Deck 13: Experimental Design and Analysis of Variance
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Deck 13: Experimental Design and Analysis of Variance
1
The F ratio in a completely randomized ANOVA is given by
A) MSTR/MSE.
B) MST/MSE.
C) MSE/MSTR.
D) MSE/MST.
A) MSTR/MSE.
B) MST/MSE.
C) MSE/MSTR.
D) MSE/MST.
MSTR/MSE.
2
In an analysis of variance problem if SST = 120 and SSTR = 80, then SSE is
A) 200.
B) 40.
C) 80.
D) 120.
A) 200.
B) 40.
C) 80.
D) 120.
40.
3
In an ANOVA procedure, a term that means the same as the term "variable" is
A) factor.
B) treatment.
C) replication.
D) within-variance.
A) factor.
B) treatment.
C) replication.
D) within-variance.
factor.
4
In the ANOVA, treatments refer to
A) experimental units.
B) different levels of a factor.
C) the dependent variables.
D) statistical applications.
A) experimental units.
B) different levels of a factor.
C) the dependent variables.
D) statistical applications.
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5
An ANOVA procedure is applied to data obtained from 6 samples where each sample contains 20 observations.The critical value of F occurs with
A) 6 numerator and 20 denominator degrees of freedom.
B) 5 numerator and 20 denominator degrees of freedom.
C) 5 numerator and 114 denominator degrees of freedom.
D) 6 numerator and 114 denominator degrees of freedom.
A) 6 numerator and 20 denominator degrees of freedom.
B) 5 numerator and 20 denominator degrees of freedom.
C) 5 numerator and 114 denominator degrees of freedom.
D) 6 numerator and 114 denominator degrees of freedom.
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6
In the analysis of variance procedure (ANOVA), "factor" refers to
A) the dependent variable.
B) the independent variable.
C) different levels of a treatment.
D) the critical value of F.
A) the dependent variable.
B) the independent variable.
C) different levels of a treatment.
D) the critical value of F.
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7
The mean square is the sum of squares divided by
A) the total number of observations.
B) its corresponding degrees of freedom.
C) its corresponding degrees of freedom minus one.
D) the total number of replications.
A) the total number of observations.
B) its corresponding degrees of freedom.
C) its corresponding degrees of freedom minus one.
D) the total number of replications.
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8
The independent variable of interest in an ANOVA procedure is called a
A) partition.
B) treatment.
C) response.
D) factor.
A) partition.
B) treatment.
C) response.
D) factor.
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9
In order to determine whether or not the means of two populations are equal,
A) a t test must be performed.
B) an analysis of variance must be performed.
C) either a t test or an analysis of variance can be performed.
D) a chi-square test can be performed.
A) a t test must be performed.
B) an analysis of variance must be performed.
C) either a t test or an analysis of variance can be performed.
D) a chi-square test can be performed.
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10
In factorial designs, the response produced when the treatments of one factor interact with the treatments of another in influencing the response variable is known as
A) main effect.
B) replication.
C) interaction.
D) error.
A) main effect.
B) replication.
C) interaction.
D) error.
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11
In an analysis of variance problem involving 3 treatments and 10 observations per treatment, SSE = 399.6.The MSE for this situation is
A) 133.2.
B) 13.32.
C) 14.8.
D) 30.0.
A) 133.2.
B) 13.32.
C) 14.8.
D) 30.0.
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12
An experimental design where the experimental units are randomly assigned to the treatments is known as
A) factor block design.
B) random factor design.
C) completely randomized design.
D) randomized treatment design.
A) factor block design.
B) random factor design.
C) completely randomized design.
D) randomized treatment design.
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13
When an analysis of variance is performed on samples drawn from k populations, the mean square due to treatments (MSTR) is
A) SSTR/nT.
B) SSTR/(nT - 1).
C) SSTR/k.
D) SSTR/(k - 1).
A) SSTR/nT.
B) SSTR/(nT - 1).
C) SSTR/k.
D) SSTR/(k - 1).
