Question 1

In an experiment on the effect of garlic on blood lipid concentrations, adult volunteers with slightly elevated cholesterol levels were randomly assigned to one of four treatments taken daily for six months: raw garlic, garlic powder, garlic extract, or a placebo. The participants' LDL levels (low-density lipoprotein, or "bad" cholesterol, in mg/dL) were assessed at the end of the six-month study period. Summary statistics and a partial ANOVA table for this study are shown here.

$\begin{array} { l c c c } 
\text { Treatment } & \text { Mean } & \text { Standard Deviation } & \text { Sample Size } n \
\text { Raw garlic } & 142 & 22 & 49 \
\text { Garlic powder } & 137 & 25 & 47 \
\text { Garlic extract } & 137 & 22 & 48 \
\text { Placebo } & 133 & 21 & 48
\end{array}$

$$\begin{array}{l}
\begin{array} { l l r } 
\text {Source }&\text {df }&\text {Sums of Squares }&\text {Mean Square }&\text {F-Ratio }\
\text { Treatment}&&&659.46\
\text {Error }&&95,457.00
\end{array}
\end{array}$$
 The research question is: Do the data provide evidence that the treatments affect the mean LDL level in this population? What is the value of the F statistic to test this hypothesis?

Accepted Answer

A) 1.30 
B) 3.43 
C) 36.19 
D) 144.75

Question 2

A medical research team is interested in determining whether a new drug has an effect on creatine kinase (CK), which is often assayed in blood tests as an indicator of myocardial infarction. A random selection of 20 patients from a pool of possible subjects is selected, and each subject is given the medication. The subjects' CK levels are observed initially, after 3 weeks, and again after 6 weeks. The purpose is to study the CK levels over time. Here is a summary of the findings:
In this example, what should we notice?
 $$\begin{array} { | c | c | c | } 
\hline \text { Time (weeks) } & \begin{array} { c } 
\text { Mean CK Level } \
\text { (U/L) }
\end{array} & \text { Standard Deviation (U/L) } \
\hline 0 & 121 & 20.37 \
\hline 3 & 106 & 16.09 \
\hline 6 & 100 & 10.21 \
\hline
\end{array}$$

Accepted Answer

A) The data show very strong evidence of a violation of the assumption that the three populations have the same standard deviation. 
B) ANOVA cannot be used on these data because the sample sizes are much too small. 
C) The assumption that the data are independent for the three time points is unreasonable because the same subjects were observed each time. 
D) There is no reason not to use ANOVA in this situation.

Question 3

How much corn should be planted per acre for a farmer to get the highest yield? Too few plants will give a low yield, while too many plants will result in plants competing for moisture and nutrients, resulting in a lower yield. Four levels of planting density are to be studied: 12,000, 16,000, 20,000, and 24,000 plants per acre. The experimenters had 12 acres available for the study; 3 acres were assigned at random to each of the planting densities. The data follow.
$$\begin{array} { | c | ccc| } 
\hline \text { Plants (per acre) } & \text { Yield (bushels per acre) } \
\hline 12,000 & 150.1 & 113.0 & 118.4 \
\hline 16,000 & 166.9 & 120.7 & 135.2 \
\hline 20,000 & 165.3 & 130.1 & 139.6 \
\hline 24,000 & 134.7 & 138.4 & 156.1 \
\hline
\end{array}$$ Assume the data can be considered four independent SRSs, one from each of the four populations of planting densities, and that the distribution of the yields is Normal. A partial ANOVA table produced by Minitab follows, along with the means and standard deviations of the yields for the four groups.
One-Way ANOVA: Yield Versus Density
Source DF SS MS F P
Density 589
Error $\quad 356$
Total
Density      N Mean StDev
$12,000 \quad 3127.1720 .04$
$16,000 \quad 3140.9323 .63$
$20,000 \quad 3145.0018 .21$
$24,000 \quad 3143.0711 .44$ What is the null hypothesis for the ANOVA?

