Deck 14: Analysis of Variance
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Deck 14: Analysis of Variance
1
The F-test used in one-way ANOVA is an extension of the t-test of 1 - 2.
True
2
The F-test in ANOVA tests whether or not the population variances are equal.
False
3
The F-test in ANOVA is an expansion of the t-test for two independent population means.
True
4
In ANOVA, a criterion by which the populations are classified is called a factor.
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5
The analysis of variance (ANOVA) technique analyzes the variance of the data to determine whether differences exist between the population means.
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6
In one-way ANOVA, the total variation SS(Total) is partitioned into two sources of variation: the sum of squares for treatments (SST) and the sum of squares for error (SSE).
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7
The sum of squares for error is also known as the between-treatments variation.
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8
The sum of squares for treatments, SST, achieves its smallest value (zero) when all the sample means are equal.
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9
If the numerator (MST) degrees of freedom is 3 and the denominator (MSE) degrees of freedom is 18, the total number of observations must equal 21.
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10
The sum of squares for error (SSE) measures the amount of variation that is explained by the ANOVA model, while the sum of squares for treatments (SST) measures the amount of variation that remains unexplained.
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11
Conducting t-tests for each pair or population means is statistically equivalent to conducting one F-test comparing all the population means.
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12
When the F-test is used for ANOVA, the rejection region is always in the right tail.
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13
The analysis of variance (ANOVA) tests hypotheses about population variances and requires all the population means to be equal.
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14
In one-way ANOVA, the test statistic is defined as the ratio of the mean square for error (MSE) and the mean square for treatments (MST), namely, F = MSE / MST.
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15
We use the analysis of variance (ANOVA) technique to compare two or more population means.
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16
The F-statistic in a one-way ANOVA represents the variation within the treatments divided by the variation between the treatments.
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17
The distribution of the test statistic for analysis of variance is the F-distribution.
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18
The sum of squares for treatments (SST) is the variation attributed to the differences between the treatment means, while the sum of squares for error (SSE) measures the within-treatment variation.
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19
We can use the F-test to determine whether 1 is greater than 2.
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20
The within-treatments variation provides a measure of the amount of variation in the response variables that is caused by the treatments.
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21
Which of the following is not a required condition for one-way ANOVA?
A) The sample sizes must be equal.
B) The populations must all be normally distributed.
C) The population variances must be equal.
D) The samples for each treatment must be selected randomly and independently.
A) The sample sizes must be equal.
B) The populations must all be normally distributed.
C) The population variances must be equal.
D) The samples for each treatment must be selected randomly and independently.
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22
The number of degrees of freedom for the denominator in one-way ANOVA test involving 4 population means with 15 observations sampled from each population is:
A) 60
B) 19
C) 56
D) 45
A) 60
B) 19
C) 56
D) 45
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23
In one-way ANOVA, the amount of total variation that is unexplained is measured by the:
A) sum of squares for treatments.
B) degrees of freedom.
C) total sum of squares.
D) sum of squares for error.
A) sum of squares for treatments.
B) degrees of freedom.
C) total sum of squares.
D) sum of squares for error.
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24
One-way ANOVA is applied to independent samples taken from three normally distributed populations with equal variances.Which of the following is the null hypothesis for this procedure?
A) 1 + 2 + 3 = 0
B) 1 + 2 + 3 0
C) 1 = 2 = 3 = 0
D) 1 = 2 = 3
A) 1 + 2 + 3 = 0
B) 1 + 2 + 3 0
C) 1 = 2 = 3 = 0
D) 1 = 2 = 3
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25
One-way 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 that
A) all population means are equal.
B) all population means differ.
C) at least two population means are equal.
D) at least two population means differ.
A) all population means are equal.
B) all population means differ.
C) at least two population means are equal.
D) at least two population means differ.
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26
In one-way ANOVA, the term
refers to the:
A) weighted average of the sample means.
B) sum of the sample means divided by the total number of observations.
C) sum of the population means.
D) sum of the sample means.

