Deck 14: Analysis of Variance

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Question
The F-test used in one-way ANOVA is an extension of the t-test of μ1 − μ2.
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Question
When the F-test is used for ANOVA,the rejection region is always in the right tail.
Question
We can use the F-test to determine whether μ1 is greater than μ2.
Question
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.
Question
The within-treatments variation provides a measure of the amount of variation in the response variables that is caused by the treatments.
Question
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.
Question
We use the analysis of variance (ANOVA)technique to compare two or more population means.
Question
The F-statistic in a one-way ANOVA represents the variation within the treatments divided by the variation between the treatments.
Question
The analysis of variance (ANOVA)technique analyzes the variance of the data to determine whether differences exist between the population means.
Question
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.
Question
The distribution of the test statistic for analysis of variance is the F-distribution.
Question
The F-test in ANOVA tests whether or not the population variances are equal.
Question
Conducting t-tests for each pair or population means is statistically equivalent to conducting one F-test comparing all the population means.
Question
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).
Question
The F-test in ANOVA is an expansion of the t-test for two independent population means.
Question
The analysis of variance (ANOVA)tests hypotheses about population variances and requires all the population means to be equal.
Question
The sum of squares for error is also known as the between-treatments variation.
Question
The sum of squares for treatments,SST,achieves its smallest value (zero)when all the sample means are equal.
Question
In ANOVA,a criterion by which the populations are classified is called a factor.
Question
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.
Question
In one-way analysis of variance,between-treatments variation is measured by the:

A)SSE
B)SS(Total)
C)SST
D)standard deviation
Question
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.
Question
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.
Question
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
Question
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.
Question
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.
Question
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.
Question
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
Question
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
Question
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.
Question
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.
Question
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.
Question
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.
Question
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.
Question
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)μ1 + μ2 + μ3 = 0
B)μ1 + μ2 + μ3 ≠ 0
C)μ1 = μ2 = μ3 = 0
D)μ1 = μ2 = μ3
Question
In one-way ANOVA,the term <strong>In one-way ANOVA,the term   refers to the:</strong> 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. <div style=padding-top: 35px> 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.
Question
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
Question
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.
Question
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.
Question
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.
Question
SST measures the variation ____________________ treatments.
Question
SSE measures the variation ____________________ treatments.
Question
If we square the t-statistic for two means,the result is the ____________________-statistic.
Question
The alternative hypothesis of ANOVA is that ____________________ population means are different.
Question
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)
Question
The ANOVA procedure tests to determine whether differences exist between two or more population ____________________.
Question
The null hypothesis of ANOVA is that all the population means are ____________________.
Question
Consider the following partial ANOVA table: <strong>Consider the following partial ANOVA table:   The numerator and denominator degrees of freedom for the F-test (identified by asterisks)are</strong> A)4 and 15 B)3 and 16 C)15 and 4 D)16 and 3 <div style=padding-top: 35px> 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
Question
The F-test statistic in ANOVA is equal to MS____________________ divided by MS____________________
and H0 is rejected for ____________________ values of F.
Question
The F-test statistic in a one-way ANOVA is equal to:

A)MST/MSE
B)SST/SSE
C)MSE/MST
D)SSE/SST
Question
If SST explains a significant portion of the total variation,we conclude that the population means ____________________ (do/do not)differ.
Question
Consider the following ANOVA table: <strong>Consider the following ANOVA table:   The total number of observations is:</strong> A)25 B)29 C)30 D)32 <div style=padding-top: 35px> The total number of observations is:

A)25
B)29
C)30
D)32
Question
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.
Question
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.
Question
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.
Question
In ANOVA the populations are classified according to one or more criterion,called ____________________.
Question
Consider the following ANOVA table: <strong>Consider the following ANOVA table:   The number of treatments is</strong> A)13 B)5 C)3 D)12 <div style=padding-top: 35px> The number of treatments is

A)13
B)5
C)3
D)12
Question
The F-test in ANOVA requires that the random variable be ____________________ distributed with equal ____________________.
Question
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.
Question
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.
Question
Fisher's least significant difference (LSD)multiple comparison method is flawed because