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14
The number of times each experimental condition is observed in a factorial design is known as
A) partition.
B) replication.
C) blocking.
D) factor.
A) partition.
B) replication.
C) blocking.
D) factor.
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15
In ANOVA, which of the following is not affected by whether or not the population means are equal?
A)
B) between-treatments estimate of 2
C) within-treatments estimate of 2
D) ratio of between- and within-treatments estimate of 2
A)
B) between-treatments estimate of 2
C) within-treatments estimate of 2
D) ratio of between- and within-treatments estimate of 2
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16
The critical F value with 6 numerator and 60 denominator degrees of freedom at α = .05 is
A) 3.74.
B) 2.25.
C) 2.37.
D) 1.96.
A) 3.74.
B) 2.25.
C) 2.37.
D) 1.96.
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17
The ANOVA procedure is a statistical approach for determining whether or not the means of
A) two samples are equal.
B) two or more samples are equal.
C) two populations are equal.
D) three or more populations are equal.
A) two samples are equal.
B) two or more samples are equal.
C) two populations are equal.
D) three or more populations are equal.
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18
The required condition for using an ANOVA procedure on data from several populations is that the
A) selected samples are dependent on each other.
B) response variables from samples are all uniform.
C) sampled populations have equal variances.
D) sampled populations have equal means.
A) selected samples are dependent on each other.
B) response variables from samples are all uniform.
C) sampled populations have equal variances.
D) sampled populations have equal means.
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19
In an analysis of variance where the total sample size for the experiment is nT and the number of populations is k, the mean square due to error is
A) SSE/(nT - k).
B) SSTR/(nT - k).
C) SSE/(k - 1).
D) SSTR/k.
A) SSE/(nT - k).
B) SSTR/(nT - k).
C) SSE/(k - 1).
D) SSTR/k.
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20
An ANOVA procedure is used for data that was obtained from four sample groups each comprised of five observations.The degrees of freedom for the critical value of F are
A) 3 and 20.
B) 3 and 16.
C) 4 and 17.
D) 3 and 19.
A) 3 and 20.
B) 3 and 16.
C) 4 and 17.
D) 3 and 19.
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21
Consider the following ANOVA table.
The sum of squares due to error equals
A) 13833.6.
B) 2073.6.
C) 5760.
D) 6000.
The sum of squares due to error equals
A) 13833.6.
B) 2073.6.
C) 5760.
D) 6000.
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22
Which of the following is not a required assumption for the analysis of variance?
A) The random variable of interest for each population has a normal probability distribution.
B) The variance associated with the random variable must be the same for all populations.
C) At least 2 populations are under consideration.
D) Populations under consideration have equal means.
A) The random variable of interest for each population has a normal probability distribution.
B) The variance associated with the random variable must be the same for all populations.
C) At least 2 populations are under consideration.
D) Populations under consideration have equal means.
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23
In an analysis of variance, one estimate of σ2 is based upon the differences between the treatment means and the
A) means of each sample.
B) overall sample mean.
C) sum of observations.
D) population means.
A) means of each sample.
B) overall sample mean.
C) sum of observations.
D) population means.
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24
Consider the following information.
The mean square due to treatments (MSTR) equals
A) 400.
B) 500.
C) 1687.5.
D) 2250.
The mean square due to treatments (MSTR) equals
A) 400.
B) 500.
C) 1687.5.
D) 2250.
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25
In a completely randomized design involving three treatments, the following information is provided:
The overall mean (the grand mean) for all the treatments is
A) 7.00.
B) 6.67.
C) 7.25.
D) 4.89.
The overall mean (the grand mean) for all the treatments is
A) 7.00.
B) 6.67.
C) 7.25.
D) 4.89.
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26
The process of allocating the total sum of squares and degrees of freedom to the various components is called
A) factoring.
B) blocking.
C) replicating.
D) partitioning.
A) factoring.
B) blocking.
C) replicating.
D) partitioning.
