variation that is not part of the ANOVA model.
سؤال
Given the following ANOVA table (some information is missing), find the critical value of F.05.  Source  Sum of Squares df Mean Square FFos  Treatment 744.004 Error 751.5015 Total 1,495.5019\begin{array} { l c c c c c } \hline \text { Source } & \text { Sum of Squares } & d f & \text { Mean Square } & F & F _ { \text {os } } \\\hline \text { Treatment } & 744.00 & 4 & & \\\text { Error } & 751.50 & 15 & & \\\hline \text { Total } & 1,495.50 & 19 & & \\\hline\end{array}

A)3.06
B)2.90
C)2.36
D)3.41
سؤال
In a one-factor ANOVA, the total sum of squares is equal to:

A)the sum of squares within groups plus the sum of squares between groups.
B)the sum of squares within groups times the sum of squares between groups.
C)the sum of squares within groups divided by the sum of squares between groups.
D)the means of all the groups squared.
سؤال
Using one-factor ANOVA with 30 observations we find at α = .05 that we cannot reject the null hypothesis of equal means. We increase the sample size from 30 observations to 60 observations and obtain the same value for the sample F-test statistic. Which is correct?

A)We might now be able to reject the null hypothesis.
B)We surely must reject H0 for 60 observations.
C)We cannot reject H0 since we obtained the same F-value.
D)It is impossible to get the same F-value for n = 60 as for n = 30.
سؤال
Prof. Gristmill sampled exam scores for five randomly chosen students from each of his two sections of ACC 200. His sample results are shown.  Day Class 9358748582 Night Class 9181856073\begin{array}{|l|l|l|l|l|l|}\hline \text { Day Class }& 93 & 58 & 74 & 85 & 82 \\\hline \text { Night Class }&91 & 81 & 85 & 60 & 73 \\\hline\end{array} He could test the population means for equality using:

A)a t-test for two means from independent samples.
B)a t-test for two means from paired (related) samples.
C)a one-factor ANOVA.
D)either a one-factor ANOVA or a two-tailed t-test.
سؤال
One-factor analysis of variance:

A)requires that the number of observations in each group be identical.
B)has less power when the number of observations per group is not identical.
C)is extremely sensitive to slight departures from normality.
D)is a generalization of the t-test for paired observations.
سؤال
Systolic blood pressure of randomly selected HMO patients was recorded on a particular Wednesday, with the results shown here:  Patient Age Group  Under 2020 to 2930 to 4950 and Over 105110122139113101114115108112128136114127124124123123125123\begin{array}{l}\text { Patient Age Group }\\\begin{array} { | c | c | c | c | } \hline \text { Under } 20 & 20 \text { to } 29 & 30 \text { to } 49 & 50 \text { and Over } \\\hline 105 & 110 & 122 & 139 \\113 & 101 & 114 & 115 \\108 & 112 & 128 & 136 \\114 & 127 & 124 & 124 \\123 & 123 & 125 & 123 \\\hline\end{array}\end{array} The appropriate hypothesis test is:

A)one-factor ANOVA.
B)two-factor ANOVA.
C)three-factor ANOVA.
D)four-factor ANOVA.
سؤال
Given the following ANOVA table (some information is missing), find the F statistic.  Source  Sum of Squares df Mean Square F Treatment 744.004 Error 751.5015 Total 1,495.5019\begin{array} { l c c c c } \hline \text { Source } & \text { Sum of Squares } & d f & \text { Mean Square } & F \\\hline \text { Treatment } & 744.00 & 4 & & \\\text { Error } & 751.50 & 15 & \\\hline \text { Total } & 1,495.50 & 19 & \\\hline\end{array}

A)3.71
B)0.99
C)0.497
D)4.02
سؤال
Systolic blood pressure of randomly selected HMO patients was recorded on a particular Wednesday, with the results shown here. An ANOVA test was performed using these data.  Patient Age Group  Under 2020 to 2930 to 4950 and Over 105110122139113101114115108112128136114127124124123123125123\begin{array}{l}\text { Patient Age Group }\\\begin{array} { | c | c | c | c | } \hline \text { Under } 20 & 20 \text { to } 29 & 30 \text { to } 49 & 50 \text { and Over } \\\hline 105 & 110 & 122 & 139 \\113 & 101 & 114 & 115 \\108 & 112 & 128 & 136 \\114 & 127 & 124 & 124 \\123 & 123 & 125 & 123 \\\hline\end{array}\end{array} Degrees of freedom for the between-treatments sum of squares would be:

A)3.
B)19.
C)17.
D)depends on ?.
سؤال
Refer to the following partial ANOVA results from Excel (some information is missing). ANOVA Table  Source  SS df MS F Treatment 44,75711,189 Error 89,025551,619 Total 133,78259\begin{array} { | l | c | c | c | c | } \hline \text { Source } & \text { SS } & d f & \text { MS } & F \\\hline \text { Treatment } & 44,757 & & 11,189 & \\\hline \text { Error } & 89,025 & 55 & 1,619 & \\\hline \text { Total } & 133,782 & 59 & & \\\hline\end{array} The p-value for the F-test would be:

A)much less than .05.
B)slightly less than .05.
C)slightly greater than .05.
D)much greater than .05.Fcalc = 11,189/1619 = 6.91, while F.05 = 2.56 using df = (4, 50) in AppendixF.
سؤال
Refer to the following MegaStat output (some information is missing). The sample size was n = 65 in a one-factor ANOVA.  Mon  Fri  Tue  Thu  Wed 147.4197.0198.1220.8223.6 Mon 147.4 Fri 197.03.02 Tue 198.13.080.07 Thu 220.84.461.451.38 Wed 223.64.641.621.550.17\begin{array}{|l|l|l|l|l|l|l|}\hline&&\text { Mon }&\text { Fri }&\text { Tue }&\text { Thu }& \text { Wed }\\&&147.4&197.0&198.1&220.8& 223.6\\\hline \text { Mon }&147.4&& & & & \\\hline \text { Fri }&197.0&3.02 & & & & \\\hline \text { Tue }&198.1&3.08 & 0.07 & & & \\\hline \text { Thu }&220.8&4.46 & 1.45 & 1.38 & & \\\hline \text { Wed }&223.6&4.64 & 1.62 & 1.55 & 0.17 & \\\hline\end{array} Which pairs of days differ significantly? Note: This question requires access to a Tukey table.

