Question 1

Several employees have submitted different methods of assembling a product. Sample data for each method are:   How many treatments are there?

Accepted Answer

A)  3 
B)  4 
C)  12 
D)  0 
A)  3 
B)  4 
C)  12 
D)  0 B

Question 2

An electronics company wants to compare the quality of their cell phones to the cell phones from three competitors. They sample 10 phones from each company and count the number of defects for each phone. If ANOVA is used to compare the average number of defects, the treatments would be defined as:

Accepted Answer

A)  the number of cell phones sampled. 
B)  the average number of defects. 
C)  the total number of phones. 
D)  the four companies. 
A)  the number of cell phones sampled. 
B)  the average number of defects. 
C)  the total number of phones. 
D)  the four companies. D

Question 3

An electronics company wants to compare the quality of their cell phones to the cell phones from three competitors. They sample 10 phones from each company and count the number of defects for each phone. Using Excel, what test is used to compare the average number of defects?

Accepted Answer

A)  ANOVA: Single Factor. 
B)  ANOVA: Two-Factor with Replication. 
C)  ANOVA: Two-Factor without Replication. 
D)  F-Test Two Sample for Variances. 
E)  t-Test: Paired Two Sample for Means. 
A)  ANOVA: Single Factor. 
B)  ANOVA: Two-Factor with Replication. 
C)  ANOVA: Two-Factor without Replication. 
D)  F-Test Two Sample for Variances. 
E)  t-Test: Paired Two Sample for Means. A

Question 4

i. To employ ANOVA, the populations need not have equal standard deviations. ii. To employ ANOVA, the populations being studied need not be normally distributed.
iii. A technique that is efficient when simultaneously comparing more than two population means is known as analysis of deviation.

Accepted Answer

A)  (i), (ii), and (iii) are all correct statements. 
B)  (i) and (ii) are correct statements but not (iii). 
C)  (i) and (iii) are correct statements but not (ii). 
D)  (ii) and (iii) are correct statements but not (i). 
E)  (i), (ii), and (iii) are all false statements. 
A)  (i), (ii), and (iii) are all correct statements. 
B)  (i) and (ii) are correct statements but not (iii). 
C)  (i) and (iii) are correct statements but not (ii). 
D)  (ii) and (iii) are correct statements but not (i). 
E)  (i), (ii), and (iii) are all false statements.

Question 5

If the confidence interval for the difference in treatment means contains zero, then:

Accepted Answer

A)  we can conclude that there is a significant difference in the selected treatment means. 
B)  we can conclude that there is no significant difference in the selected treatment means. 
C)  we can conclude that the selected treatment means both equal zero. 
D)  we can conclude that at least one of the selected treatment means equals zero. 
A)  we can conclude that there is a significant difference in the selected treatment means. 
B)  we can conclude that there is no significant difference in the selected treatment means. 
C)  we can conclude that the selected treatment means both equal zero. 
D)  we can conclude that at least one of the selected treatment means equals zero.

Question 6

i. One characteristic of the F distribution is that computed F can only range between -1 and +1. ii. The shape of the F distribution is determined by the degrees of freedom for the F-statistic, one for the numerator and one for the denominator.
iii. The F distribution's curve is positively skewed.

Accepted Answer

A)  (i), (ii), and (iii) are all correct statements. 
B)  (i) and (ii) are correct statements but not (iii). 
C)  (i) and (iii) are correct statements but not (ii). 
D)  (ii) and (iii) are correct statements but not (i). 
E)  (i), (ii), and (iii) are all false statements. 
A)  (i), (ii), and (iii) are all correct statements. 
B)  (i) and (ii) are correct statements but not (iii). 
C)  (i) and (iii) are correct statements but not (ii). 
D)  (ii) and (iii) are correct statements but not (i). 
E)  (i), (ii), and (iii) are all false statements.

Question 7

Suppose we select 20 observations from each of five treatments. The appropriate degrees of freedom are:

Accepted Answer

A)  4 and 95 
B)  5 and 20 
C)  4 and 19 
D)  4 and 20 
A)  4 and 95 
B)  5 and 20 
C)  4 and 19 
D)  4 and 20

Question 8

Which of the following is NOT a characteristic of the F distribution?

Accepted Answer

A)  It is based on two sets of degrees of freedom. 
B)  It is positively skewed. 
C)  As the values of X increase, the F curve approaches the X-axis and eventually equals zero. 
D)  It is asymptotic. 
A)  It is based on two sets of degrees of freedom. 
B)  It is positively skewed. 
C)  As the values of X increase, the F curve approaches the X-axis and eventually equals zero. 
D)  It is asymptotic.

Question 9

Two accounting professors decided to compare the variation of their grading procedures. To accomplish this, they each graded the same 10 exams with the following results:   What is the critical value of F at the 0.05 level of significance?