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27
An ANOVA procedure is used for data obtained from four populations.Four samples, each comprised of 30 observations, were taken from the four populations.The numerator and denominator (respectively) degrees of freedom for the critical value of F are
A) 3 and 30.
B) 4 and 30.
C) 3 and 119.
D) 3 and 116.
A) 3 and 30.
B) 4 and 30.
C) 3 and 119.
D) 3 and 116.
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28
An experimental design that permits simultaneous statistical conclusions about two or more factors is a
A) randomized block design.
B) factorial design.
C) completely randomized design.
D) multiple block design.
A) randomized block design.
B) factorial design.
C) completely randomized design.
D) multiple block design.
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29
Consider the following ANOVA table.
The null hypothesis is to be tested at the 5% level of significance.The null hypothesis
A) should be rejected.
B) should not be rejected.
C) should be revised.
D) should not be tested.
The null hypothesis is to be tested at the 5% level of significance.The null hypothesis
A) should be rejected.
B) should not be rejected.
C) should be revised.
D) should not be tested.
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30
Consider the following ANOVA table.
The mean square due to treatments equals
A) 288.
B) 518.4.
C) 1200.
D) 8294.4.
The mean square due to treatments equals
A) 288.
B) 518.4.
C) 1200.
D) 8294.4.
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31
An ANOVA procedure is used for data obtained from five populations.Five samples, each comprised of 20 observations, were taken from the five populations.The numerator and denominator (respectively) degrees of freedom for the critical value of F are
A) 5 and 20.
B) 4 and 20.
C) 4 and 99.
D) 4 and 95.
A) 5 and 20.
B) 4 and 20.
C) 4 and 99.
D) 4 and 95.
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32
Consider the following information.
SSTR = 6750
H0: μ1 = μ2 = μ3 = μ4
SSE = 8000
Ha: At least one mean is different
The null hypothesis is to be tested at the 5% level of significance.The null hypothesis
A) should be rejected.
B) should not be rejected.
C) was designed incorrectly.
D) cannot be tested.
SSTR = 6750
H0: μ1 = μ2 = μ3 = μ4
SSE = 8000
Ha: At least one mean is different
The null hypothesis is to be tested at the 5% level of significance.The null hypothesis
A) should be rejected.
B) should not be rejected.
C) was designed incorrectly.
D) cannot be tested.
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33
In a completely randomized design involving four treatments, the following information is provided.
The overall mean (the grand mean) for all treatments is
A) 40.0.
B) 37.3.
C) 48.3.
D) 37.0.
The overall mean (the grand mean) for all treatments is
A) 40.0.
B) 37.3.
C) 48.3.
D) 37.0.
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34
Consider the following ANOVA table.
The test statistic to test the null hypothesis equals
A) .432.
B) 1.8.
C) 4.17.
D) 28.8.
The test statistic to test the null hypothesis equals
A) .432.
B) 1.8.
C) 4.17.
D) 28.8.
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35
Consider the following information.
If n = 5, the mean square due to error (MSE) equals
A) 400.
B) 500.
C) 1687.5.
D) 2250.
If n = 5, the mean square due to error (MSE) equals
A) 400.
B) 500.
C) 1687.5.
D) 2250.
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36
Consider the following ANOVA table.
The null hypothesis for this ANOVA problem is
A) 1 = 2 = 3 = 4.
B) 1 = 2 = 3 = 4 = 5.
C) 1 = 2 = 3 = 4 = 5 = 6.
D) 1 = 2 = ... = 20.
The null hypothesis for this ANOVA problem is
A) 1 = 2 = 3 = 4.
B) 1 = 2 = 3 = 4 = 5.
C) 1 = 2 = 3 = 4 = 5 = 6.
D) 1 = 2 = ... = 20.
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37
Consider the following information.
The test statistic to test the null hypothesis equals
A) .22.
B) .84.
C) 4.22.
D) 4.50.
The test statistic to test the null hypothesis equals
A) .22.
B) .84.
C) 4.22.
D) 4.50.
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38
Consider the following information.