A)(Mon, Thu) and (Mon, Wed) only.
B)(Mon, Wed) only.
C)(Mon, Thu) only.
D)(Mon, Thu) and (Mon, Wed) and (Mon, Fri) and (Mon, Tue).
سؤال
Refer to the following partial ANOVA results from Excel (some information is missing). ANOVA table  Source  SS dfMSFp-value  Treatment 717.403.0442 Error 70.675 Total 1,848.2019\begin{array} { | l | c | c | c | c | c | } \hline \text { Source } & \text { SS } & d f & M S & F & p \text {-value } \\\hline \text { Treatment } & 717.40 & 3 & & & .0442 \\\hline \text { Error } & & & 70.675 & & \\\hline \text { Total } & 1,848.20 & 19 & & & \\\hline\end{array} The 5 percent critical value for the F test is:

A)2.46.
B)3.24.
C)3.38.
D)impossible to ascertain from the given information.Error df = 19 - 3 = 16, so F.05 = 3.24 using df = (3, 16) in AppendixF.
سؤال
Systolic blood pressure of randomly selected HMO patients was recorded on a particular Wednesday, with the results shown here. An ANOVA test was performed using these data.  Patient Age Group  Under 20  20 to 2930 to 49 50 and Over 105110122139113101114115108112128136114127124124123123125123\begin{array}{l}\text { Patient Age Group }\\\begin{array} { | c | c | c | c | } \hline \text { Under 20 } & \text { 20 to } 29 & 30 \text { to } 49 & \text { 50 and Over } \\\hline 105 & 110 & 122 & 139 \\113 & 101 & 114 & 115 \\108 & 112 & 128 & 136 \\114 & 127 & 124 & 124 \\123 & 123 & 125 & 123 \\\hline\end{array}\end{array} What are the degrees of freedom for the error sum of squares?

A)3
B)19
C)16
D)It depends on ?.
سؤال
What is the .05 critical value of Hartley's test statistic for a one-factor ANOVA with n1 = 5, n2 = 8, n3 = 7, n4 = 8, n5 = 6, n6 = 8? Note: This question requires access to a Hartley table.

A)10.8
B)11.8
C)13.7
D)15.0
سؤال
Refer to the following MegaStat output (some information is missing). The sample size was n = 24 in a one-factor ANOVA.  Tukev simultaneous comparison t-values \text { Tukev simultaneous comparison t-values }
 Med 4 Med 1 Med 3 Med 2133.2140.7141.7145.2 Med 4133.2 Med 1140.71.78 Med 3141.72.010.24 Med 2145.22.841.070.83\begin{array}{|c|c|c|c|c|c|}\hline &&\text { Med } 4 & \text { Med } 1& \text { Med } 3& \text { Med } 2\\&&133.2 & 140.7 & 141.7 & 145.2 \\\hline\text { Med } 4 &133.2&& & & \\\hline \text { Med } 1&140.7&1.78 & & & \\\hline \text { Med } 3&141.7&2.01 & 0.24 & & \\\hline\text { Med } 2&145.2&2.84 & 1.07 & 0.83 & \\\hline\end{array} Which pairs of meds differ at ? = .05? Note: This question requires access to a Tukey table.

A)Med 1, Med 2
B)Med 2, Med 4
C)Med 3, Med 4
D)None of them.
سؤال
Refer to the following partial ANOVA results from Excel (some information is missing). ANOVA Table  Source  SS dfMSF Treatment 44,75711,189 Error 89,025551,619 Total 133,78259\begin{array} { | l | c | c | c | c | } \hline \text { Source } & \text { SS } & d f & M S & F \\\hline \text { Treatment } & 44,757 & & 11,189 & \\\hline \text { Error } & 89,025 & 55 & 1,619 & \\\hline \text { Total } & 133,782 & 59 & & \\\hline\end{array} The F statistic is:

A)2.88.
B)4.87.
C)5.93.
D)6.91.
سؤال
After performing a one-factor ANOVA test, John noticed that the sample standard deviations for his four groups were, respectively, 33, 24, 79, and 35. John should:

A)feel confident in his ANOVA test.
B)use Hartley's test to check his assumptions.
C)use an independent samples t-test instead of ANOVA.
D)use a paired t-test instead of ANOVA.
سؤال
Sound levels are measured at random moments under typical driving conditions for various full-size truck models. The Excel ANOVA results are shown below.  Mean  Std Dev  n  Big Bruin 57.448.9446 Gran Conto 56.56.3618 MaxRanger 64.513.9837 Oso Grande 54.154.16610 Overall 57.736.82431 Source  SS d.f.  MS  Between 462.43154.1 Within 934.702734.6 Total 1,397.1030\begin{array} { l r r r } & \text { Mean } & \text { Std Dev } & \text { n } \\\text { Big Bruin } & 57.44 & 8.944 & 6 \\\text { Gran Conto } & 56.5 & 6.361 & 8 \\\text { MaxRanger } & 64.51 & 3.983 & 7 \\\text { Oso Grande } & 54.15 & 4.166 & 10 \\\text { Overall } & 57.73 & 6.824 & 31 \\\\\text { Source } &\text { SS} &\text { d.f. } &\text { MS } \\\text { Between } & 462.4 &3 & 154.1 \\\text { Within } & 934.70 & 27 &34.6 \\\text { Total } & 1,397.10 & 30 & \end{array} The test statistic to compare the five means simultaneously is:

A)2.96.
B)15.8.
C)5.56.
D)4.45.
سؤال
What are the degrees of freedom for Tukey's test statistic for a one-factor ANOVA with n1 = 6, n2 = 6, n3 = 6?

A)3, 6
B)6, 3
C)6, 15
D)3, 15
سؤال
Refer to the following partial ANOVA results from Excel (some information is missing). ANOVA table  Source  SS dfMSFp-value  Treatment 717.403.0442 Error 70.675 Total 1,848.2019\begin{array} { | l | c | c | c | c | c | } \hline \text { Source } & \text { SS } & d f & M S & F & p \text {-value } \\\hline \text { Treatment } & 717.40 & 3 & & & .0442 \\\hline \text { Error } & & & 70.675 & & \\\hline \text { Total } & 1,848.20 & 19 & & & \\\hline\end{array} The MS (mean square) for the treatments is:

A)239.13.
B)106.88.
C)1,130.8.
D)impossible to ascertain from the information given.
سؤال
What are the degrees of freedom for Hartley's test statistic for a one-factor ANOVA with n1 = 5, n2 = 8, n3 = 7, n4 = 8, n5 = 6, n6 = 8?