A) weighted average of the sample means.
B) sum of the sample means divided by the total number of observations.
C) sum of the population means.
D) sum of the sample means.
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27
The value of the test statistic in a completely randomized design for ANOVA is F = 6.29.The degrees of freedom for the numerator and denominator are 5 and 10, respectively.Using an F table, the most accurate statements to be made about the p-value is that it is:
A) greater than 0.05
B) between 0.001 and 0.010.
C) between 0.010 and 0.025.
D) between 0.025 and 0.050.
A) greater than 0.05
B) between 0.001 and 0.010.
C) between 0.010 and 0.025.
D) between 0.025 and 0.050.
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28
In a single-factor 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 larger than MSE.
B) MST is much smaller than MSE.
C) MST is equal to MSE.
D) None of these choices.
A) MST is much larger than MSE.
B) MST is much smaller than MSE.
C) MST is equal to MSE.
D) None of these choices.
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29
In one-way analysis of variance, between-treatments variation is measured by the:
A) SSE
B) SS(Total)
C) SST
D) standard deviation
A) SSE
B) SS(Total)
C) SST
D) standard deviation
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30
In the one-way ANOVA where there are k treatments and n observations, the degrees of freedom for the F-statistic are equal to, respectively:
A) n and k.
B) k and n.
C) n - k and k - 1.
D) k - 1 and n - k.
A) n and k.
B) k and n.
C) n - k and k - 1.
D) k - 1 and n - k.
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31
The analysis of variance is a procedure that allows statisticians to compare two or more population:
A) proportions.
B) means.
C) variances.
D) standard deviations.
A) proportions.
B) means.
C) variances.
D) standard deviations.
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32
The test statistic of the single-factor ANOVA equals:
A) sum of squares for treatments / sum of squares for error.
B) sum of squares for error / sum of squares for treatments.
C) mean square for treatments / mean square for error.
D) mean square for error / mean square for treatments.
A) sum of squares for treatments / sum of squares for error.
B) sum of squares for error / sum of squares for treatments.
C) mean square for treatments / mean square for error.
D) mean square for error / mean square for treatments.
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33
In ANOVA, the F-test is the ratio of two sample variances.In the one-way ANOVA (completely randomized design), the variance used as a numerator of the ratio is:
A) mean square for treatments.
B) mean square for error.
C) total sum of squares.
D) None of these choices.
A) mean square for treatments.
B) mean square for error.
C) total sum of squares.
D) None of these choices.
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34
In the one-way ANOVA where k is the number of treatments and n is the number of observations in all samples, the number of degrees of freedom for error is:
A) k -1
B) n - 1
C) n -k
D) n - k + 1
A) k -1
B) n - 1
C) n -k
D) n - k + 1
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35
The distribution of the test statistic for analysis of variance is the:
A) normal distribution.
B) Student t-distribution.
C) F-distribution.
D) None of these choices.
A) normal distribution.
B) Student t-distribution.
C) F-distribution.
D) None of these choices.
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36
Which of the following is a required condition for ANOVA?
A) The populations are normally distributed.
B) The population variances are equal.
C) The samples are independent.
D) All of these choices are required conditions for ANOVA.
A) The populations are normally distributed.
B) The population variances are equal.
C) The samples are independent.
D) All of these choices are required conditions for ANOVA.
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37
In a one-way 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 variation.
B) within-treatments variation.
C) between-treatments variation.
D) None of these choices.
A) the total variation.
B) within-treatments variation.
C) between-treatments variation.
D) None of these choices.
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38
How does conducting multiple t-tests compare to conducting a single F-test?
A) Multiple t-tests increases the chance of a Type I error.
B) Multiple t-tests decreases the chance of a Type I error.
C) Multiple t-tests does not affect the chance of a Type I error.
D) This comparison cannot be made without knowing the number of populations.
A) Multiple t-tests increases the chance of a Type I error.
B) Multiple t-tests decreases the chance of a Type I error.
C) Multiple t-tests does not affect the chance of a Type I error.
D) This comparison cannot be made without knowing the number of populations.
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39
In the one-way 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) n- k
B) k - 1
C) n - 1
D) n - k + 1
A) n- k
B) k - 1
C) n - 1
D) n - k + 1
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40
In a completely randomized design for ANOVA, the numerator and denominator degrees of freedom are 4 and 25, respectively.The total number of observations must equal:
A) 24
B) 25
C) 29
D) 30
A) 24
B) 25
C) 29
D) 30
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41
SST measures the variation ____________________ treatments.
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42
In one-way analysis of variance, within-treatments variation is measured by:
A) sum of squares for error.
B) sum of squares for treatments.
C) total sum of squares.
D) standard deviation.
A) sum of squares for error.
B) sum of squares for treatments.
C) total sum of squares.
D) standard deviation.
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43
Consider the following ANOVA table:
The number of treatments is
A) 13
B) 5
C) 3
D) 12