SSTR = 6750
H0: μ1 = μ2 = μ3 = μ4
SSE = 8000
Ha: At least one mean is different
The null hypothesis is to be tested at the 5% level of significance.The p-value is
A) less than .01.
B) between .01 and .025.
C) between .025 and .05.
D) greater than .10.
SSTR = 6750
H0: μ1 = μ2 = μ3 = μ4
SSE = 8000
Ha: At least one mean is different
The null hypothesis is to be tested at the 5% level of significance.The p-value is
A) less than .01.
B) between .01 and .025.
C) between .025 and .05.
D) greater than .10.
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39
The critical F value with 8 numerator and 29 denominator degrees of freedom at α = .01 is
A) 2.28.
B) 3.20.
C) 3.33.
D) 3.64.
A) 2.28.
B) 3.20.
C) 3.33.
D) 3.64.
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40
Consider the following ANOVA table.
The null hypothesis is to be tested at the 5% level of significance.The p-value is
A) greater than .10.
B) between .05 to .10.
C) between .025 to .05.
D) less than .01.
The null hypothesis is to be tested at the 5% level of significance.The p-value is
A) greater than .10.
B) between .05 to .10.
C) between .025 to .05.
D) less than .01.
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41
To test whether or not there is a difference between treatments A, B, and C, a sample of 12 observations has been randomly assigned to the 3 treatments.You are given the results below.
The null hypothesis is to be tested at the 1% level of significance.The null hypothesis
A) should be rejected.
B) should not be rejected.
C) should be revised.
D) should not be tested.
The null hypothesis is to be tested at the 1% level of significance.The null hypothesis
A) should be rejected.
B) should not be rejected.
C) should be revised.
D) should not be tested.
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42
To test whether or not there is a difference between treatments A, B, and C, a sample of 12 observations has been randomly assigned to the 3 treatments.You are given the results below.
The null hypothesis for this ANOVA problem is
A) 1 = 2.
B) 1 = 2 = 3.
C) 1 = 2 = 3 = 4.
D) 1 = 2 = ... = 12.
The null hypothesis for this ANOVA problem is
A) 1 = 2.
B) 1 = 2 = 3.
C) 1 = 2 = 3 = 4.
D) 1 = 2 = ... = 12.
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43
In a completely randomized experimental design involving five treatments, 13 observations were recorded for each of the five treatments (a total of 65 observations).Also, the design provided the following information.
SSTR = 200 (Sum of Squares Due to Treatments)
SST = 800 (Total Sum of Squares)
The mean square due to error (MSE) is
A) 50.
B) 10.
C) 200.
D) 600.
SSTR = 200 (Sum of Squares Due to Treatments)
SST = 800 (Total Sum of Squares)
The mean square due to error (MSE) is
A) 50.
B) 10.
C) 200.
D) 600.
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44
Part of an ANOVA table is shown below. The number of degrees of freedom corresponding to between-treatments is
A) 18.
B) 2.
C) 4.
D) 3.
A) 18.
B) 2.
C) 4.
D) 3.
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45
In a completely randomized experimental design involving five treatments, 13 observations were recorded for each of the five treatments (a total of 65 observations).Also, the design provided the following information.
SSTR = 200 (Sum of Squares Due to Treatments)
SST = 800 (Total Sum of Squares)
If, at a 5% level of significance, we want to determine whether or not the means of the five populations are equal, the critical value of F is
A) 2.53.
B) 19.48.
C) 4.98.
D) 5.69.
SSTR = 200 (Sum of Squares Due to Treatments)
SST = 800 (Total Sum of Squares)
If, at a 5% level of significance, we want to determine whether or not the means of the five populations are equal, the critical value of F is
A) 2.53.
B) 19.48.
C) 4.98.
D) 5.69.
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46
Part of an ANOVA table is shown below. The mean square due to error (MSE) is
A) 60.
B) 15.
C) 18.
D) 20.
A) 60.
B) 15.
C) 18.
D) 20.
