Question 1

Find the critical value F0 for a one-tailed test using α = 0.05, with 8 numerator degrees of freedom and 15 denominator degrees of freedom

Accepted Answer

A) 3.20 
B) 3.22 
C) 4.10 
D) 2.64 
A) 3.20 
B) 3.22 
C) 4.10 
D) 2.64

Question 2

Given that the sum of squares for treatments (SST)for an ANOVA F-test is 9,000 and there are four total treatments, find the mean square for treatments (MST).

Accepted Answer

A) 2,250 
B) 1,500 
C) 3,000 
D) 1,800 
A) 2,250 
B) 1,500 
C) 3,000 
D) 1,800

Question 3

In a study to determine the least amount of time necessary to clean an SUV while maintaining a high quality standard, the owner of a chain of car washes designed an experiment where 20
Employees were divided into four groups, each with five members. Each member of each group
Was assigned an SUV to clean within a certain time limit. The time limits for the groups were 20
Minutes, 25 minutes, 30 minutes, and 35 minutes. After the time limits for each group had expired,
The owner inspected each SUV and rated the quality of the cleaning job on a scale of 1 to 10. What
Are the factor levels for this study?

Accepted Answer

A) the quality ratings: 1, 2, 3, 4, 5, 6, 7, 8, 9, 10 
B) the number of employees in each group: 5 
C) the number of groups: 4 
D) the time limits: 20 min, 25 min, 30 min, 35 min 
A) the quality ratings: 1, 2, 3, 4, 5, 6, 7, 8, 9, 10 
B) the number of employees in each group: 5 
C) the number of groups: 4 
D) the time limits: 20 min, 25 min, 30 min, 35 min

Question 4

A scientist is hoping to compare the mean levels of DDT toxin found in three species of fish in a local river. He randomly samples 50 of each species to use in the analysis. For each fish, he
Measures the amount of DDT toxin present. Ideally he will be able to rank the species based on the
Mean level of toxin found in each of the three species. How many factors are present in this study?

Accepted Answer

A) 3 
B) 50 
C) 6 
D) 1 
A) 3 
B) 50 
C) 6 
D) 1

Question 5

Psychologists have found that people are generally reluctant to transmit bad news to their peers. This phenomenon has been named the ʺMUM effect.ʺ To investigate the cause of the MUM effect,
Undergraduates at a university participated in an experiment. Each subject was asked to
Administer an IQ test to another student and then provide the test taker with his or her percentile
Score. (Unknown to the subject, the test taker was a bogus student who was working with the
Researchers.)The experimenters manipulated two factors, subject visibility and success of test
Taker, each at two levels. Subject visibility was either visible or not visible to the test taker. Success
Of test taker was either top 20% or bottom 20%. Ten subjects were randomly assigned to each of the
2 x 2 = 4 experimental conditions. Then the time (in seconds)between the end of the test and the
Delivery of the percentile score from the subject to the test taker was measured. (This variable is
Called the latency to feedback.)What type of experimental design was employed in this study?

Accepted Answer

A) completely randomized design with four treatments 
B) randomized block design with four treatments and 10 blocks 
C) 2 x 2 factorial design with 10 replications 
D) 4 x 20 factorial design with no replications 
A) completely randomized design with four treatments 
B) randomized block design with four treatments and 10 blocks 
C) 2 x 2 factorial design with 10 replications 
D) 4 x 20 factorial design with no replications

Question 6

307 diamonds were sampled and randomly sorted into three groups of diamonds. These diamonds were randomly assigned to one of the three organizations, or groups (HRD, GIA, or IGI), that
Certify the appraisal of diamonds. A study was conducted to determine if the average size of
Diamonds reported by these three certification groups differ. A completely randomized design was
Used and the resulting ANOVA table is shown below. One-Way AOV for CARAT by CERT
$\begin{array}{llcccc}
\text { Source } & \text { DF } & \text { SS } & \text { MS } & \text { F } & \text { P } \
\hline \text { CERT } & 2 & 8.3265 & 4.16326 & 83.21 & 0.0000 \
\text { Error } & 305 & 15.2604 & 0.05003 & & \
\text { Total } & 307 & 23.5869 & & &
\end{array}$

Specify the null hypothesis for a test to compare the mean size of a diamond for the three certification groups (HRD, GIA, and IGI).

Accepted Answer

A)  At least two of the population mean carat weights differ for the three certification groups. 
B)  $\mathrm { H } _ { 0 } : \mu _ { \mathrm { HRD } } = \mu _ { \mathrm { GIA } } = \mu _ { \mathrm { IGI } }$, where $\mu _ { \mathrm { j } } =$ mean carat weight for certification group i 
C)  $\mathrm { H } _ { 0 } : \mu = 0$, where $\mu =$ mean carat weight. 
D)  $\mathrm { H } _ { 0 } : \mu _ { \mathrm { HRD } } = \mu _ { \mathrm { GIA } } = \mu _ { \mathrm { IGI } } = 0$, where $\mu _ { \mathrm { j } } =$ mean carat weight for certification group i 
A)  At least two of the population mean carat weights differ for the three certification groups. 
B)  $\mathrm { H } _ { 0 } : \mu _ { \mathrm { HRD } } = \mu _ { \mathrm { GIA } } = \mu _ { \mathrm { IGI } }$, where $\mu _ { \mathrm { j } } =$ mean carat weight for certification group i 
C)  $\mathrm { H } _ { 0 } : \mu = 0$, where $\mu =$ mean carat weight. 
D)  $\mathrm { H } _ { 0 } : \mu _ { \mathrm { HRD } } = \mu _ { \mathrm { GIA } } = \mu _ { \mathrm { IGI } } = 0$, where $\mu _ { \mathrm { j } } =$ mean carat weight for certification group i

Question 7

Complete the ANOVA table. $$\begin{array} { l c c c c } 
\hline \text { Source } & \text { df } & \text { SS } & \text { MS } & F \
\hline \text { Treatments } & 3 & 857.1 & & \
\text { Error } & 8 & 372.8 & & \
\hline \text { Total } & & & & \
\hline
\end{array}$$

Accepted Answer

The answer of Complete the ANOVA table. \[\begin{array} { l...