A)7, 6
B)6, 6
C)6, 41
سؤال
Refer to the following partial ANOVA results from Excel (some information is missing). ANOVA Table  Source  SS dfMSF Treatment 44,75711,189 Error 89,025551,619 Total 133,78259\begin{array} { | l | c | c | c | c | } \hline \text { Source } & \text { SS } & d f & M S & F \\\hline \text { Treatment } & 44,757 & & 11,189 & \\\hline \text { Error } & 89,025 & 55 & 1,619 & \\\hline \text { Total } & 133,782 & 59 & & \\\hline\end{array} The number of treatment groups is:

A)5.
B)4.
C)3.
D)impossible to ascertain from given.
سؤال
Refer to the following MegaStat output (some information is missing). The sample size was n = 65 in a one-factor ANOVA.  Mon  Fri  Tue  Thu  Wed 147.4197.0198.1220.8223.6 Mon 147.4 Fri 197.03.02 Tue 198.13.080.07 Thu 220.84.461.451.38 Wed 223.64.641.621.550.17\begin{array}{|l|l|l|l|l|l|l|}\hline&&\text { Mon }&\text { Fri }&\text { Tue }&\text { Thu }& \text { Wed }\\&&147.4&197.0&198.1&220.8& 223.6\\\hline \text { Mon }&147.4&& & & & \\\hline \text { Fri }&197.0&3.02 & & & & \\\hline \text { Tue }&198.1&3.08 & 0.07 & & & \\\hline \text { Thu }&220.8&4.46 & 1.45 & 1.38 & & \\\hline \text { Wed }&223.6&4.64 & 1.62 & 1.55 & 0.17 & \\\hline\end{array} At ? = .05, which is the critical value of the test statistic for a two-tailed test for a significant difference in means that are to be compared simultaneously? Note: This question requires a Tukey table.

A)2.81
B)2.54
C)2.33
D)1.96
سؤال
Sound levels are measured at random moments under typical driving conditions for various full-size truck models. The ANOVA results are shown below.  Mean  Std Dev  n  Big Bruin 57.448.9446 Gran Conto 56.56.3618 MaxRanger 64.513.9837 Oso Grande 54.154.16610 Overall 57.736.82431 Source  SS  d.f.  MS  Between 462.43154.1 Within 934.702734.6 Total 1,397.1030\begin{array} { l r r r } & \text { Mean } & \text { Std Dev } & \text { n } \\\text { Big Bruin } & 57.44 & 8.944 & 6 \\\text { Gran Conto } & 56.5 & 6.361 & 8 \\\text { MaxRanger } & 64.51 & 3.983 & 7 \\\text { Oso Grande } & 54.15 & 4.166 & 10 \\\text { Overall } & 57.73 & 6.824 & 31 \\& & & \\\text { Source } & \text { SS } & \text { d.f. } & \text { MS } \\\text { Between } & 462.4 & 3 & 154.1 \\\text { Within } & 934.70 & 27 & 34.6 \\\text { Total } & 1,397.10 & 30 &\end{array} The test statistic for Hartley's test for homogeneity of variance is:

A)2.25.
B)5.04.
C)4.61.
D)4.45.
سؤال
Refer to the following MegaStat output (some information is missing). The sample size was n = 24 in a one-factor ANOVA.  Tukev simultaneous comparison t-values \text { Tukev simultaneous comparison t-values }
 Med 4 Med 1 Med 3 Med 2133.2140.7141.7145.2 Med 4133.2 Med 1140.71.78 Med 3141.72.010.24 Med 2145.22.841.070.83\begin{array}{|c|c|c|c|c|c|}\hline &&\text { Med } 4 & \text { Med } 1& \text { Med } 3& \text { Med } 2\\&&133.2 & 140.7 & 141.7 & 145.2 \\\hline\text { Med } 4 &133.2&& & & \\\hline \text { Med } 1&140.7&1.78 & & & \\\hline \text { Med } 3&141.7&2.01 & 0.24 & & \\\hline\text { Med } 2&145.2&2.84 & 1.07 & 0.83 & \\\hline\end{array} At ? = .05, what is the critical value of the Tukey test statistic for a two-tailed test for a significant difference in means that are to be compared simultaneously? Note: This question requires access to a Tukey table.

A)2.07
B)2.80
C)2.76
D)1.96
سؤال
What is the .05 critical value of Tukey's test statistic for a one-factor ANOVA with n1 = 6, n2 = 6, n3 = 6? Note: This question requires access to a Tukey table.

A)3.67
B)2.60
C)3.58
D)2.75
سؤال
Refer to the following partial ANOVA results from Excel (some information is missing). ANOVA table  Source  SS dfMSFp-value  Treatment 717.403.0442 Error 70.675 Total 1,848.2019\begin{array} { | l | c | c | c | c | c | } \hline \text { Source } & \text { SS } & d f & M S & F & p \text {-value } \\\hline \text { Treatment } & 717.40 & 3 & & & .0442 \\\hline \text { Error } & & & 70.675 & & \\\hline \text { Total } & 1,848.20 & 19 & & & \\\hline\end{array} The F statistic is:

A)4.87.
B)3.38.
C)5.93.
D)6.91.
سؤال
Refer to the following partial ANOVA results from Excel (some information is missing). ANOVA Table  Source  SS dfMSF Treatment 44,75711,189 Error 89,025551,619 Total 133,78259\begin{array} { | l | c | c | c | c | } \hline \text { Source } & \text { SS } & d f & M S & F \\\hline \text { Treatment } & 44,757 & & 11,189 & \\\hline \text { Error } & 89,025 & 55 & 1,619 & \\\hline \text { Total } & 133,782 & 59 & & \\\hline\end{array} Using Appendix F, the 5 percent critical value for the F-test is approximately:

A)3.24.
B)6.91.
C)2.56.
D)2.06.Treatment df = 59 - 55 = 4, so F.05 = 2.56 using df = (4, 50) in AppendixF.
سؤال
Refer to the following partial ANOVA results from Excel (some information is missing). ANOVA table  Source  SS dfMSFp-value  Treatment 717.403.0442 Error 70.675 Total 1,848.2019\begin{array} { | l | c | c | c | c | c | } \hline \text { Source } & \text { SS } & d f & M S & F & p \text {-value } \\\hline \text { Treatment } & 717.40 & 3 & & & .0442 \\\hline \text { Error } & & & 70.675 & & \\\hline \text { Total } & 1,848.20 & 19 & & & \\\hline\end{array} Our decision about the hypothesis of equal treatment means is that the null hypothesis:

