By the end of this chapter, the student should be able to:
Interpret the F probability distribution as the number of groups and the sample size change.
Discuss two uses for the F distribution, ANOVA and the test of two variances.
Conduct and interpret ANOVA.
Conduct and interpret hypothesis tests of two variances (optional).
Many statistical applications in psychology, social science, business administration, and the natural sciences involve several groups. For example, an environmentalist is interested in knowing if the average amount of pollution varies in several bodies of water. A sociologist is interested in knowing if the amount of income a person earns varies according to his or her upbringing. A consumer looking for a new car might compare the average gas mileage of several models.
For hypothesis tests involving more than two averages, statisticians have developed a method called Analysis of Variance” (abbreviated ANOVA). In this chapter, you will study the simplest form of ANOVA called single factor or oneway ANOVA. You will also study the F distribution, used for ANOVA, and the test of two variances. This is just a very brief overview of ANOVA. You will study this topic in much greater detail in future statistics courses.
ANOVA, as it is presented here, relies heavily on a calculator or computer.
For further information about ANOVA, use the online link ANOVA. Use the back button to return here. (The url is http://en.wikipedia.org/wiki/Analysis_of_variance.)
The purpose of an ANOVA test is to determine the existence of a statistically significant difference among several group means. The test actually uses variances to help determine if the means are equal or not.
In order to perform an ANOVA test, there are three basic assumptions to be fulfilled:
Each population from which a sample is taken is assumed to be normal.
Each sample is randomly selected and independent.
The populations are assumed to have equal standard deviations (or variances).
The null hypothesis is simply that all the group population means are the same. The alternate hypothesis is that at least one pair of means is different. For example, if there are k groups:
H _{ o } : μ _{ 1 } = μ _{ 2 } = μ _{ 3 } = … = μ _{ k }
H _{ a } : At least two of the group means μ _{1} , μ _{2} , μ _{3} , … , μ _{ k } are not equal.
Also referred to as ANOVA. A method of testing whether or not the means of three or more populations are equal. The method is applicable if:
All populations of interest are normally distributed.
The populations have equal standard deviations.
Samples (not necessarily of the same size) are randomly and independently selected from each population.
The test statistic for analysis of variance is the Fratio.
Mean of the squared deviations from the mean. Square of the standard deviation. For a set of data, a deviation can be represented as where x is a value of the data and is the sample mean. The sample variance is equal to the sum of the squares of the deviations divided by the difference of the sample size and 1.
The distribution used for the hypothesis test is a new one. It is called the F distribution, named after Sir Ronald Fisher, an English statistician. The F statistic is a ratio (a fraction). There are two sets of degrees of freedom; one for the numerator and one for the denominator.
For example, if F follows an F distribution and the degrees of freedom for the numerator are 4 and the degrees of freedom for the denominator are 10, then F ~ F _{ 4 , 10 } .
To calculate the F ratio, two estimates of the variance are made.
Variance between samples: An estimate of σ ^{2} that is the variance of the sample means. If the samples are different sizes, the variance between samples is weighted to account for the different sample sizes. The variance is also called variation due to treatment or explained variation.
Variance within samples: An estimate of σ ^{2} that is the average of the sample variances (also known as a pooled variance). When the sample sizes are different, the variance within samples is weighted. The variance is also called the variation due to error or unexplained variation.
SS_{between} = the sum of squares that represents the variation among the different samples.
SS_{within} = the sum of squares that represents the variation within samples that is due to chance.
To find a “sum of squares” means to add together squared quantities which, in some cases, may be weighted. We used sum of squares to calculate the sample variance and the sample standard deviation in Descriptive Statistics.
MS means “mean square.” MS_{between} is the variance between groups and MS_{within} is the variance within groups.
Calculation of Sum of Squares and Mean Square
k = the number of different groups
n _{ j } = the size of the group
s _{ j } = the sum of the values in the group
N = total number of all the values combined. (total sample size: ∑n _{ j } )
x = one value: ∑x = ∑s _{ j }
Sum of squares of all values from every group combined: ∑x ^{2}
Between group variability:
Total sum of squares:
Explained variation sum of squares representing variation among the different samples
Unexplained variation sum of squares representing variation within samples due to chance:
df’s for different groups (df’s for the numerator):
Equation for errors within samples (df’s for the denominator):
Mean square (variance estimate) explained by the different groups:
Mean square (variance estimate) that is due to chance (unexplained):
MS_{between} and MS_{within} can be written as follows:
The ANOVA test depends on the fact that MS_{between} can be influenced by population differences among means of the several groups. Since MS_{within} compares values of each group to its own group mean, the fact that group means might be different does not affect MS_{within} .