A) 13
B) 5
C) 3
D) 12
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44
The F-test statistic in ANOVA is equal to MS____________________ divided by MS____________________ and H0 is rejected for ____________________ values of F.
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45
Consider the following ANOVA table:
The total number of observations is:
A) 25
B) 29
C) 30
D) 32

A) 25
B) 29
C) 30
D) 32
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46
If SST explains a significant portion of the total variation, we conclude that the population means ____________________ (do/do not) differ.
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47
The F-test statistic in a one-way ANOVA is equal to:
A) MST/MSE
B) SST/SSE
C) MSE/MST
D) SSE/SST
A) MST/MSE
B) SST/SSE
C) MSE/MST
D) SSE/SST
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48
If we square the t-statistic for two means, the result is the ____________________-statistic.
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49
The alternative hypothesis of ANOVA is that ____________________ population means are different.
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50
In one-way analysis of variance, if all the sample means are equal, then the:
A) total sum of squares is zero.
B) sum of squares for treatments is zero.
C) sum of squares for error is zero.
D) sum of squares for error equals sum of squares for treatments.
A) total sum of squares is zero.
B) sum of squares for treatments is zero.
C) sum of squares for error is zero.
D) sum of squares for error equals sum of squares for treatments.
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51
SSE measures the variation ____________________ treatments.
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52
The ANOVA procedure tests to determine whether differences exist between two or more population ____________________.
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53
In which case can an F-test be used to compare two population means?
A) For one tail tests only.
B) For two tail tests only.
C) For either one or two tail tests.
D) None of these choices.
A) For one tail tests only.
B) For two tail tests only.
C) For either one or two tail tests.
D) None of these choices.
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54
Which of the following statements is true?
A) F = t2
B) The F-test can be used instead of a two tail t-test when you compare two population means.
C) Doing three t-tests is statistically equivalent to doing one F-test when you compare three population means.
D) All of these choices are true.
A) F = t2
B) The F-test can be used instead of a two tail t-test when you compare two population means.
C) Doing three t-tests is statistically equivalent to doing one F-test when you compare three population means.
D) All of these choices are true.
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55
Consider the following partial ANOVA table:
The numerator and denominator degrees of freedom for the F-test (identified by asterisks) are
A) 4 and 15
B) 3 and 16
C) 15 and 4
D) 16 and 3

A) 4 and 15
B) 3 and 16
C) 15 and 4
D) 16 and 3
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56
The null hypothesis of ANOVA is that all the population means are ____________________.
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57
The F-test in ANOVA requires that the random variable be ____________________ distributed with equal ____________________.
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58
Which of the following components in an ANOVA table is not additive?
A) Sum of squares
B) Degrees of freedom
C) Mean squares
D) All of these choices are additive.
A) Sum of squares
B) Degrees of freedom
C) Mean squares
D) All of these choices are additive.
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59
The numerator and denominator degrees of freedom for the F-test in a one-way ANOVA are, respectively,
A) (n- k) and (k - 1)
B) (k - 1) and (n - k)
C) (k - n) and (n - 1)
D) (n- 1) and (k - n)
A) (n- k) and (k - 1)
B) (k - 1) and (n - k)
C) (k - n) and (n - 1)
D) (n- 1) and (k - n)
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60
In ANOVA the populations are classified according to one or more criterion, called ____________________.
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61
The Bonferroni adjustment decreases the experimentwise Type I error rate, but it increases the probability of a Type II error.
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62
In Fisher's least significant difference (LSD) multiple comparison method, the LSD value will be the same for all pairs of means if:
A) all sample sizes are the same.
B) all sample means are the same.
C) all population means are the same.
D) None of these choices.
A) all sample sizes are the same.
B) all sample means are the same.
C) all population means are the same.
D) None of these choices.
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63
Sub Sandwich Customers
The marketing manager of a Sub Shop 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 Sub Shops, hamburger emporiums, and fast-food chicken restaurants are different.The ages of eight randomly selected customers of each of the restaurants are recorded and listed below.From previous analyses we know that the ages are normally distributed with equal variances for each group.

-{Sub Sandwich Customers Narrative} 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?
The marketing manager of a Sub Shop 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 Sub Shops, hamburger emporiums, and fast-food chicken restaurants are different.The ages of eight randomly selected customers of each of the restaurants are recorded and listed below.From previous analyses we know that the ages are normally distributed with equal variances for each group.