Accepted Answer

A) The population mean yield is the same for all four planting densities. 
B) The population mean yield is increasing as the planting density gets larger. 
C) The population mean yield is decreasing as the planting density gets larger. 
D) The population mean yield first increases and then decreases as the planting density gets larger.

Question 4

How much corn should be planted per acre for a farmer to get the highest yield? Too few plants will give a low yield, while too many plants will result in plants competing for moisture and nutrients, resulting in a lower yield. Four levels of planting density are to be studied: 12,000, 16,000, 20,000, and 24,000 plants per acre. The experimenters had 12 acres available for the study; 3 acres were assigned at random to each of the planting densities. The data follow.
$$\begin{array} { | c | c c l | } 
\hline \text { Plants (per acre) } & \text { Yield (bushels per acre) } \
\hline 12,000 & 150.1 & 113.0 & 118.4 \
\hline 16,000 & 166.9 & 120.7 & 135.2 \
\hline 20,000 & 165.3 & 130.1 & 139.6 \
\hline 24,000 & 134.7 & 138.4 & 156.1 \
\hline
\end{array}$$ Assume the data can be considered four independent SRSs, one from each of the four populations of planting densities, and that the distribution of the yields is Normal. A partial ANOVA table produced by Minitab follows, along with the means and standard deviations of the yields for the four groups.

$$\begin{array}{l}
\text { One-Way ANOVA: Yield Versus Density }\
\
\begin{array} { l r r r r r } 
\text { Source } & \text { DF } & \text { SS } & \text { MS } & \text { F } & \text { P } \
\text { Density } & & 589 & & & \
\text { Error } & & & 356 & & \
\text { Total } & & & & &
\end{array}\
\
\begin{array} { l c r r } 
\text { Density } & \mathrm { N } & \text { Mean } & \text { StDev } \
12,000 & 3 & 127.17 & 20.04 \
16,000 & 3 & 140.93 & 23.63 \
20,000 & 3 & 145.00 & 18.21 \
24,000 & 3 & 143.07 & 11.44
\end{array}
\end{array}$$
 What is the degrees of freedom for the density (i.e., between-group) component?

Accepted Answer

A) 2 
B) 3 
C) 8 
D) 11

Question 5

Does fibrolytic enzyme level in feedlot cattle affect cattle weight gain, on average? Researchers randomly assigned 20 feedlot cattle to four groups, with each group being fed a diet containing a different enzyme level. Cattle weight gain (in kilograms) was recorded over the time of the study. There was no obvious deviation from Normality for these data. The findings are summarized below:
$$\begin{array} { | l | l | l | l | } 
\hline \text { Enzyme level } & n & \text { mean } & \text { std. dev. } \
\hline \text { None } & 5 & 51.8 & 4.9 \
\hline \text { Low } & 5 & 55.0 & 7.8 \
\hline \text { Medium } & 5 & 67.4 & 9.6 \
\hline \text { High } & 5 & 48.8 & 7.3 \
\hline
\end{array}$$ What is the alternative hypothesis for the appropriate ANOVA?

Accepted Answer

A) The population mean weight gain is the same for all four feed conditions. 
B) The population mean weight gain is different for each of the four feed conditions. 
C) The population mean weight gain is not the same for the four feed conditions. 
D) The population mean weight gain is highest with a medium enzyme level.

Question 6

Does fibrolytic enzyme level in feedlot cattle affect cattle weight gain, on average? Researchers randomly assigned 20 feedlot cattle to four groups, with each group being fed a diet containing a different enzyme level. Cattle weight gain (in kilograms) was recorded over the time of the study. There was no obvious deviation from Normality for these data. The findings are summarized below:
When checking the assumptions for this ANOVA test, what should we conclude?
 $$\begin{array} { | l | l | l | l | } 
\hline \text { Enzyme level } & n & \text { mean } & \text { std. dev. } \
\hline \text { None } & 5 & 51.8 & 4.9 \
\hline \text { Low } & 5 & 55.0 & 7.8 \
\hline \text { Medium } & 5 & 67.4 & 9.6 \
\hline \text { High } & 5 & 48.8 & 7.3 \
\hline
\end{array}$$

Accepted Answer

A) The assumptions are all reasonably met. 
B) The assumption of equal population standard deviations is not met. 
C) The assumption of independent random samples is not met. 
D) None of the assumptions is met.