The null hypothesis says that all groups are samples from populations having the same normal distribution. The alternate hypothesis says that at least two of the sample groups come from populations with different normal distributions. If the null hypothesis is true, MS_{between} and MS_{within} should both estimate the same value.
The null hypothesis says that all the group population means are equal. The hypothesis of equal means implies that the populations have the same normal distribution because it is assumed that the populations are normal and that they have equal variances.
If MS_{between} and MS_{within} estimate the same value (following the belief that H _{ o } is true), then the Fratio should be approximately equal to 1. Only sampling errors would contribute to variations away from 1. As it turns out, MS_{between} consists of the population variance plus a variance produced from the differences between the samples. MS_{within} is an estimate of the population variance. Since variances are always positive, if the null hypothesis is false, MS_{between} will be larger than MS_{within} . The Fratio will be larger than 1.
The above calculations were done with groups of different sizes. If the groups are the same size, the calculations simplify somewhat and the F ratio can be written as:
where …
the variance of the sample means
n = the sample size of each group
the mean of the sample variances (pooled variance)
The ANOVA hypothesis test is always righttailed because larger Fvalues are way out in the right tail of the Fdistribution curve and tend to make us reject H _{ o } .
The notation for the F distribution is F ~ F _{ df(num) , df(denom) }
where df(num) = df_{between} and df(denom) = df_{ within }
The mean for the F distribution is
The curve is not symmetrical but skewed to the right.
There is a different curve for each set of dfs .
The F statistic is greater than or equal to zero.
As the degrees of freedom for the numerator and for the denominator get larger, the curve approximates the normal.
Other uses for the F distribution include comparing two variances and TwoWay Analysis of Variance. Comparing two variances is discussed at the end of the chapter. TwoWay Analysis is mentioned for your information only.
Figure 13.1.
Example 13.1.
OneWay ANOVA: Four sororities took a random sample of sisters regarding their grade averages for the past term. The results are shown below:
GRADE AVERAGES FOR FOUR SORORITIES  

Sorority 1  Sorority 2  Sorority 3  Sorority 4 
2.17  2.63  2.63  3.79 
1.85  1.77  3.78  3.45 
2.83  3.25  4.00  3.08 
1.69  1.86  2.55  2.26 
3.33  2.21  2.45  3.18 
Problem
Using a significance level of 1%, is there a difference in grade averages among the sororities?
Solution
Let μ _{1} , μ _{2} , μ _{3} , μ _{4} be the population means of the sororities. Remember that the null hypothesis claims that the sorority groups are from the same normal distribution. The alternate hypothesis says that at least two of the sorority groups come from populations with different normal distributions. Notice that the four sample sizes are each size 5.
H _{ a } : Not all of the means μ _{1} , μ _{2} , μ _{3} , μ _{4} are equal.
Distribution for the test: F _{ 3 , 16 }
where k = 4 groups and N = 20 samples in total
df(num) = k – 1 = 4 – 1 = 3
df(denom) = N – k = 20 – 4 = 16
Calculate the test statistic: F = 2.23
Graph:
Figure 13.2.
Probability statement: pvalue = P ( F > 2.23 ) = 0.1241
Compare α and the pvalue : α < pvalue
Make a decision: Since α < pvalue , you cannot reject H _{ o } .
This means that the population averages appear to be the same.
Conclusion: There is not sufficient evidence to conclude that there is a difference among the grade averages for the sororities.
TI83+ or TI 84: Put the data into lists L1, L2, L3, and L4. Press STAT
and
arrow over to TESTS
. Arrow down to F:ANOVA
. Press ENTER
and Enter
(L1,L2,L3,L4
). The F statistic is 2.2303 and the pvalue
is 0.1241.
df(numerator) = 3 (under "Factor"
) and df(denominator) = 16 (under Error
).
Example 13.2.
A fourth grade class is studying the environment. One of the assignments is to grow bean plants in different soils. Tommy chose to grow his bean plants in soil found outside his classroom mixed with dryer lint. Tara chose to grow her bean plants in potting soil bought at the local nursery. Nick chose to grow his bean plants in soil from his mother’s garden. No chemicals were used on the plants, only water. They were grown inside the classroom next to a large window. Each child grew 5 plants. At the end of the growing period, each plant was measured, producing the following data (in inches):
Tommy’s Plants  Tara’s Plants  Nick’s Plants 

24  25  23 
21  31  27 
23  23  22 
30  20  30 
23  28  20 
Problem 1.