-{Sub Sandwich Customers Narrative} 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?
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64
GMAT Scores
A recent college graduate is in the process of deciding which one of three graduate schools he should apply to.He decides to judge the quality of the schools on the basis of the Graduate Management Admission Test (GMAT) scores of those who are accepted into the school.A random sample of six students in each school produced the following GMAT scores.Assume that the data are normally distributed with equal variances for each school.

-{GMAT Scores Narrative} Can he infer at the 10% significance level that the GMAT scores differ among the three schools?
A recent college graduate is in the process of deciding which one of three graduate schools he should apply to.He decides to judge the quality of the schools on the basis of the Graduate Management Admission Test (GMAT) scores of those who are accepted into the school.A random sample of six students in each school produced the following GMAT scores.Assume that the data are normally distributed with equal variances for each school.

-{GMAT Scores Narrative} Can he infer at the 10% significance level that the GMAT scores differ among the three schools?
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65
Fisher's least significant difference (LSD) multiple comparison method is flawed because
A) it will increase ; the probability of committing a Type II error
B) it will increase ; the probability of committing a Type I error
C) it will increase both and , the probabilities of committing Type I and Type II errors, respectively.
D) None of these choices.
A) it will increase ; the probability of committing a Type II error
B) it will increase ; the probability of committing a Type I error
C) it will increase both and , the probabilities of committing Type I and Type II errors, respectively.
D) None of these choices.
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66
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.


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67
Arthritis Pain Formulas
A pharmaceutical manufacturer has been researching new medications formulas to provide quicker relief of arthritis pain.Their laboratories have produced three different medications and they want to determine if the different medications produce different responses.Fifteen people who complained of arthritis pains were recruited for an experiment; five were randomly assigned to each medication.Each person was asked to take the medicine and report the length of time until some relief was felt (minutes).The results are shown below.(Assume normal distributions with equal variances.)

-{Arthritis Pain Formulas Narrative} Set up the ANOVA Table.Use = 0.05 to determine the critical value.
A pharmaceutical manufacturer has been researching new medications formulas to provide quicker relief of arthritis pain.Their laboratories have produced three different medications and they want to determine if the different medications produce different responses.Fifteen people who complained of arthritis pains were recruited for an experiment; five were randomly assigned to each medication.Each person was asked to take the medicine and report the length of time until some relief was felt (minutes).The results are shown below.(Assume normal distributions with equal variances.)

-{Arthritis Pain Formulas Narrative} Set up the ANOVA Table.Use = 0.05 to determine the critical value.
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68
Tukey's multiple comparison method is based on the Studentized range statistic.
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69
Fisher's least significant difference method (LSD) substitutes the pooled variance estimator from the equal variances t-test with the MSE from ANOVA.
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70
Which of the following statements about multiple comparison methods is false?
A) They are to be use once the F-test in ANOVA has been rejected.
B) They are used to determine which particular population means differ.
C) There are many different multiple comparison methods but all yield the same conclusions.
D) All of these choices are true.
A) They are to be use once the F-test in ANOVA has been rejected.
B) They are used to determine which particular population means differ.
C) There are many different multiple comparison methods but all yield the same conclusions.
D) All of these choices are true.
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71
GMAT Scores
A recent college graduate is in the process of deciding which one of three graduate schools he should apply to.He decides to judge the quality of the schools on the basis of the Graduate Management Admission Test (GMAT) scores of those who are accepted into the school.A random sample of six students in each school produced the following GMAT scores.Assume that the data are normally distributed with equal variances for each school.

-{GMAT Scores Narrative} Set up the ANOVA Table.Use = 0.05 to determine the critical value.
A recent college graduate is in the process of deciding which one of three graduate schools he should apply to.He decides to judge the quality of the schools on the basis of the Graduate Management Admission Test (GMAT) scores of those who are accepted into the school.A random sample of six students in each school produced the following GMAT scores.Assume that the data are normally distributed with equal variances for each school.

-{GMAT Scores Narrative} Set up the ANOVA Table.Use = 0.05 to determine the critical value.
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72
Sub Sandwich Customers
The marketing manager of a Sub Shop 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 Sub Shops, hamburger emporiums, and fast-food chicken restaurants are different.The ages of eight randomly selected customers of each of the restaurants are recorded and listed below.From previous analyses we know that the ages are normally distributed with equal variances for each group.