Accepted Answer

A)  5.85 
B)  5.35 
C)  3.18 
D)  4.03 
A)  5.85 
B)  5.35 
C)  3.18 
D)  4.03

Question 10

In ANOVA analysis, when the null hypothesis is rejected, we can find which means are different by:

Accepted Answer

A)  constructing confidence intervals. 
B)  adding another treatment. 
C)  doing an additional ANOVA. 
D)  doing attest. 
A)  constructing confidence intervals. 
B)  adding another treatment. 
C)  doing an additional ANOVA. 
D)  doing attest.

Question 11

Two accounting professors decided to compare the variation of their grading procedures. To accomplish this, they each graded the same 10 exams with the following results:   
What is H1?

Accepted Answer

A)   $\sigma$21=  $\sigma$22 
B)   $\sigma$21$\neq$ $\sigma$22 
C)  µ1=µ2 
D)  µ1$\neq$µ2 
A)   $\sigma$21=  $\sigma$22 
B)   $\sigma$21$\neq$ $\sigma$22 
C)  µ1=µ2 
D)  µ1$\neq$µ2

Question 12

i. The statistical technique used to test the equality of three or more population means is called analysis of variance (ANOVA). ii. To employ ANOVA, the populations should have approximately equal standard deviations.
iii. The least number of sources of variation in ANOVA is two.

Accepted Answer

A)  (i), (ii), and (iii) are all correct statements. 
B)  (i) and (ii) are correct statements but not (iii). 
C)  (i) and (iii) are correct statements but not (ii). 
D)  (ii) and (iii) are correct statements but not (i). 
E)  (i), (ii), and (iii) are all false statements. 
A)  (i), (ii), and (iii) are all correct statements. 
B)  (i) and (ii) are correct statements but not (iii). 
C)  (i) and (iii) are correct statements but not (ii). 
D)  (ii) and (iii) are correct statements but not (i). 
E)  (i), (ii), and (iii) are all false statements.

Question 13

A preliminary study of hourly wages paid to unskilled employees in three metropolitan areas was conducted. Seven employees were included from Area A, 9 from Area B and 12 from Area C. The test statistic was computed to be 4.91. What can we conclude at the 0.05 level?

Accepted Answer

A)  Mean hourly wages of unskilled employees all areas are equal. 
B)  Mean hourly wages in at least 2 metropolitan areas are different. 
C)  More degrees of freedom are needed. 
A)  Mean hourly wages of unskilled employees all areas are equal. 
B)  Mean hourly wages in at least 2 metropolitan areas are different. 
C)  More degrees of freedom are needed.

Question 14

Given the following Analysis of Variance table for three treatments each with six observations.   What is the mean square for treatments?

Accepted Answer

A)  71.2 
B)  71.4 
C)  558 
D)  534 
A)  71.2 
B)  71.4 
C)  558 
D)  534

Question 15

Suppose that an automobile manufacturer designed a radically new lightweight engine and wants to recommend the grade of gasoline to use. The four grades are: below regular, regular, premium, and super premium. The test car made three trial runs on the test track using each of the four grades. Is there a difference in the performance between the four grades of gas?   If you were to use Excel to assist in your solution to this problem, which test would you use?

Accepted Answer

A)  ANOVA: Single Factor. 
B)  ANOVA: Two-Factor with Replication. 
C)  ANOVA: Two-Factor without Replication. 
D)  F-Test Two Sample for Variances. 
E)  t-Test: Paired Two Sample for Means. 
A)  ANOVA: Single Factor. 
B)  ANOVA: Two-Factor with Replication. 
C)  ANOVA: Two-Factor without Replication. 
D)  F-Test Two Sample for Variances. 
E)  t-Test: Paired Two Sample for Means.

Question 16

i. For the population means, the alternate hypothesis used in the analysis of variance test states thatµ1=µ2=µ3. ii. For an ANOVA test, rejection of the null hypothesis does not identify which populations differ significantly.
iii. If the computed value of F is 4.01 and the critical value is 2.67, we would conclude that all the population means are equal.

Accepted Answer

A)  (i), (ii), and (iii) are all correct statements. 
B)  (ii) is a correct statement but not (i) or (iii). 
C)  (iii) is a correct statement but not (i) or (ii). 
D)  (ii) and (iii) are correct statements but not (i). 
E)  (i), (ii), and (iii) are all false statements. 
A)  (i), (ii), and (iii) are all correct statements. 
B)  (ii) is a correct statement but not (i) or (iii). 
C)  (iii) is a correct statement but not (i) or (ii). 
D)  (ii) and (iii) are correct statements but not (i). 
E)  (i), (ii), and (iii) are all false statements.