Question 7

Researchers examined a new treatment for advanced ovarian cancer in a mouse model. They created a nanoparticle-based delivery system for a suicide-gene therapy to be delivered directly to the tumor cells. The mice were randomly assigned to have their tumor injected with either the gene -- nanoparticle combination, the gene alone, or some buffer solution (placebo). The following table shows the tumor fold-increases after two weeks in a total of 29 mice. For reference, a fold-increase of 1 represents no change; a 2 represents a doubling in volume of the tumor.
$$\begin{array} { | r | r | r | } 
\hline \text { Buffer }  &  \text { Gene }  &  \text { Gene+Nano}  \
\hline 9.1 & 5.6 & 4.1 \
8.1 & 5.3 & 3.5 \
7.8 & 5.6 & 2.1 \
7.0 & 4.3 & 2.1 \
6.8 & 3.9 & 1.8 \
5.4 & 3.4 & 1.8 \
5.4 & 2.7 & 1.2 \
4.1 & 3.0 & 1.1 \
3.8 & 3.3 & 1.1 \
3.3 & & 1.4 \
\hline
\end{array}$$ We want to test the null hypothesis that there is no difference in population mean tumor fold-increase for the three treatments. What are the correct summary statistics for these data? (Use technology.)

Accepted Answer

A) 
$$\begin{array} { | r | c | c | c | } 
\hline& \text { Buffer } & \text { Gene } & \text { Gene+Nano } \
\hline \text { mean } & 6.08 & 4.12 & 2.02 \
\text { standard deviation } & 1.98 & 1.14 & 1.02 \
\hline
\end{array}$$ 
B) 
$$\begin{array} { | r | c | c | c | } 
\hline& \text { Buffer } & \text { Gene } & \text { Gene+Nano } \
\hline \text { mean } & 5.08 & 4.12 & 2.02 \
\text { standard deviation } & 3.92 & 1.29 & 1.04 \
\hline
\end{array}$$ 
C) 
$$\begin{array} { | r | c | c | c | } 
\hline& \text { Buffer } & \text { Gene } & \text { Gene+Nano } \
\hline \text { mean } & 6.08 & 4.12 & 2.02 \
\text { standard deviation } & 1.24 & 0.43 & 0.33 \
\hline
\end{array}$$ 
D)  $$\begin{array} { | r | c | c | c | } 
\hline& \text { Buffer } & \text { Gene } & \text { Gene+Nano } \
\hline \text { mean } & 6.08 & 4.12 & 2.02 \
\text { standard deviation } & 0.63 & 0.38 & 0.32 \
\hline
\end{array}$$

Question 8

Researchers examined a new treatment for advanced ovarian cancer in a mouse model. They created a nanoparticle-based delivery system for a suicide-gene therapy to be delivered directly to the tumor cells. The mice were randomly assigned to have their tumor injected with either the gene -- nanoparticle combination, the gene alone, or some buffer solution (placebo). The following table shows the tumor fold-increases after two weeks in a total of 29 mice. For reference, a fold-increase of 1 represents no change; a 2 represents a doubling in volume of the tumor.
- $$\begin{array} { | r | r | r | } 
\hline  \text { Buffer } &  \text { Gene } & \text { Gene+Nano } \
\hline 9.1 & 5.6 & 4.1 \
8.1 & 5.3 & 3.5 \
7.8 & 5.6 & 2.1 \
7.0 & 4.3 & 2.1 \
6.8 & 3.9 & 1.8 \
5.4 & 3.4 & 1.8 \
5.4 & 2.7 & 1.2 \
4.1 & 3.0 & 1.1 \
3.8 & 3.3 & 1.1 \
3.3 & & 1.4 \
\hline
\end{array}$$ We want to test the null hypothesis that there is no difference in population mean tumor fold-increase for the three treatments. The ANOVA procedure relies on three important assumptions to be valid:
(1) The data are independent random samples.
(2) The populations are Normally distributed or the sample sizes are large enough.
(3) The populations have the same standard deviation .
$$\sigma$$ In the case of this ANOVA test, which of the following statements about this scenario is correct?