Question 8

A partially completed ANOVA table for a completely randomized design is shown here. $\begin{array}{lcccc}
\text { Source } & \text { df } & \text { SS } & \text { MS } & F \
\hline \text { Time } & & 25.2 & \
\text { Error } & 11 & & \
\hline \text { Total } & 13 & 86.4 & \
\hline
\end{array}$

a. Complete the ANOVA table.
b. How many treatments are involved in the experiment?
c. Do the data provide sufficient evidence to indicate a difference among the population means? Test using $\alpha = .05$.

Accepted Answer

None

Question 9

A certain HMO is attempting to show the benefits of managed care to an insurance company. The HMO believes that certain types of doctors are more cost-effective than others. One theory is that
Certification level is an important factor in measuring the cost-effectiveness of physicians. To
Investigate this, the HMO obtained independent random samples of 25 physicians from each of the
Three certification levels-Board certified (C); Uncertified, board eligible (E); and Uncertified, board
Ineligible (I)-and recorded the total per-member, per-month charges for each (a total of 25
Physicians). In order to compare the mean charges for the three groups, the data will be subjected
To an analysis of variance. Write the null hypothesis tested by the ANOVA.

Accepted Answer

A)  $H _ { 0 } : \beta _ { 1 } = \beta _ { 2 } = \beta _ { 3 } = 0$ 
B)  $H _ { 0 } : p _ { 1 } = p _ { 2 } = p _ { 3 }$ 
C)  $H _ { 0 } : \mu _ { \mathrm { C } } = \mu _ { \mathrm { E } } = \mu _ { \mathrm { I } } = 0$ 
D)  $H _ { 0 } : \mu _ { \mathrm { C } } = \mu _ { \mathrm { E } } = \mu _ { \mathrm { I } }$ 
A)  $H _ { 0 } : \beta _ { 1 } = \beta _ { 2 } = \beta _ { 3 } = 0$ 
B)  $H _ { 0 } : p _ { 1 } = p _ { 2 } = p _ { 3 }$ 
C)  $H _ { 0 } : \mu _ { \mathrm { C } } = \mu _ { \mathrm { E } } = \mu _ { \mathrm { I } } = 0$ 
D)  $H _ { 0 } : \mu _ { \mathrm { C } } = \mu _ { \mathrm { E } } = \mu _ { \mathrm { I } }$

Question 10

Given that the mean square for treatments (MST)for an ANOVA F-test is 5,000 and the mean square for error (MSE)is 3,750, find the value of the test statistic F.

Accepted Answer

A) 1.25 
B) .800 
C) 1.33 
D) .750 
A) 1.25 
B) .800 
C) 1.33 
D) .750

Question 11

Consider a completely randomized design with five treatments. How many pairwise comparisons of treatments are made in a Bonferroni analysis?

Accepted Answer

A) 5! = 120 
B) 10 
C) 5 
D) 20 
A) 5! = 120 
B) 10 
C) 5 
D) 20

Find the critical value F0 for a one-tailed test using α = 0.05, with 8 numerator degrees of freedom and 15 denominator degrees of freedom

Given that the sum of squares for treatments (SST)for an ANOVA F-test is 9,000 and there are four total treatments, find the mean square for treatments (MST).

Complete the ANOVA table. Source df SS MS F Treatments 3 857.1 Error 8 372.8 Total

Given that the mean square for treatments (MST)for an ANOVA F-test is 5,000 and the mean square for error (MSE)is 3,750, find the value of the test statistic F.

Consider a completely randomized design with five treatments. How many pairwise comparisons of treatments are made in a Bonferroni analysis?

Statistics, Data, and Statistical Thinking

Methods for Describing Sets of Data

Probability

Discrete Random Variables

Continuous Random Variables

Sampling Distributions

Inferences Based on a Single Sample: Estimation With Confidence Intervals

Inferences Based on a Single

Inferences Based on Two Samples: Confidence Intervals and Tests of Hypotheses

Simple Linear Regression

Multiple Regression and Model Building

Categorical Data Analysis

Nonparametric Statistics Available Online

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Exam 10: Analysis of Variance: Comparing More Than Two Means

Find the critical value F0 for a one-tailed test using α = 0.05, with 8 numerator degrees of freedom and 15 denominator degrees of freedom

Given that the sum of squares for treatments (SST)for an ANOVA F-test is 9,000 and there are four total treatments, find the mean square for treatments (MST).

Complete the ANOVA table. Source df SS MS F Treatments 3 857.1 Error 8 372.8 Total

Given that the mean square for treatments (MST)for an ANOVA F-test is 5,000 and the mean square for error (MSE)is 3,750, find the value of the test statistic F.

Consider a completely randomized design with five treatments. How many pairwise comparisons of treatments are made in a Bonferroni analysis?

Statistics, Data, and Statistical Thinking

Methods for Describing Sets of Data

Probability

Discrete Random Variables

Continuous Random Variables

Sampling Distributions

Inferences Based on a Single Sample: Estimation With Confidence Intervals

Inferences Based on a Single

Inferences Based on Two Samples: Confidence Intervals and Tests of Hypotheses

Simple Linear Regression

Multiple Regression and Model Building

Categorical Data Analysis

Nonparametric Statistics Available Online

Filters