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47
Part of an ANOVA table is shown below. The number of degrees of freedom corresponding to within-treatments is
A) 22.
B) 4.
C) 5.
D) 18.
A) 22.
B) 4.
C) 5.
D) 18.
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48
In a completely randomized experimental design involving five treatments, 13 observations were recorded for each of the five treatments (a total of 65 observations).The following information is provided.
SSTR = 200 (Sum of Squares Due to Treatments)
SST = 800 (Total Sum of Squares)
The test statistic is
A) .2.
B) 5.0.
C) 3.75.
D) 15.
SSTR = 200 (Sum of Squares Due to Treatments)
SST = 800 (Total Sum of Squares)
The test statistic is
A) .2.
B) 5.0.
C) 3.75.
D) 15.
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49
In a completely randomized experimental design involving five treatments, 13 observations were recorded for each of the five treatments (a total of 65 observations).Also, the design provided the following information.
SSTR = 200 (Sum of Squares Due to Treatments)
SST = 800 (Total Sum of Squares)
The mean square due to treatments (MSTR) is
A) 40.00.
B) 10.00.
C) 50.00.
D) 12.00.
SSTR = 200 (Sum of Squares Due to Treatments)
SST = 800 (Total Sum of Squares)
The mean square due to treatments (MSTR) is
A) 40.00.
B) 10.00.
C) 50.00.
D) 12.00.
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50
To test whether or not there is a difference between treatments A, B, and C, a sample of 12 observations has been randomly assigned to the 3 treatments.You are given the results below.
The null hypothesis is to be tested at the 1% level of significance.The p-value is
A) greater than .1.
B) between .05 to .10.
C) less than .01.
D) between .01 to .025.
The null hypothesis is to be tested at the 1% level of significance.The p-value is
A) greater than .1.
B) between .05 to .10.
C) less than .01.
D) between .01 to .025.
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51
To test whether or not there is a difference between treatments A, B, and C, a sample of 12 observations has been randomly assigned to the 3 treatments.You are given the results below.
The mean square due to treatments (MSTR) equals
A) 1.872.
B) 5.86.
C) 34.
D) 36.
The mean square due to treatments (MSTR) equals
A) 1.872.
B) 5.86.
C) 34.
D) 36.
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52
Part of an ANOVA table is shown below. The test statistic is
A) 2.25.
B) 6.00.
C) 2.67.
D) 3.00.
A) 2.25.
B) 6.00.
C) 2.67.
D) 3.00.
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53
To test whether or not there is a difference between treatments A, B, and C, a sample of 12 observations has been randomly assigned to the 3 treatments.You are given the results below.
The test statistic to test the null hypothesis equals
A) .944.
B) 1.06.
C) 3.13.
D) 19.231.
The test statistic to test the null hypothesis equals
A) .944.
B) 1.06.
C) 3.13.
D) 19.231.
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54
To test whether or not there is a difference between treatments A, B, and C, a sample of 12 observations has been randomly assigned to the 3 treatments.You are given the results below.
The mean square due to error (MSE) equals
A) 1.872.
B) 5.86.
C) 34.
D) 36.
The mean square due to error (MSE) equals
A) 1.872.
B) 5.86.
C) 34.
D) 36.
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55
In a completely randomized experimental design involving five treatments, 13 observations were recorded for each of the five treatments (a total of 65 observations).Also, the design provided the following information. SSTR (Sum of Squares Due to Treatments)
SST (Total Sum of Squares)
The sum of squares due to error (SSE) is
A) 1000.
B) 600.
C) 200.
D) 1600.
SST (Total Sum of Squares)
The sum of squares due to error (SSE) is
A) 1000.
B) 600.
C) 200.
D) 1600.
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56
In a completely randomized experimental design involving five treatments, 13 observations were recorded for each of the five treatments (a total of 65 observations).Also, the design provided the following information.
SSTR = 200 (Sum of Squares Due to Treatments)
SST = 800 (Total Sum of Squares)
The number of degrees of freedom corresponding to within-treatments is
A) 60.