Accepted Answer

A) Assumption (1) is the only assumption not satisfied. 
B) Assumption (2) is the only assumption not satisfied. 
C) Assumption (3) is the only assumption not satisfied. 
D) All the test assumptions are satisfied.

Question 9

Researchers examined a new treatment for advanced ovarian cancer in a mouse model. They created a nanoparticle-based delivery system for a suicide-gene therapy to be delivered directly to the tumor cells. The mice were randomly assigned to have their tumor injected with either the gene -- nanoparticle combination, the gene alone, or some buffer solution (placebo). The following table shows the tumor fold-increases after two weeks in a total of 29 mice. For reference, a fold-increase of 1 represents no change; a 2 represents a doubling in volume of the tumor.
- $$\begin{array} { | r | r | r | } 
\hline \text { Buffer } &\text { Gene } & \text { Gene+Nano } \
\hline 9.1 & 5.6 & 4.1 \
8.1 & 5.3 & 3.5 \
7.8 & 5.6 & 2.1 \
7.0 & 4.3 & 2.1 \
6.8 & 3.9 & 1.8 \
5.4 & 3.4 & 1.8 \
5.4 & 2.7 & 1.2 \
4.1 & 3.0 & 1.1 \
3.8 & 3.3 & 1.1 \
3.3 & & 1.4 \
\hline
\end{array}$$ We want to test the null hypothesis that there is no difference in population mean tumor fold-increase for the three treatments. Based on this ANOVA test, and using a significance level of 0.05, what should you conclude?

Accepted Answer

A) The population mean fold-increases for gene therapy, combination therapy, and placebo are significantly different from one another. 
B) The population mean fold-increases for gene therapy, combination therapy, and placebo are not significantly different. 
C) The population mean fold-increase for the combination (gene + nano) therapy is significantly less than the population mean fold-increase with the placebo (buffer). 
D) There is significant evidence that the population mean fold-increases with the gene therapy, combination therapy, and placebo are not all the same.

Question 10

Researchers wish to examine the effectiveness of a new weight-loss pill. A total of 200 obese adults are randomly assigned to one of four conditions: weight-loss pill alone, weight-loss pill with a low-fat diet, placebo pill alone, or placebo pill with a low-fat diet. The weight loss after six months of treatment is recorded in pounds for each subject. To analyze these data, which inference procedure would you use?

Accepted Answer

A) z test 
B) t test 
C) ANOVA F test 
D) Chi-square test

Question 11

In an experiment on the effect of garlic on blood lipid concentrations, adult volunteers with slightly elevated cholesterol levels were randomly assigned to one of four treatments taken daily for six months: raw garlic, garlic powder, garlic extract, or a placebo. The participants' LDL levels (low-density lipoprotein, or "bad" cholesterol, in mg/dL) were assessed at the end of the six-month study period. Summary statistics and a partial ANOVA table for this study are shown here.

$\begin{array} { l c c c } 
\text { Treatment } & \text { Mean } & \text { Standard Deviation } & \text { Sample Size } n \
\text { Raw garlic } & 142 & 22 & 49 \
\text { Garlic powder } & 137 & 25 & 47 \
\text { Garlic extract } & 137 & 22 & 48 \
\text { Placebo } & 133 & 21 & 48
\end{array}$

$$\begin{array}{l}
\begin{array} { l l r } 
\text {Source }&\text {df }&\text {Sums of Squares }&\text {Mean Square }&\text {F-Ratio }\
\text { Treatment}&&&659.46\
\text {Error }&&95,457.00
\end{array}
\end{array}$$
 The research question is: Do the data provide evidence that the treatments affect the mean LDL level in this population? What is the degrees of freedom in the numerator for this test?