-{Sub Sandwich Customers Narrative} Set up the ANOVA Table.Use = 0.05 to determine the critical value.
The marketing manager of a Sub Shop 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 Sub Shops, hamburger emporiums, and fast-food chicken restaurants are different.The ages of eight randomly selected customers of each of the restaurants are recorded and listed below.From previous analyses we know that the ages are normally distributed with equal variances for each group.

-{Sub Sandwich Customers Narrative} Set up the ANOVA Table.Use = 0.05 to determine the critical value.
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73
TV News Viewing Habits
A statistician employed by a television rating service wanted to determine if there were differences in television viewing habits among three different cities in New York.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.The results are shown below.(Assume normal distributions with equal variances.)

-{TV News Viewing Habits Narrative} Can she infer at the 5% significance level that differences in hours of television watching exist among the three cities?
A statistician employed by a television rating service wanted to determine if there were differences in television viewing habits among three different cities in New York.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.The results are shown below.(Assume normal distributions with equal variances.)

-{TV News Viewing Habits Narrative} Can she infer at the 5% significance level that differences in hours of television watching exist among the three cities?
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74
The Bonferroni adjustment to Fisher's Least Significant Difference (LSD) multiple comparison method is made by dividing the specified experimentwise Type I error rate by the number of pairs of population means.
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75
Gold Funds
An investor studied the percentage rates of return of three different gold funds.Random samples of percentage rates of return for four periods were taken from each fund.The results appear in the table below:

-{Gold Funds Narrative} Set up the ANOVA Table.Use = 0.05 to determine the critical value.
An investor studied the percentage rates of return of three different gold funds.Random samples of percentage rates of return for four periods were taken from each fund.The results appear in the table below:

-{Gold Funds Narrative} Set up the ANOVA Table.Use = 0.05 to determine the critical value.
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76
Tukey's multiple comparison method is more powerful than Fisher's LSD Method at finding differences in pairwise population means.
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77
TV News Viewing Habits
A statistician employed by a television rating service wanted to determine if there were differences in television viewing habits among three different cities in New York.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.The results are shown below.(Assume normal distributions with equal variances.)

-{TV News Viewing Habits Narrative} Set up the ANOVA Table.Use = 0.05 to determine the critical value.
A statistician employed by a television rating service wanted to determine if there were differences in television viewing habits among three different cities in New York.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.The results are shown below.(Assume normal distributions with equal variances.)

-{TV News Viewing Habits Narrative} Set up the ANOVA Table.Use = 0.05 to determine the critical value.
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78
Arthritis Pain Formulas
A pharmaceutical manufacturer has been researching new medications formulas to provide quicker relief of arthritis pain.Their laboratories have produced three different medications and they want to determine if the different medications produce different responses.Fifteen people who complained of arthritis pains were recruited for an experiment; five were randomly assigned to each medication.Each person was asked to take the medicine and report the length of time until some relief was felt (minutes).The results are shown below.(Assume normal distributions with equal variances.)

-{Arthritis Pain Formulas Narrative} Do these data provide sufficient evidence to indicate that differences in the average time of relief exist among the three medications? Use = 0.05.
A pharmaceutical manufacturer has been researching new medications formulas to provide quicker relief of arthritis pain.Their laboratories have produced three different medications and they want to determine if the different medications produce different responses.Fifteen people who complained of arthritis pains were recruited for an experiment; five were randomly assigned to each medication.Each person was asked to take the medicine and report the length of time until some relief was felt (minutes).The results are shown below.(Assume normal distributions with equal variances.)

-{Arthritis Pain Formulas Narrative} Do these data provide sufficient evidence to indicate that differences in the average time of relief exist among the three medications? Use = 0.05.
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79
Multiple comparison methods are used to determine whether or not any differences occur amongst a group of population means.
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80
Gold Funds
An investor studied the percentage rates of return of three different gold funds.Random samples of percentage rates of return for four periods were taken from each fund.The results appear in the table below:

-{Gold Funds Narrative} Test at the 5% significance level to determine whether the mean percentage rates for the three funds differ.
An investor studied the percentage rates of return of three different gold funds.Random samples of percentage rates of return for four periods were taken from each fund.The results appear in the table below:

-{Gold Funds Narrative} Test at the 5% significance level to determine whether the mean percentage rates for the three funds differ.
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