A)cannot be rejected at ? = .05.
B)can be rejected at ? = .05.
C)can be rejected for any typical value of ?.
D)cannot be assessed from the given information.
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Deck 11: Analysis of Variance
1
ANOVA is a procedure intended to compare the means of several groups (treatments).
True
2
Three-factor ANOVA is required if we have three treatment groups (i.e., three data columns).
False
3
In a 3×4 randomized block (two-factor unreplicated) ANOVA, we have 12 treatment groups.
True
4
Sample sizes must be equal in one-factor ANOVA.
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5
ANOVA is a procedure intended to compare the variances of several groups (treatments).
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6
Tukey's test is similar to a two-sample t-test except that it pools the variances for all c samples.
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7
Hartley's test is the largest sample mean divided by the smallest sample mean.
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8
Interaction plots that show parallel lines would suggest interaction effects.
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9
Comparison of c means in one-factor ANOVA can equivalently be done by using c individual t-tests on c pairs of means at the same α.
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10
ANOVA assumes normal populations.
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11
Hartley's test measures the equality of the means for several groups.
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12
Tukey's test compares pairs of treatment means in an ANOVA.
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13
One-factor ANOVA with two groups is equivalent to a two-tailed t-test.
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14
ANOVA assumes equal variances within each treatment group.
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15
Interaction plots that show crossing lines indicate likely interactions.
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16
If you have four factors (call them A, B, C, and D) in an ANOVA experiment with replication, you could have a maximum of four different two-factor interactions.
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17
Tukey's test is not needed if we have the overall F statistic for the ANOVA.
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18
One-factor ANOVA stacked data for five groups will be arranged in five separate columns.
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19
Hartley's test is to check for unequal variances for c groups.
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20
In a two-factor ANOVA with three columns and four rows, there can be more than two interaction effects.
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21
Which of the following is not a characteristic of the F distribution?

A)It is always right-skewed.
B)It describes the ratio of two variances.
C)It is a family based on two sets of degrees of freedom.
D)It is negative when s12 is smaller than s22.
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22
Levene's test for homogeneity of variance is attractive because it does not depend on the assumption of normality.
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23
In a one-factor ANOVA, the computed value of F will be negative:

A)when there is no difference in the treatment means.
B)when there is no difference within the treatments.
C)when the SST (total) is larger than SSE (error).
D)under no circumstances.
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24
Tukey's test with seven groups would entail 21 comparisons of means.
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25
Which Excel function gives the right-tail p-value for an ANOVA test with a test statistic Fcalc = 4.52, n = 29 observations, and c = 4 groups?

A)=F.DIST.RT(4.52, 3, 25)
B)=F.INV(4.52, 4, 28)
C)=F.DIST(4.52, 4, 28)
D)=F.INV(4.52, 3, 25)
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26
ANOVA is robust to violations of the equal-variance assumption as long as group sizes are equal.
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27
Which is not an assumption of ANOVA?

A)Normality of the treatment populations.
B)Homogeneous treatment variances.
C)Independent sample observations.
D)Equal population sizes for groups.
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28
In an ANOVA, the SSE (error) sum of squares reflects:

A)the effect of the combined factor(s).
B)the overall variation in Y that is to be explained.
C)the variation that is not explained by the factors.
D)the combined effect of treatments and sample size.
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29
Which is not assumed in ANOVA?

A)Observations are independent.
B)Populations are normally distributed.
C)Variances of all treatment groups are the same.
D)Population variances are known.
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30
In an ANOVA, when would the F-test statistic be zero?

A)When there is no difference in the variances
B)When the treatment means are the same
C)When the observations are normally distributed
D)The F-test statistic cannot ever be zero
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31
It is desirable, but not necessary, that sample sizes be equal in a one-factor ANOVA.
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32
Tukey's test for five groups would require 10 comparisons of means.
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33
Tukey's test pools all the sample variances.
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34
Degrees of freedom for the between-group variation in a one-factor ANOVA with n1 = 8, n2 = 5, n3 = 7, n4 = 9 would be:

A)28.
B)3.
C)29.
D)4.
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35
To test the null hypothesis H0: μ1 = μ2 = μ3 using samples from normal populations with unknown but equal variances, we:

A)cannot safely use ANOVA.
B)can safely employ ANOVA.
C)would prefer three separate t-tests.
D)would need three-factor ANOVA.
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36
Which is the Excel function to find the critical value of F for α = .05, df1 = 3, df2 = 25?

A)=F.DIST(.05, 2, 24)
B)=F.INV.RT(.05, 3, 25)
C)=F.DIST(.05, 3, 25)
D)=F.INV(.05, 2, 24)
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37
Degrees of freedom for the between-group variation in a one-factor ANOVA with n1 = 5, n2 = 6, n3 = 7 would be:

A)18.
B)17.
C)6.
D)2.
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38
Variation "within" the ANOVA treatments represents:

A)random variation.
B)differences between group means.
C)differences between group variances.
D)the effect of sample size.
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39
Analysis of variance is a technique used to test for:

A)equality of two or more variances.
B)equality of two or more means.
C)equality of a population mean and a given value.
D)equality of more than two variances.
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40
ANOVA is used to compare:

A)proportions of several groups.
B)variances of several groups.
C)means of several groups.
D)both means and variances.
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41
Refer to the following partial ANOVA results from Excel (some information is missing).  Source of Variation SSdfMSF Between groups 210.2778 Within groups 148374.15 Total 2113.833\begin{array} { | l | c | c | c | c | } \hline \text { Source of Variation } & S S & d f & M S & F \\\hline \text { Between groups } & & & 210.2778 & \\\hline \text { Within groups } & 1483 & & 74.15 & \\\hline \text { Total } & 2113.833 & & & \\\hline\end{array} The number of treatment groups is:

A)4.
B)3.
C)2.
D)1.
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42
Refer to the following partial ANOVA results from Excel (some information is missing).  Source of Variation SS df MSF Between groups 210.2778 Within groups 148374.15 Total 2113.833\begin{array} { | l | c | c | c | c | } \hline \text { Source of Variation } & S S & \text { df } & M S & F \\\hline \text { Between groups } & & & 210.2778 & \\\hline \text { Within groups } & 1483 & & 74.15 & \\\hline \text { Total } & 2113.833 & & & \\\hline\end{array} Degrees of freedom for the F-test are:

A)5, 22.
B)4, 21.
C)3, 20.
D)impossible to determine.
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43
The Internal Revenue Service wishes to study the time required to process tax returns in three regional centers. A random sample of three tax returns is chosen from each of three centers. The time (in days) required to process each return is recorded as shown below.  East  West  Midwest 494754395249455156\begin{array} { | c | c | c | } \hline \text { East } & \text { West } & \text { Midwest } \\\hline 49 & 47 & 54 \\39 & 52 & 49 \\45 & 51 & 56 \\\hline\end{array} The test to use to compare the means for all three groups would require:

A)three-factor ANOVA.
B)one-factor ANOVA.
C)repeated two-sample test of means.
D)two-factor ANOVA with replication.
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44
Refer to the following partial ANOVA results from Excel (some information is missing).  Source of Variation SSdfMSF Between groups 210.2788 Within groups 148374.15 Total 2113.833\begin{array} { | l | c | c | c | c | } \hline \text { Source of Variation } & S S & d f & M S & F \\\hline \text { Between groups } & & & 210.2788 & \\\hline \text { Within groups } & 1483 & & 74.15 & \\\hline \text { Total } & 2113.833 & & & \\\hline\end{array} The F-test statistic is:

A)2.84.
B)3.56.
C)2.80.
D)2.79.
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45
For this one-factor ANOVA (some information is missing), how many treatment groups were there?  Source  Sum of Squares df Mean Square F Treatment 654218 Error 3,456128 Total 4,110\begin{array} { l c c c c } \hline \text { Source } & \text { Sum of Squares } & d f & \text { Mean Square } & F \\\hline \text { Treatment } & 654 && 218 & \\\text { Error } & 3,456 && 128 \\\hline \text { Total } & 4,110 & \\\hline\end{array}

A)Cannot be determined
B)3
C)4
D)2
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46
Refer to the following partial ANOVA results from Excel (some information is missing).  Source of Variation  SS df MS FP-value  Between groups 210.27780.064139 Within groups 148374.15 Total 2113.833\begin{array} { | l | c | c | c | c | c | } \hline \text { Source of Variation } & \text { SS } & d f & \text { MS } & F & P \text {-value } \\\hline \text { Between groups } & & & 210.2778 & & 0.064139 \\\hline \text { Within groups } & 1483 & & 74.15 & & \\\hline \text { Total } & 2113.833 & & & & \\\hline\end{array} At ? = 0.05, the difference between group means is:

A)highly significant.
B)barely significant.
C)not quite significant.
D)clearly insignificant.
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47
Refer to the following partial ANOVA results from Excel (some information is missing).  Source of Variation SSdfMSF Between groups 210.2778 Within groups 148374.15 Total 2113.833\begin{array} { | l | c | c | c | c | } \hline \text { Source of Variation } & S S & d f & M S & F \\\hline \text { Between groups } & & & 210.2778 & \\\hline \text { Within groups } & 1483 & & 74.15 & \\\hline \text { Total } & 2113.833 & & & \\\hline\end{array} SS for between-groups variation will be:

A)129.99.
B)630.83.
C)1233.4.
D)Can't tell.
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48
Refer to the following partial ANOVA results from Excel (some information is missing).  Source of Variation  SS dfMSFP-value F crit  Between groups 210.27780.064139 Within groups 148374.15 Total 2113.833\begin{array} { | l | c | c | c | c | c | c | } \hline \text { Source of Variation } & \text { SS } & d f & M S & F & P \text {-value } & F \text { crit } \\\hline \text { Between groups } & & & 210.2778 & & 0.064139 & \\\hline \text { Within groups } & 1483 & & 74.15 & & & \\\hline \text { Total } & 2113.833 & & & & & \\\hline\end{array} The critical value of F at ? = 0.05 is:

A)1.645.
B)2.84.
C)3.10.
D)4.28.
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49
Refer to the following partial ANOVA results from Excel (some information is missing).  Source of Variation SSdfMSF Between groups 210.2778 Within groups 148374.15 Total 2113.833\begin{array} { | l | c | c | c | c | } \hline \text { Source of Variation } & S S & d f & M S & F \\\hline \text { Between groups } & & & 210.2778 & \\\hline \text { Within groups } & 1483 & & 74.15 & \\\hline \text { Total } & 2113.833 & & & \\\hline\end{array} Degrees of freedom for between-groups variation are:

A)3.
B)4.
C)5.
D)Can't tell from given information.
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50
For this one-factor ANOVA (some information is missing), what is the F-test statistic?  Source  Sum of Squares df Mean Square F Treatment 654218 Error 3,456128 Total 4,110\begin{array} { l c c c c } \hline \text { Source } & \text { Sum of Squares } & d f & \text { Mean Square } & F \\\hline \text { Treatment } & 654 && 218 & \\\text { Error } & 3,456 & & 128 & \\\hline \text { Total } & 4,110 & & \\\hline\end{array}

A)0.159
B)2.833
C)1.703
D)Cannot be determined
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51
Identify the degrees of freedom for the treatment and error in this one-factor ANOVA (blanks indicate missing information).  Source  Sum of Squares df Mean Square  Treatment 993331.0 Error 1,00250.1 Total 1,99523\begin{array} { l c c c } \hline \text { Source } & \text { Sum of Squares } & d f & \text { Mean Square } \\\hline \text { Treatment } & 993 & & 331.0 \\\text { Error } & 1,002 & & 50.1 \\\hline \text { Total } & 1,995 & 23 & \\\hline\end{array}

A)4, 24
B)3, 20
C)5, 23
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52
The within-treatment variation reflects:

A)variation among individuals of the same group.
B)variation between individuals in different groups.
C)variation explained by factors included in the ANOVA model.
D)variation that is not part of the ANOVA model.
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53
Given the following ANOVA table (some information is missing), find the critical value of F.05.  Source  Sum of Squares df Mean Square FFos  Treatment 744.004 Error 751.5015 Total 1,495.5019\begin{array} { l c c c c c } \hline \text { Source } & \text { Sum of Squares } & d f & \text { Mean Square } & F & F _ { \text {os } } \\\hline \text { Treatment } & 744.00 & 4 & & \\\text { Error } & 751.50 & 15 & & \\\hline \text { Total } & 1,495.50 & 19 & & \\\hline\end{array}

A)3.06
B)2.90
C)2.36
D)3.41
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54
In a one-factor ANOVA, the total sum of squares is equal to:

A)the sum of squares within groups plus the sum of squares between groups.
B)the sum of squares within groups times the sum of squares between groups.
C)the sum of squares within groups divided by the sum of squares between groups.
D)the means of all the groups squared.
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55
Using one-factor ANOVA with 30 observations we find at α = .05 that we cannot reject the null hypothesis of equal means. We increase the sample size from 30 observations to 60 observations and obtain the same value for the sample F-test statistic. Which is correct?