Question 17

A manufacturer of automobile transmissions uses three different processes. The management ordered a study of the production costs to see if there is a difference among the three processes. A summary of the findings is shown below.   What is the calculated F?

Accepted Answer

A)  0.086 
B)  1.168 
C)  11.56 
D)  13.50 
A)  0.086 
B)  1.168 
C)  11.56 
D)  13.50

Question 18

A random sample of 30 executives from companies with assets over $1 million was selected and asked for their annual income and level of education. The ANOVA comparing the average income among three levels of education rejected the null hypothesis. The Mean Square Error (MSE) was 243.7. The following table summarized the results:   
When comparing the mean annual incomes for executives with Undergraduate and Master's Degree or more, the following 95% confidence interval can be constructed:

Accepted Answer

A)  2.0  $\pm$ 2.052 * 6.52 
B)  2.0  $\pm$ 3.182 * 6.51 
C)  2.0  $\pm$ 2.052 * 42.46 
A)  2.0  $\pm$ 2.052 * 6.52 
B)  2.0  $\pm$ 3.182 * 6.51 
C)  2.0  $\pm$ 2.052 * 42.46

Question 19

A manufacturer of automobile transmissions uses three different processes. The management ordered a study of the production costs to see if there is a difference among the three processes. A summary of the findings is shown below.   What is the critical value of F at the 1% level of significance?

Accepted Answer

A)  99.46 
B)  5.49 
C)  5.39 
D)  4.61 
A)  99.46 
B)  5.49 
C)  5.39 
D)  4.61

Question 20

i. If the computed value of F is 4.01 and the critical value is 2.67, we would conclude that all the population means are equal. ii. Not rejecting the null hypothesis in ANOVA indicates that the population means are equal.
iii. The null hypothesis for an ANOVA isµ1-µ2=µ3.

Accepted Answer

A)  (i), (ii), and (iii) are all correct statements. 
B)  (i) and (ii) are correct statements but not (iii). 
C)  (i) and (iii) are correct statements but not (ii). 
D)  (ii) and (iii) are correct statements but not (i). 
E)  (i), (ii), and (iii) are all false statements. 
A)  (i), (ii), and (iii) are all correct statements. 
B)  (i) and (ii) are correct statements but not (iii). 
C)  (i) and (iii) are correct statements but not (ii). 
D)  (ii) and (iii) are correct statements but not (i). 
E)  (i), (ii), and (iii) are all false statements.

Several employees have submitted different methods of assembling a product. Sample data for each method are: How many treatments are there?

An electronics company wants to compare the quality of their cell phones to the cell phones from three competitors. They sample 10 phones from each company and count the number of defects for each phone. If ANOVA is used to compare the average number of defects, the treatments would be defined as:

An electronics company wants to compare the quality of their cell phones to the cell phones from three competitors. They sample 10 phones from each company and count the number of defects for each phone. Using Excel, what test is used to compare the average number of defects?

i. To employ ANOVA, the populations need not have equal standard deviations. ii. To employ ANOVA, the populations being studied need not be normally distributed. iii. A technique that is efficient when simultaneously comparing more than two population means is known as analysis of deviation.

If the confidence interval for the difference in treatment means contains zero, then:

i. One characteristic of the F distribution is that computed F can only range between -1 and +1. ii. The shape of the F distribution is determined by the degrees of freedom for the F-statistic, one for the numerator and one for the denominator. iii. The F distribution's curve is positively skewed.

Suppose we select 20 observations from each of five treatments. The appropriate degrees of freedom are:

Which of the following is NOT a characteristic of the F distribution?

Two accounting professors decided to compare the variation of their grading procedures. To accomplish this, they each graded the same 10 exams with the following results: What is the critical value of F at the 0.05 level of significance?

In ANOVA analysis, when the null hypothesis is rejected, we can find which means are different by:

Two accounting professors decided to compare the variation of their grading procedures. To accomplish this, they each graded the same 10 exams with the following results: What is H₁?

i. The statistical technique used to test the equality of three or more population means is called analysis of variance (ANOVA). ii. To employ ANOVA, the populations should have approximately equal standard deviations. iii. The least number of sources of variation in ANOVA is two.

A preliminary study of hourly wages paid to unskilled employees in three metropolitan areas was conducted. Seven employees were included from Area A, 9 from Area B and 12 from Area C. The test statistic was computed to be 4.91. What can we conclude at the 0.05 level?

Given the following Analysis of Variance table for three treatments each with six observations. What is the mean square for treatments?

A manufacturer of automobile transmissions uses three different processes. The management ordered a study of the production costs to see if there is a difference among the three processes. A summary of the findings is shown below. What is the calculated F?