A)it will increase β; the probability of committing a Type II error
B)it will increase α; the probability of committing a Type I error
C)it will increase both α and β,the probabilities of committing Type I and Type II errors,respectively.
D)None of these choices.
Question
Tukey's multiple comparison method is more powerful than Fisher's LSD Method at finding differences in pairwise population means.
Question
In a completely randomized design,12 experimental units were assigned to the first treatment,15 units to the second treatment,and 18 units to the third treatment.A partial ANOVA table is shown below: In a completely randomized design,12 experimental units were assigned to the first treatment,15 units to the second treatment,and 18 units to the third treatment.A partial ANOVA table is shown below:   ​ a.Fill in the blanks (identified by asterisks)in the above ANOVA table. b.Test at the 5% significance level to determine if differences exist among the three treatment means.<div style=padding-top: 35px>
a.Fill in the blanks (identified by asterisks)in the above ANOVA table.
b.Test at the 5% significance level to determine if differences exist among the three treatment means.
Question
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 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 α = 0.05 to determine the critical value.<div style=padding-top: 35px> ​ ​
{TV News Viewing Habits Narrative} Set up the ANOVA Table.Use α = 0.05 to determine the critical value.
Question
Multiple comparison methods are used to determine whether or not any differences occur amongst a group of population means.
Question
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 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 α = 0.05 to determine the critical value.<div style=padding-top: 35px> ​ ​
{GMAT Scores Narrative} Set up the ANOVA Table.Use α = 0.05 to determine the critical value.
Question
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 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 α = 0.05.<div style=padding-top: 35px> ​ ​
{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 α = 0.05.
Question
Tukey's multiple comparison method is based on the Studentized range statistic.
Question
Fisher's least significant difference method (LSD)substitutes the pooled variance estimator from the equal variances t-test with the MSE from ANOVA.
Question
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. 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.  <div style=padding-top: 35px>
Question
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 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 α = 0.05 to determine the critical value.<div style=padding-top: 35px> ​ ​
{Gold Funds Narrative} Set up the ANOVA Table.Use α = 0.05 to determine the critical value.
Question
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.
Question
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.
Question
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 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 α = 0.05 to determine the critical value.<div style=padding-top: 35px> ​ ​
{Sub Sandwich Customers Narrative} Set up the ANOVA Table.Use α = 0.05 to determine the critical value.
Question
The Bonferroni adjustment decreases the experimentwise Type I error rate,but it increases the probability of a Type II error.
Question
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 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.<div style=padding-top: 35px> ​ ​
{Gold Funds Narrative} Test at the 5% significance level to determine whether the mean percentage rates for the three funds differ.
Question
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 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 α = 0.05 to determine the critical value.<div style=padding-top: 35px> ​ ​
{Arthritis Pain Formulas Narrative} Set up the ANOVA Table.Use α = 0.05 to determine the critical value.
Question
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 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?<div style=padding-top: 35px> ​ ​
{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?
Question
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 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?<div style=padding-top: 35px> ​ ​
{GMAT Scores Narrative} Can he infer at the 10% significance level that the GMAT scores differ among the three schools?
Question
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 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?<div style=padding-top: 35px> ​ ​
{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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Deck 14: Analysis of Variance
1
The F-test used in one-way ANOVA is an extension of the t-test of μ1 − μ2.
True
2
When the F-test is used for ANOVA,the rejection region is always in the right tail.
True
3
We can use the F-test to determine whether μ1 is greater than μ2.
False
4
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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5
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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6
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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7
We use the analysis of variance (ANOVA)technique to compare two or more population means.
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8
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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9
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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10
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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11
The distribution of the test statistic for analysis of variance is the F-distribution.
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12
The F-test in ANOVA tests whether or not the population variances are equal.
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13
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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14
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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15
The F-test in ANOVA is an expansion of the t-test for two independent population means.
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16
The analysis of variance (ANOVA)tests hypotheses about population variances and requires all the population means to be equal.
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17
The sum of squares for error is also known as the between-treatments variation.
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18
The sum of squares for treatments,SST,achieves its smallest value (zero)when all the sample means are equal.
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19
In ANOVA,a criterion by which the populations are classified is called a factor.
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20
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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21
In one-way analysis of variance,between-treatments variation is measured by the:

A)SSE
B)SS(Total)
C)SST
D)standard deviation
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22
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.
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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.
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24
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
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25
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.
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26
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.
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27
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.
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28
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
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29
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
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30
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.
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31
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.
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32
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.
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33
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.
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34
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.
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35
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)μ1 + μ2 + μ3 = 0
B)μ1 + μ2 + μ3 ≠ 0
C)μ1 = μ2 = μ3 = 0
D)μ1 = μ2 = μ3
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36
In one-way ANOVA,the term <strong>In one-way ANOVA,the term   refers to the:</strong> 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. 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.
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37
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
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38
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.
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39
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.
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40
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.
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41
SST measures the variation ____________________ treatments.
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42
SSE measures the variation ____________________ treatments.
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43
If we square the t-statistic for two means,the result is the ____________________-statistic.
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44
The alternative hypothesis of ANOVA is that ____________________ population means are different.
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45
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)
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46
The ANOVA procedure tests to determine whether differences exist between two or more population ____________________.
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47
The null hypothesis of ANOVA is that all the population means are ____________________.
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48
Consider the following partial ANOVA table: <strong>Consider the following partial ANOVA table:   The numerator and denominator degrees of freedom for the F-test (identified by asterisks)are</strong> A)4 and 15 B)3 and 16 C)15 and 4 D)16 and 3 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
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49
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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50
The F-test statistic in a one-way ANOVA is equal to:

A)MST/MSE
B)SST/SSE
C)MSE/MST
D)SSE/SST
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51
If SST explains a significant portion of the total variation,we conclude that the population means ____________________ (do/do not)differ.
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52
Consider the following ANOVA table: <strong>Consider the following ANOVA table:   The total number of observations is:</strong> A)25 B)29 C)30 D)32 The total number of observations is:

A)25
B)29
C)30
D)32
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53
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.
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54
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.
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55
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.
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56
In ANOVA the populations are classified according to one or more criterion,called ____________________.
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57
Consider the following ANOVA table: <strong>Consider the following ANOVA table:   The number of treatments is</strong> A)13 B)5 C)3 D)12 The number of treatments is

A)13
B)5
C)3
D)12
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58
The F-test in ANOVA requires that the random variable be ____________________ distributed with equal ____________________.
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59
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.
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60
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.
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61
Fisher's least significant difference (LSD)multiple comparison method is flawed because

A)it will increase β; the probability of committing a Type II error
B)it will increase α; the probability of committing a Type I error
C)it will increase both α and β,the probabilities of committing Type I and Type II errors,respectively.
D)None of these choices.
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62
Tukey's multiple comparison method is more powerful than Fisher's LSD Method at finding differences in pairwise population means.
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63
In a completely randomized design,12 experimental units were assigned to the first treatment,15 units to the second treatment,and 18 units to the third treatment.A partial ANOVA table is shown below: In a completely randomized design,12 experimental units were assigned to the first treatment,15 units to the second treatment,and 18 units to the third treatment.A partial ANOVA table is shown below:   ​ a.Fill in the blanks (identified by asterisks)in the above ANOVA table. b.Test at the 5% significance level to determine if differences exist among the three treatment means.
a.Fill in the blanks (identified by asterisks)in the above ANOVA table.
b.Test at the 5% significance level to determine if differences exist among the three treatment means.
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64
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 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 α = 0.05 to determine the critical value. ​ ​
{TV News Viewing Habits Narrative} Set up the ANOVA Table.Use α = 0.05 to determine the critical value.
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65
Multiple comparison methods are used to determine whether or not any differences occur amongst a group of population means.
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66
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 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 α = 0.05 to determine the critical value. ​ ​
{GMAT Scores Narrative} Set up the ANOVA Table.Use α = 0.05 to determine the critical value.
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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 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 α = 0.05. ​ ​
{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 α = 0.05.
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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
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. 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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71
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 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 α = 0.05 to determine the critical value. ​ ​
{Gold Funds Narrative} Set up the ANOVA Table.Use α = 0.05 to determine the critical value.
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72
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.
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73
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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74
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 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 α = 0.05 to determine the critical value. ​ ​
{Sub Sandwich Customers Narrative} Set up the ANOVA Table.Use α = 0.05 to determine the critical value.
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75
The Bonferroni adjustment decreases the experimentwise Type I error rate,but it increases the probability of a Type II error.
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76
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 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. ​ ​
{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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77
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 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 α = 0.05 to determine the critical value. ​ ​
{Arthritis Pain Formulas Narrative} Set up the ANOVA Table.Use α = 0.05 to determine the critical value.
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78
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 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? ​ ​
{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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79
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 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? ​ ​
{GMAT Scores Narrative} Can he infer at the 10% significance level that the GMAT scores differ among the three schools?
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80
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 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? ​ ​
{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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