Does it appear that the three media in which the bean plants were grown produce the same average height? Test at a 3% level of significance.
Solution
This time, we will perform the calculations that lead to the F’ statistic. Notice that each group has the same number of plants so we will use the formula .
First, calculate the sample mean and sample variance of each group.
Tommy’s Plants  Tara’s Plants  Nick’s Plants  

Sample Mean  24.2  25.4  24.4 
Sample Variance  11.7  18.3  16.3 
Next, calculate the variance of the three group means (Calculate the variance of 24.2, 25.4, and 24.4). Variance of the group means = 0.413 =
Then where n = 5 is the sample size (number of plants each child grew).
Calculate the average of the three sample variances (Calculate the average of 11.7, 18.3, and 16.3). Average of the sample variances = 15.433
Then .
The F statistic (or F ratio) is
The dfs for the numerator = the number of groups – 1 = 3 – 1 = 2
The dfs for the denominator = the total number of samples – the number of groups = 15 – 3 = 12
The distribution for the test is F _{ 2 , 12 } and the F statistic is F = 0.134
The pvalue is P ( F > 0.134 ) = 0.8759 .
Decision: Since α = 0.03 and the pvalue = 0.8759 , do not reject H _{ o } . (Why?)
Conclusion: With a 3% the level of significance, from the sample data, the evidence is not sufficient to conclude that the average heights of the bean plants are not different. Of the three media tested, it appears that it does not matter which one the bean plants are grown in.
(This experiment was actually done by three classmates of the son of one of the authors.)
Another fourth grader also grew bean plants but this time in a jellylike mass. The heights were (in inches) 24, 28, 25, 30, and 32.
Problem 2. (Go to Solution)
Do an ANOVA test on the 4 groups. You may use your calculator or computer to perform the test. Are the heights of the bean plants different? Use a solution sheet.
Randomly divide the class into four groups of the same size. Have each member of each group record the number of states in the United States he or she has visited. Run an ANOVA test to determine if the average number of states visited in the four groups are the same. Test at a 1% level of significance. Use one of the solution sheets at the end of the chapter (after the homework).
Solution to Exercise 2. (Return to Problem)
F = 0.9496
pvalue = 0.4401
The heights of the bean plants are the same.
Another of the uses of the F distribution is testing two variances. It is often desirable to compare two variances rather than two averages. For instance, college administrators would like two college professors grading exams to have the same variation in their grading. In order for a lid to fit a container, the variation in the lid and the container should be the same. A supermarket might be interested in the variability of checkout times for two checkers.
In order to perform a F test of two variances, it is important that the following are true:
The populations from which the two samples are drawn are normally distributed.
The two populations are independent of each other.
Suppose we sample randomly from two independent normal populations. Let σ _{1} ^{2} and σ _{2} ^{2} be the population variances and s _{1} ^{2} and s _{2} ^{2} be the sample variances. Let the sample sizes be n _{1} and n _{2} . Since we are interested in comparing the two sample variances, we use the F ratio
F has the distribution F ~
where n _{1} – 1 are the degrees of freedom for the numerator and n _{2} – 1 are the degrees of freedom for the denominator.
If the null hypothesis is σ _{1} ^{2} = σ _{2} ^{2} , then the FRatio becomes .
If the two populations have equal variances, then s _{1} ^{2} and s _{2} ^{2} are close in value and F = is close to 1. But if the two population variances are very different, s _{1} ^{2} and s _{2} ^{2} tend to be very different, too. Choosing s _{1} ^{2} as the larger sample variance causes the ratio to be greater than 1. If s _{1} ^{2} and s _{2} ^{2} are far apart, then F = is a large number.
Therefore, if F is close to 1, the evidence favors the null hypothesis (the two population variances are equal). But if F is much larger than 1, then the evidence is against the null hypothesis.
A test of two variances may be left, right, or twotailed.
Example 13.3.
Two college instructors are interested in whether or not there is any variation in the way they grade math exams. They each grade the same set of 30 exams. The first instructor’s grades have a variance of 52.3. The second instructor’s grades have a variance of 89.9.
Problem
Test the claim that the first instructor’s variance is smaller. (In most colleges, it is desirable for the variances of exam grades to be nearly the same among instructors.) The level of significance is 10%.