A)We might now be able to reject the null hypothesis.
B)We surely must reject H0 for 60 observations.
C)We cannot reject H0 since we obtained the same F-value.
D)It is impossible to get the same F-value for n = 60 as for n = 30.
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56
Prof. Gristmill sampled exam scores for five randomly chosen students from each of his two sections of ACC 200. His sample results are shown.  Day Class 9358748582 Night Class 9181856073\begin{array}{|l|l|l|l|l|l|}\hline \text { Day Class }& 93 & 58 & 74 & 85 & 82 \\\hline \text { Night Class }&91 & 81 & 85 & 60 & 73 \\\hline\end{array} He could test the population means for equality using:

A)a t-test for two means from independent samples.
B)a t-test for two means from paired (related) samples.
C)a one-factor ANOVA.
D)either a one-factor ANOVA or a two-tailed t-test.
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57
One-factor analysis of variance:

A)requires that the number of observations in each group be identical.
B)has less power when the number of observations per group is not identical.
C)is extremely sensitive to slight departures from normality.
D)is a generalization of the t-test for paired observations.
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58
Systolic blood pressure of randomly selected HMO patients was recorded on a particular Wednesday, with the results shown here:  Patient Age Group  Under 2020 to 2930 to 4950 and Over 105110122139113101114115108112128136114127124124123123125123\begin{array}{l}\text { Patient Age Group }\\\begin{array} { | c | c | c | c | } \hline \text { Under } 20 & 20 \text { to } 29 & 30 \text { to } 49 & 50 \text { and Over } \\\hline 105 & 110 & 122 & 139 \\113 & 101 & 114 & 115 \\108 & 112 & 128 & 136 \\114 & 127 & 124 & 124 \\123 & 123 & 125 & 123 \\\hline\end{array}\end{array} The appropriate hypothesis test is:

A)one-factor ANOVA.
B)two-factor ANOVA.
C)three-factor ANOVA.
D)four-factor ANOVA.
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59
Given the following ANOVA table (some information is missing), find the F statistic.  Source  Sum of Squares df Mean Square F Treatment 744.004 Error 751.5015 Total 1,495.5019\begin{array} { l c c c c } \hline \text { Source } & \text { Sum of Squares } & d f & \text { Mean Square } & F \\\hline \text { Treatment } & 744.00 & 4 & & \\\text { Error } & 751.50 & 15 & \\\hline \text { Total } & 1,495.50 & 19 & \\\hline\end{array}

A)3.71
B)0.99
C)0.497
D)4.02
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60
Systolic blood pressure of randomly selected HMO patients was recorded on a particular Wednesday, with the results shown here. An ANOVA test was performed using these data.  Patient Age Group  Under 2020 to 2930 to 4950 and Over 105110122139113101114115108112128136114127124124123123125123\begin{array}{l}\text { Patient Age Group }\\\begin{array} { | c | c | c | c | } \hline \text { Under } 20 & 20 \text { to } 29 & 30 \text { to } 49 & 50 \text { and Over } \\\hline 105 & 110 & 122 & 139 \\113 & 101 & 114 & 115 \\108 & 112 & 128 & 136 \\114 & 127 & 124 & 124 \\123 & 123 & 125 & 123 \\\hline\end{array}\end{array} Degrees of freedom for the between-treatments sum of squares would be:

A)3.
B)19.
C)17.
D)depends on ?.
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61
Refer to the following partial ANOVA results from Excel (some information is missing). ANOVA Table  Source  SS df MS F Treatment 44,75711,189 Error 89,025551,619 Total 133,78259\begin{array} { | l | c | c | c | c | } \hline \text { Source } & \text { SS } & d f & \text { MS } & F \\\hline \text { Treatment } & 44,757 & & 11,189 & \\\hline \text { Error } & 89,025 & 55 & 1,619 & \\\hline \text { Total } & 133,782 & 59 & & \\\hline\end{array} The p-value for the F-test would be:

A)much less than .05.
B)slightly less than .05.
C)slightly greater than .05.
D)much greater than .05.Fcalc = 11,189/1619 = 6.91, while F.05 = 2.56 using df = (4, 50) in AppendixF.
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62
Refer to the following MegaStat output (some information is missing). The sample size was n = 65 in a one-factor ANOVA.  Mon  Fri  Tue  Thu  Wed 147.4197.0198.1220.8223.6 Mon 147.4 Fri 197.03.02 Tue 198.13.080.07 Thu 220.84.461.451.38 Wed 223.64.641.621.550.17\begin{array}{|l|l|l|l|l|l|l|}\hline&&\text { Mon }&\text { Fri }&\text { Tue }&\text { Thu }& \text { Wed }\\&&147.4&197.0&198.1&220.8& 223.6\\\hline \text { Mon }&147.4&& & & & \\\hline \text { Fri }&197.0&3.02 & & & & \\\hline \text { Tue }&198.1&3.08 & 0.07 & & & \\\hline \text { Thu }&220.8&4.46 & 1.45 & 1.38 & & \\\hline \text { Wed }&223.6&4.64 & 1.62 & 1.55 & 0.17 & \\\hline\end{array} Which pairs of days differ significantly? Note: This question requires access to a Tukey table.