Accepted Answer

A) 3 
B) 4 
C) 6 
D) 46

Question 12

A study randomly assigned adult subjects to one of three exercise treatments: (1) a single long exercise period five days per week; (2) several ten-minute exercise periods five days per week; and (3) several ten-minute periods five days per week using a home treadmill. The study report contains the following summary statistics about weight loss (in kilograms) after six months of treatment:
$$\begin{array} { | l | c | c | c | } 
\hline\text { Treatment } & \text { Mean } & \text { Std. Dev. } & \boldsymbol { n } \
\hline \text { Long periods } & 10.2 & 4.2 & 37 \
\hline \text { Multiple short periods } & 9.3 & 4.5 & 36 \
\hline \text { Multiple short periods with treadmill } & 10.2 & 5.2 & 42 \
\hline
\end{array}$$ Here is a partial ANOVA table based on these data:
​
$$\begin{array}{l}
\begin{array} { l | l r } 
&\text {Source }&\text {df  }&\text {Sums of Squares }&\text {Mean Square}&\text { F-Ratio}&\
\hline\text {group }&&&20.032\
\text { Error} && & &21.8967\
\hline\text {Total }&
\end{array}
\end{array}$$ What is the alternative hypothesis for the ANOVA?

Accepted Answer

A) The population mean weight loss after six months is greatest with multiple short periods of exercise using a treadmill. 
B) The population mean weight loss after six months is different for each of the three exercise conditions. 
C) The population mean weight loss after six months is not the same for the three exercise conditions. 
D) The population mean weight loss after six months is the same for all three exercise conditions.

Question 13

Heliconia is a genus of tropical plant with different varieties often fertilized by distinct species of hummingbirds. Researchers measured the flower length (in millimeters) of independent random samples of three varieties of Heliconia to see if the three varieties differ significantly in flower length. To analyze these data, which inference procedure would you use?

Accepted Answer

A) z test 
B) t test 
C) ANOVA F test 
D) Chi-square test

Question 14

Researchers examined a new treatment for advanced ovarian cancer in a mouse model. They created a nanoparticle-based delivery system for a suicide-gene therapy to be delivered directly to the tumor cells. The mice were randomly assigned to have their tumor injected with either the gene -- nanoparticle combination, the gene alone, or some buffer solution (placebo). The following table shows the tumor fold-increases after two weeks in a total of 29 mice. For reference, a fold-increase of 1 represents no change; a 2 represents a doubling in volume of the tumor.
- $$\begin{array} { | r | r | r | } 
\hline  \text { Buffer } & \text { Gene } & \text { Gene+Nano } \
\hline 9.1 & 5.6 & 4.1 \
8.1 & 5.3 & 3.5 \
7.8 & 5.6 & 2.1 \
7.0 & 4.3 & 2.1 \
6.8 & 3.9 & 1.8 \
5.4 & 3.4 & 1.8 \
5.4 & 2.7 & 1.2 \
4.1 & 3.0 & 1.1 \
3.8 & 3.3 & 1.1 \
3.3 & & 1.4 \
\hline
\end{array}$$ We want to test the null hypothesis that there is no difference in population mean tumor fold-increase for the three treatments. What is the mean square for groups to test this hypothesis? (Use technology.)