B) 59.
C) 5.
D) 4.
SSTR = 200 (Sum of Squares Due to Treatments)
SST = 800 (Total Sum of Squares)
The number of degrees of freedom corresponding to within-treatments is
A) 60.
B) 59.
C) 5.
D) 4.
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57
Part of an ANOVA table is shown below. The mean square due to treatments (MSTR) is
A) 20.
B) 60.
C) 18.
D) 15.
A) 20.
B) 60.
C) 18.
D) 15.
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58
Part of an ANOVA table is shown below. The mean square due to treatments (MSTR) is
A) 36.
B) 16.
C) 64.
D) 15.
A) 36.
B) 16.
C) 64.
D) 15.
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59
In a completely randomized experimental design involving five treatments, 13 observations were recorded for each of the five treatments (a total of 65 observations).The following information is provided.
SSTR = 200 (Sum of Squares Due to Treatments)
SST = 800 (Total Sum of Squares)
If we want to determine whether or not the means of the five populations are equal, the p-value is
A) greater than .10.
B) between .025 to .05.
C) between .01 to .025.
D) less than .01.
SSTR = 200 (Sum of Squares Due to Treatments)
SST = 800 (Total Sum of Squares)
If we want to determine whether or not the means of the five populations are equal, the p-value is
A) greater than .10.
B) between .025 to .05.
C) between .01 to .025.
D) less than .01.
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60
In a completely randomized experimental design involving five treatments, 13 observations were recorded for each of the five treatments (a total of 65 observations).Also, the design provided the following information.
SSTR = 200 (Sum of Squares Due to Treatments)
SST = 800 (Total Sum of Squares)
The number of degrees of freedom corresponding to between-treatments is
A) 60.
B) 59.
C) 5.
D) 4.
SSTR = 200 (Sum of Squares Due to Treatments)
SST = 800 (Total Sum of Squares)
The number of degrees of freedom corresponding to between-treatments is
A) 60.
B) 59.
C) 5.
D) 4.
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Unlock Deck
k this deck
61
The process of using the same or similar experimental units for all treatments is called
A) factoring.
B) blocking.
C) replicating.
D) partitioning.
A) factoring.
B) blocking.
C) replicating.
D) partitioning.
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62
Part of an ANOVA table is shown below. At a 5% level of significance, if we want to determine whether or not the means of the populations are equal, the conclusion of the test is that
A) all means are equal.
B) some means may be equal.
C) not all means are equal.
D) some means will never be equal.
A) all means are equal.
B) some means may be equal.
C) not all means are equal.
D) some means will never be equal.
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63
Part of an ANOVA table is shown below. If we want to determine whether or not the means of the populations are equal, the p-value is
A) greater than .1.
B) between .05 to .1.
C) between .025 to .05.
D) less than .01.
A) greater than .1.
B) between .05 to .1.
C) between .025 to .05.
D) less than .01.
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64
If we are testing for the equality of three population means, we should use the
A) test statistic t.
B) test statistic z.
C) test statistic F.
D) test statistic χ2.
A) test statistic t.
B) test statistic z.
C) test statistic F.
D) test statistic χ2.
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65
A completely randomized design is useful when the experimental units are
A) homogeneous.
B) heterogeneous.
C) clustered.
D) stratified.
A) homogeneous.
B) heterogeneous.
C) clustered.
D) stratified.
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66
In a factorial experiment, if there are x levels of factor A and y levels of factor B, there is a total of
A) x + y treatment combinations.
B) (x + y)/2 treatment combinations.
C) 2(x + y) treatment combinations.
D) xy treatment combinations.
A) x + y treatment combinations.
B) (x + y)/2 treatment combinations.
C) 2(x + y) treatment combinations.
D) xy treatment combinations.
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67
In testing for the equality of k population means, the number of treatments is
A) k.
B) k - 1.
C) nT.
D) nT - k.
A) k.
B) k - 1.
C) nT.
D) nT - k.
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