A manufacturer of automobile transmissions uses three different processes. The management ordered a study of the production costs to see if there is a difference among the three processes. A summary of the findings is shown below. What is the critical value of F at the 1% level of significance?

i. If the computed value of F is 4.01 and the critical value is 2.67, we would conclude that all the population means are equal. ii. Not rejecting the null hypothesis in ANOVA indicates that the population means are equal. iii. The null hypothesis for an ANOVA isµ₁-µ₂=µ₃.

What Is Statistics

Describing Data: Frequency Tables, Frequency Distributions, and Graphic Presentation

Describing Data: Numerical Measures

A Survey of Probability Concepts

Discrete Probability Distributions

Continuous Probability Distributions

Sampling Methods and the Central Limit Theorem

Estimation and Confidence Intervals

One-Sample Tests of Hypothesis

Two-Sample Tests of Hypothesis

Linear Regression and Correlation

Multiple Regression and Correlation Analysis

Chi-Square Applications

Index Numbers

Time Series and Forecasting

An Introduction to Decision Theory

Filters

Exam 11: Analysis of Variance

Several employees have submitted different methods of assembling a product. Sample data for each method are: How many treatments are there?

An electronics company wants to compare the quality of their cell phones to the cell phones from three competitors. They sample 10 phones from each company and count the number of defects for each phone. If ANOVA is used to compare the average number of defects, the treatments would be defined as:

An electronics company wants to compare the quality of their cell phones to the cell phones from three competitors. They sample 10 phones from each company and count the number of defects for each phone. Using Excel, what test is used to compare the average number of defects?

i. To employ ANOVA, the populations need not have equal standard deviations. ii. To employ ANOVA, the populations being studied need not be normally distributed. iii. A technique that is efficient when simultaneously comparing more than two population means is known as analysis of deviation.

If the confidence interval for the difference in treatment means contains zero, then:

i. One characteristic of the F distribution is that computed F can only range between -1 and +1. ii. The shape of the F distribution is determined by the degrees of freedom for the F-statistic, one for the numerator and one for the denominator. iii. The F distribution's curve is positively skewed.

Suppose we select 20 observations from each of five treatments. The appropriate degrees of freedom are:

Which of the following is NOT a characteristic of the F distribution?

Two accounting professors decided to compare the variation of their grading procedures. To accomplish this, they each graded the same 10 exams with the following results: What is the critical value of F at the 0.05 level of significance?

In ANOVA analysis, when the null hypothesis is rejected, we can find which means are different by:

Two accounting professors decided to compare the variation of their grading procedures. To accomplish this, they each graded the same 10 exams with the following results: What is H1?

i. The statistical technique used to test the equality of three or more population means is called analysis of variance (ANOVA). ii. To employ ANOVA, the populations should have approximately equal standard deviations. iii. The least number of sources of variation in ANOVA is two.

A preliminary study of hourly wages paid to unskilled employees in three metropolitan areas was conducted. Seven employees were included from Area A, 9 from Area B and 12 from Area C. The test statistic was computed to be 4.91. What can we conclude at the 0.05 level?

Given the following Analysis of Variance table for three treatments each with six observations. What is the mean square for treatments?

A manufacturer of automobile transmissions uses three different processes. The management ordered a study of the production costs to see if there is a difference among the three processes. A summary of the findings is shown below. What is the calculated F?

A manufacturer of automobile transmissions uses three different processes. The management ordered a study of the production costs to see if there is a difference among the three processes. A summary of the findings is shown below. What is the critical value of F at the 1% level of significance?

i. If the computed value of F is 4.01 and the critical value is 2.67, we would conclude that all the population means are equal. ii. Not rejecting the null hypothesis in ANOVA indicates that the population means are equal. iii. The null hypothesis for an ANOVA isµ1-µ2=µ3.

What Is Statistics

Describing Data: Frequency Tables, Frequency Distributions, and Graphic Presentation

Describing Data: Numerical Measures

A Survey of Probability Concepts

Discrete Probability Distributions

Continuous Probability Distributions

Sampling Methods and the Central Limit Theorem

Estimation and Confidence Intervals

One-Sample Tests of Hypothesis

Two-Sample Tests of Hypothesis

Linear Regression and Correlation

Multiple Regression and Correlation Analysis

Chi-Square Applications

Index Numbers

Time Series and Forecasting

An Introduction to Decision Theory

Filters

Two accounting professors decided to compare the variation of their grading procedures. To accomplish this, they each graded the same 10 exams with the following results: What is H₁?

i. If the computed value of F is 4.01 and the critical value is 2.67, we would conclude that all the population means are equal. ii. Not rejecting the null hypothesis in ANOVA indicates that the population means are equal. iii. The null hypothesis for an ANOVA isµ₁-µ₂=µ₃.