Accepted Answer

A) MSG = 2.11 
B) MSG = 41.23 
C) MSG = 54.99 
D) MSG = 82.45

Question 15

How much corn should be planted per acre for a farmer to get the highest yield? Too few plants will give a low yield, while too many plants will result in plants competing for moisture and nutrients, resulting in a lower yield. Four levels of planting density are to be studied: 12,000, 16,000, 20,000, and 24,000 plants per acre. The experimenters had 12 acres available for the study; 3 acres were assigned at random to each of the planting densities. The data follow.
$$\begin{array} { | c | c c c | } 
\hline \text { Plants (per acre) } & \text { Yield (bushels per acre) } \
\hline 12,000 & 150.1 & 113.0 & 118.4 \
\hline 16,000 & 166.9 & 120.7 & 135.2 \
\hline 20,000 & 165.3 & 130.1 & 139.6 \
\hline 24,000 & 134.7 & 138.4 & 156.1 \
\hline
\end{array}$$ Assume the data can be considered four independent SRSs, one from each of the four populations of planting densities, and that the distribution of the yields is Normal. A partial ANOVA table produced by Minitab follows, along with the means and standard deviations of the yields for the four groups.
One-Way ANOVA: Yield Versus Density
Source DF SS MS F P
Density 589
Error $\quad 356$
Total
Density      N Mean StDev
$12,000 \quad 3127.1720 .04$
$16,000 \quad 3140.9323 .63$
$20,000 \quad 3145.0018 .21$
$24,000 \quad 3143.0711 .44$ What is the P-value for the ANOVA that tests for equality of the population means of the four densities?

Accepted Answer

A) Less than 0.01 
B) Between 0.01 and 0.025 
C) Between 0.025 and 0.05 
D) Greater than 0.05

Question 16

Researchers examined a new treatment for advanced ovarian cancer in a mouse model. They created a nanoparticle-based delivery system for a suicide-gene therapy to be delivered directly to the tumor cells. The mice were randomly assigned to have their tumor injected with either the gene -- nanoparticle combination, the gene alone, or some buffer solution (placebo). The following table shows the tumor fold-increases after two weeks in a total of 29 mice. For reference, a fold-increase of 1 represents no change; a 2 represents a doubling in volume of the tumor.
- $$\begin{array} { | r | r | r | } 
\hline \text { Buffer } & \text { Gene } & \text { Gene+Nano } \
\hline 9.1 & 5.6 & 4.1 \
8.1 & 5.3 & 3.5 \
7.8 & 5.6 & 2.1 \
7.0 & 4.3 & 2.1 \
6.8 & 3.9 & 1.8 \
5.4 & 3.4 & 1.8 \
5.4 & 2.7 & 1.2 \
4.1 & 3.0 & 1.1 \
3.8 & 3.3 & 1.1 \
3.3 & & 1.4 \
\hline
\end{array}$$ We want to test the null hypothesis that there is no difference in population mean tumor fold-increase for the three treatments. What is the sum of squares for error to test this hypothesis? (Use technology.)

Accepted Answer

A) SSE = 2.11 
B) SSE = 41.23 
C) SSE = 54.99 
D) SSE = 82.45

Question 17

Researchers examined a new treatment for advanced ovarian cancer in a mouse model. They created a nanoparticle-based delivery system for a suicide-gene therapy to be delivered directly to the tumor cells. The mice were randomly assigned to have their tumor injected with either the gene -- nanoparticle combination, the gene alone, or some buffer solution (placebo). The following table shows the tumor fold-increases after two weeks in a total of 29 mice. For reference, a fold-increase of 1 represents no change; a 2 represents a doubling in volume of the tumor.
- $$\begin{array} { | r | r | r | } 
\hline \text { Buffer }  &  \text { Gene }  & \text { Gene+Nano } \
\hline 9.1 & 5.6 & 4.1 \
8.1 & 5.3 & 3.5 \
7.8 & 5.6 & 2.1 \
7.0 & 4.3 & 2.1 \
6.8 & 3.9 & 1.8 \
5.4 & 3.4 & 1.8 \
5.4 & 2.7 & 1.2 \
4.1 & 3.0 & 1.1 \
3.8 & 3.3 & 1.1 \
3.3 & & 1.4 \
\hline
\end{array}$$ We want to test the null hypothesis that there is no difference in population mean tumor fold-increase for the three treatments. What is the P-value for this ANOVA test? (Use technology.)