A)(Mon, Thu) and (Mon, Wed) only.
B)(Mon, Wed) only.
C)(Mon, Thu) only.
D)(Mon, Thu) and (Mon, Wed) and (Mon, Fri) and (Mon, Tue).
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63
Refer to the following partial ANOVA results from Excel (some information is missing). ANOVA table  Source  SS dfMSFp-value  Treatment 717.403.0442 Error 70.675 Total 1,848.2019\begin{array} { | l | c | c | c | c | c | } \hline \text { Source } & \text { SS } & d f & M S & F & p \text {-value } \\\hline \text { Treatment } & 717.40 & 3 & & & .0442 \\\hline \text { Error } & & & 70.675 & & \\\hline \text { Total } & 1,848.20 & 19 & & & \\\hline\end{array} The 5 percent critical value for the F test is:

A)2.46.
B)3.24.
C)3.38.
D)impossible to ascertain from the given information.Error df = 19 - 3 = 16, so F.05 = 3.24 using df = (3, 16) in AppendixF.
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64
Systolic blood pressure of randomly selected HMO patients was recorded on a particular Wednesday, with the results shown here. An ANOVA test was performed using these data.  Patient Age Group  Under 20  20 to 2930 to 49 50 and Over 105110122139113101114115108112128136114127124124123123125123\begin{array}{l}\text { Patient Age Group }\\\begin{array} { | c | c | c | c | } \hline \text { Under 20 } & \text { 20 to } 29 & 30 \text { to } 49 & \text { 50 and Over } \\\hline 105 & 110 & 122 & 139 \\113 & 101 & 114 & 115 \\108 & 112 & 128 & 136 \\114 & 127 & 124 & 124 \\123 & 123 & 125 & 123 \\\hline\end{array}\end{array} What are the degrees of freedom for the error sum of squares?

A)3
B)19
C)16
D)It depends on ?.
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65
What is the .05 critical value of Hartley's test statistic for a one-factor ANOVA with n1 = 5, n2 = 8, n3 = 7, n4 = 8, n5 = 6, n6 = 8? Note: This question requires access to a Hartley table.

A)10.8
B)11.8
C)13.7
D)15.0
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66
Refer to the following MegaStat output (some information is missing). The sample size was n = 24 in a one-factor ANOVA.  Tukev simultaneous comparison t-values \text { Tukev simultaneous comparison t-values }
 Med 4 Med 1 Med 3 Med 2133.2140.7141.7145.2 Med 4133.2 Med 1140.71.78 Med 3141.72.010.24 Med 2145.22.841.070.83\begin{array}{|c|c|c|c|c|c|}\hline &&\text { Med } 4 & \text { Med } 1& \text { Med } 3& \text { Med } 2\\&&133.2 & 140.7 & 141.7 & 145.2 \\\hline\text { Med } 4 &133.2&& & & \\\hline \text { Med } 1&140.7&1.78 & & & \\\hline \text { Med } 3&141.7&2.01 & 0.24 & & \\\hline\text { Med } 2&145.2&2.84 & 1.07 & 0.83 & \\\hline\end{array} Which pairs of meds differ at ? = .05? Note: This question requires access to a Tukey table.

A)Med 1, Med 2
B)Med 2, Med 4
C)Med 3, Med 4
D)None of them.
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67
Refer to the following partial ANOVA results from Excel (some information is missing). ANOVA Table  Source  SS dfMSF Treatment 44,75711,189 Error 89,025551,619 Total 133,78259\begin{array} { | l | c | c | c | c | } \hline \text { Source } & \text { SS } & d f & M S & F \\\hline \text { Treatment } & 44,757 & & 11,189 & \\\hline \text { Error } & 89,025 & 55 & 1,619 & \\\hline \text { Total } & 133,782 & 59 & & \\\hline\end{array} The F statistic is:

A)2.88.
B)4.87.
C)5.93.
D)6.91.
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68
After performing a one-factor ANOVA test, John noticed that the sample standard deviations for his four groups were, respectively, 33, 24, 79, and 35. John should:

A)feel confident in his ANOVA test.
B)use Hartley's test to check his assumptions.
C)use an independent samples t-test instead of ANOVA.
D)use a paired t-test instead of ANOVA.
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69
Sound levels are measured at random moments under typical driving conditions for various full-size truck models. The Excel ANOVA results are shown below.  Mean  Std Dev  n  Big Bruin 57.448.9446 Gran Conto 56.56.3618 MaxRanger 64.513.9837 Oso Grande 54.154.16610 Overall 57.736.82431 Source  SS d.f.  MS  Between 462.43154.1 Within 934.702734.6 Total 1,397.1030\begin{array} { l r r r } & \text { Mean } & \text { Std Dev } & \text { n } \\\text { Big Bruin } & 57.44 & 8.944 & 6 \\\text { Gran Conto } & 56.5 & 6.361 & 8 \\\text { MaxRanger } & 64.51 & 3.983 & 7 \\\text { Oso Grande } & 54.15 & 4.166 & 10 \\\text { Overall } & 57.73 & 6.824 & 31 \\\\\text { Source } &\text { SS} &\text { d.f. } &\text { MS } \\\text { Between } & 462.4 &3 & 154.1 \\\text { Within } & 934.70 & 27 &34.6 \\\text { Total } & 1,397.10 & 30 & \end{array} The test statistic to compare the five means simultaneously is:

A)2.96.
B)15.8.
C)5.56.
D)4.45.
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70
What are the degrees of freedom for Tukey's test statistic for a one-factor ANOVA with n1 = 6, n2 = 6, n3 = 6?

A)3, 6
B)6, 3
C)6, 15
D)3, 15
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71
Refer to the following partial ANOVA results from Excel (some information is missing). ANOVA table  Source  SS dfMSFp-value  Treatment 717.403.0442 Error 70.675 Total 1,848.2019\begin{array} { | l | c | c | c | c | c | } \hline \text { Source } & \text { SS } & d f & M S & F & p \text {-value } \\\hline \text { Treatment } & 717.40 & 3 & & & .0442 \\\hline \text { Error } & & & 70.675 & & \\\hline \text { Total } & 1,848.20 & 19 & & & \\\hline\end{array} The MS (mean square) for the treatments is:

A)239.13.
B)106.88.
C)1,130.8.
D)impossible to ascertain from the information given.
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72
What are the degrees of freedom for Hartley's test statistic for a one-factor ANOVA with n1 = 5, n2 = 8, n3 = 7, n4 = 8, n5 = 6, n6 = 8?