Accepted Answer

A) Less than 0.01 
B) Between 0.01 and 0.025 
C) Between 0.025 and 0.05 
D) Greater than 0.05

Question 18

Researchers examined a new treatment for advanced ovarian cancer in a mouse model. They created a nanoparticle-based delivery system for a suicide-gene therapy to be delivered directly to the tumor cells. The mice were randomly assigned to have their tumor injected with either the gene -- nanoparticle combination, the gene alone, or some buffer solution (placebo). The following table shows the tumor fold-increases after two weeks in a total of 29 mice. For reference, a fold-increase of 1 represents no change; a 2 represents a doubling in volume of the tumor.
- $$\begin{array} { | r | r | r | } 
\hline  \text { Buffer }  &  \text { Gene }  & \text { Gene+Nano } \
\hline 9.1 & 5.6 & 4.1 \
8.1 & 5.3 & 3.5 \
7.8 & 5.6 & 2.1 \
7.0 & 4.3 & 2.1 \
6.8 & 3.9 & 1.8 \
5.4 & 3.4 & 1.8 \
5.4 & 2.7 & 1.2 \
4.1 & 3.0 & 1.1 \
3.8 & 3.3 & 1.1 \
3.3 & & 1.4 \
\hline
\end{array}$$ We want to test the null hypothesis that there is no difference in population mean tumor fold-increase for the three treatments. What is the value of the F statistic to test this hypothesis? (Use technology.)

Accepted Answer

A) 0.75 
B) 1.50 
C) 19.49 
D) 40.02

Question 19

In an experiment on the effect of garlic on blood lipid concentrations, adult volunteers with slightly elevated cholesterol levels were randomly assigned to one of four treatments taken daily for six months: raw garlic, garlic powder, garlic extract, or a placebo. The participants' LDL levels (low-density lipoprotein, or "bad" cholesterol, in mg/dL) were assessed at the end of the six-month study period. Summary statistics and a partial ANOVA table for this study are shown here.

$\begin{array}{l}
\begin{array} { l c c c } 
 \text { Treatment } & \text { Mean } & \text { Standard Deviation } & \text { Sample Size } n \
\text { Raw garlic } & 142 & 22 & 49 \
\text { Garlic powder } & 137 & 25 & 47 \
\text { Garlic extract } & 137 & 22 & 48 \
\text { Placebo } & 133 & 21 & 48
\end{array}
\end{array}$

$$\begin{array}{l}
\begin{array} { l l r } 
\text {Source }&\text {df }&\text {Sums of Squares }&\text {Mean Square }&\text {F-Ratio }\
\text { Treatment}&&&659.46\
\text {Error }&&95,457.00
\end{array}
\end{array}$$
 The research question is: Do the data provide evidence that the treatments affect the mean LDL level in this population? What is the null hypothesis for this test?

Accepted Answer

A) The population mean LDL level after six months is the same for all four treatments. 
B) The population mean LDL level after six months is higher among individuals given a placebo. 
C) The population mean LDL level after six months is lower among individuals given a placebo. 
D) The population mean LDL level after six months is highest among individuals given raw garlic and lowest among individuals given a placebo.

Question 20

Does fibrolytic enzyme level in feedlot cattle affect cattle weight gain, on average? Researchers randomly assigned 20 feedlot cattle to four groups, with each group being fed a diet containing a different enzyme level. Cattle weight gain (in kilograms) was recorded over the time of the study. There was no obvious deviation from Normality for these data. The findings are summarized below:
The numerical value of the test statistic is 5.795. What is the P-value for the ANOVA that tests for equality of the population mean weight gains under the four feed conditions?
 $$\begin{array} { | l | l | l | l | } 
\hline \text { Enzyme level } & n & \text { mean } & \text { std. dev. } \
\hline \text { None } & 5 & 51.8 & 4.9 \
\hline \text { Low } & 5 & 55.0 & 7.8 \
\hline \text { Medium } & 5 & 67.4 & 9.6 \
\hline \text { High } & 5 & 48.8 & 7.3 \
\hline
\end{array}$$

Accepted Answer

A) Less than 0.01 
B) Between 0.01 and 0.05 
C) Between 0.05 and 0.10 
D) Greater than 0.10

Exam 22: One-Way Analysis of Variance: Comparing Several Means