A)7, 6
B)6, 6
C)6, 41
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73
Refer to the following partial ANOVA results from Excel (some information is missing). ANOVA Table  Source  SS dfMSF Treatment 44,75711,189 Error 89,025551,619 Total 133,78259\begin{array} { | l | c | c | c | c | } \hline \text { Source } & \text { SS } & d f & M S & F \\\hline \text { Treatment } & 44,757 & & 11,189 & \\\hline \text { Error } & 89,025 & 55 & 1,619 & \\\hline \text { Total } & 133,782 & 59 & & \\\hline\end{array} The number of treatment groups is:

A)5.
B)4.
C)3.
D)impossible to ascertain from given.
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74
Refer to the following MegaStat output (some information is missing). The sample size was n = 65 in a one-factor ANOVA.  Mon  Fri  Tue  Thu  Wed 147.4197.0198.1220.8223.6 Mon 147.4 Fri 197.03.02 Tue 198.13.080.07 Thu 220.84.461.451.38 Wed 223.64.641.621.550.17\begin{array}{|l|l|l|l|l|l|l|}\hline&&\text { Mon }&\text { Fri }&\text { Tue }&\text { Thu }& \text { Wed }\\&&147.4&197.0&198.1&220.8& 223.6\\\hline \text { Mon }&147.4&& & & & \\\hline \text { Fri }&197.0&3.02 & & & & \\\hline \text { Tue }&198.1&3.08 & 0.07 & & & \\\hline \text { Thu }&220.8&4.46 & 1.45 & 1.38 & & \\\hline \text { Wed }&223.6&4.64 & 1.62 & 1.55 & 0.17 & \\\hline\end{array} At ? = .05, which is the critical value of the test statistic for a two-tailed test for a significant difference in means that are to be compared simultaneously? Note: This question requires a Tukey table.

A)2.81
B)2.54
C)2.33
D)1.96
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75
Sound levels are measured at random moments under typical driving conditions for various full-size truck models. The ANOVA results are shown below.  Mean  Std Dev  n  Big Bruin 57.448.9446 Gran Conto 56.56.3618 MaxRanger 64.513.9837 Oso Grande 54.154.16610 Overall 57.736.82431 Source  SS  d.f.  MS  Between 462.43154.1 Within 934.702734.6 Total 1,397.1030\begin{array} { l r r r } & \text { Mean } & \text { Std Dev } & \text { n } \\\text { Big Bruin } & 57.44 & 8.944 & 6 \\\text { Gran Conto } & 56.5 & 6.361 & 8 \\\text { MaxRanger } & 64.51 & 3.983 & 7 \\\text { Oso Grande } & 54.15 & 4.166 & 10 \\\text { Overall } & 57.73 & 6.824 & 31 \\& & & \\\text { Source } & \text { SS } & \text { d.f. } & \text { MS } \\\text { Between } & 462.4 & 3 & 154.1 \\\text { Within } & 934.70 & 27 & 34.6 \\\text { Total } & 1,397.10 & 30 &\end{array} The test statistic for Hartley's test for homogeneity of variance is:

A)2.25.
B)5.04.
C)4.61.
D)4.45.
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76
Refer to the following MegaStat output (some information is missing). The sample size was n = 24 in a one-factor ANOVA.  Tukev simultaneous comparison t-values \text { Tukev simultaneous comparison t-values }
 Med 4 Med 1 Med 3 Med 2133.2140.7141.7145.2 Med 4133.2 Med 1140.71.78 Med 3141.72.010.24 Med 2145.22.841.070.83\begin{array}{|c|c|c|c|c|c|}\hline &&\text { Med } 4 & \text { Med } 1& \text { Med } 3& \text { Med } 2\\&&133.2 & 140.7 & 141.7 & 145.2 \\\hline\text { Med } 4 &133.2&& & & \\\hline \text { Med } 1&140.7&1.78 & & & \\\hline \text { Med } 3&141.7&2.01 & 0.24 & & \\\hline\text { Med } 2&145.2&2.84 & 1.07 & 0.83 & \\\hline\end{array} At ? = .05, what is the critical value of the Tukey test statistic for a two-tailed test for a significant difference in means that are to be compared simultaneously? Note: This question requires access to a Tukey table.

A)2.07
B)2.80
C)2.76
D)1.96
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77
What is the .05 critical value of Tukey's test statistic for a one-factor ANOVA with n1 = 6, n2 = 6, n3 = 6? Note: This question requires access to a Tukey table.

A)3.67
B)2.60
C)3.58
D)2.75
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Refer to the following partial ANOVA results from Excel (some information is missing). ANOVA table  Source  SS dfMSFp-value  Treatment 717.403.0442 Error 70.675 Total 1,848.2019\begin{array} { | l | c | c | c | c | c | } \hline \text { Source } & \text { SS } & d f & M S & F & p \text {-value } \\\hline \text { Treatment } & 717.40 & 3 & & & .0442 \\\hline \text { Error } & & & 70.675 & & \\\hline \text { Total } & 1,848.20 & 19 & & & \\\hline\end{array} The F statistic is:

A)4.87.
B)3.38.
C)5.93.
D)6.91.
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Refer to the following partial ANOVA results from Excel (some information is missing). ANOVA Table  Source  SS dfMSF Treatment 44,75711,189 Error 89,025551,619 Total 133,78259\begin{array} { | l | c | c | c | c | } \hline \text { Source } & \text { SS } & d f & M S & F \\\hline \text { Treatment } & 44,757 & & 11,189 & \\\hline \text { Error } & 89,025 & 55 & 1,619 & \\\hline \text { Total } & 133,782 & 59 & & \\\hline\end{array} Using Appendix F, the 5 percent critical value for the F-test is approximately:

A)3.24.
B)6.91.
C)2.56.
D)2.06.Treatment df = 59 - 55 = 4, so F.05 = 2.56 using df = (4, 50) in AppendixF.
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Refer to the following partial ANOVA results from Excel (some information is missing). ANOVA table  Source  SS dfMSFp-value  Treatment 717.403.0442 Error 70.675 Total 1,848.2019\begin{array} { | l | c | c | c | c | c | } \hline \text { Source } & \text { SS } & d f & M S & F & p \text {-value } \\\hline \text { Treatment } & 717.40 & 3 & & & .0442 \\\hline \text { Error } & & & 70.675 & & \\\hline \text { Total } & 1,848.20 & 19 & & & \\\hline\end{array} Our decision about the hypothesis of equal treatment means is that the null hypothesis:

A)cannot be rejected at ? = .05.
B)can be rejected at ? = .05.
C)can be rejected for any typical value of ?.
D)cannot be assessed from the given information.
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