Solution
Let 1 and 2 be the subscripts that indicate the first and second instructor, respectively.
n _{1} = n _{2} = 30.
H _{ o } : σ _{1} ^{2} = σ _{2} ^{2} and H _{ a } : σ _{1} ^{2} < σ _{2} ^{2}
Calculate the test statistic: By the null hypothesis , the F statistic is
Distribution for the test: where n _{1} – 1 = 29 and n _{2} – 1 = 29.
Graph: This test is left tailed.
Draw the graph labeling and shading appropriately.
Figure 13.3.
Probability statement: pvalue = P ( F < 0.582) = 0.0755
Compare α and the pvalue: α = 0.10 .
Make a decision: Since α > pvalue, reject H _{ o } .
Conclusion: With a 10% level of significance, from the data, there is sufficient evidence to conclude that the variance in grades for the first instructor is smaller.
TI83+ and TI84: Press STAT
and arrow over to TESTS
. Arrow down to
D:2SampFTest
. Press ENTER
. Arrow to Stats
and press ENTER
. For
Sx1
, n1
, Sx2
, and n2
, enter
, 30
,
, and 30
. Press ENTER
after
each. Arrow to σ1:
and
< σ2
. Press ENTER
. Arrow down to Calculate
and
press ENTER
.
F = 0.5818 and pvalue = 0.0753. Do the procedure again
and try Draw
instead of Calculate
.
An ANOVA hypothesis test determines if several population means are equal. The distribution for the test is the F distribution with 2 different degrees of freedom.
Assumptions:
Each population from which a sample is taken is assumed to be normal.
Each sample is randomly selected and independent.
The populations are assumed to have equal standard deviations (or variances)
A Test of Two Variances hypothesis test determines if two variances are the same. The distribution for the hypothesis test is the F distribution with 2 different degrees of freedom.
Assumptions:
The populations from which the two samples are drawn are normally distributed.
The two populations are independent of each other.
The student will explore the properties of ANOVA.
Suppose a group is interested in determining whether teenagers obtain their drivers licenses at approximately the same average age across the country. Suppose that the following data are randomly collected from five teenagers in each region of the country. The numbers represent the age at which teenagers obtained their drivers licenses.
Northeast  South  West  Central  East  

16.3  16.9  16.4  16.2  17.1  
16.1  16.5  16.5  16.6  17.2  
16.4  16.4  16.6  16.5  16.6  
16.5  16.2  16.1  16.4  16.8  
________  ________  ________  ________  ________  
s ^{ 2 } =  ________  ________  ________  ________  ________ 
Exercise 13.7.1.
State the hypotheses.
H _{ o } :
H _{ a } :
Enter the data into your calculator or computer.
State the decisions and conclusions (in complete sentences) for the following preconceived levels of α .
Exercise 13.7.6.
α = 0.05
Decision:
Conclusion:
Exercise 13.7.7.
α = 0 . 01
Decision:
Conclusion:
Use a solution sheet to conduct the following hypothesis tests. The solution sheet can be found in the Table of Contents 14. Appendix.
Exercise 13.8.1. (Go to Solution)
Three students, Linda, Tuan, and Javier, are given 5 laboratory rats each for a nutritional experiment. Each rat’s weight is recorded in grams. Linda feeds her rats Formula A, Tuan feeds his rats Formula B, and Javier feeds his rats Formula C. At the end of a specified time period, each rat is weighed again and the net gain in grams is recorded. Using a significance level of 10%, test the hypothesis that the three formulas produce the same average weight gain.
Linda’s rats  Tuan’s rats  Javier’s rats 

43.5  47.0  51.2 
39.4  40.5  40.9 
41.3  38.9  37.9 
46.0  46.3  45.0 
38.2  44.2  48.6 
Exercise 13.8.2.
A grassroots group opposed to a proposed increase in the gas tax claimed that the increase would hurt workingclass people the most, since they commute the farthest to work. Suppose that the group randomly surveyed 24 individuals and asked them their daily oneway commuting mileage. The results are below:
workingclass  professional (middle incomes)  professional (wealthy) 

17.8  16.5  8.5 
26.7  17.4  6.3 
49.4  22.0  4.6 
9.4  7.4  12.6 
65.4  9.4  11.0 
47.1  2.1  28.6 
19.5  6.4  15.4 
51.2  13.9  9.3 
Exercise 13.8.3. (Go to Solution)
Refer to Exercise 13.8.1. Determine whether or not the variance in weight gain is statistically the same among Javier’s and Linda’s rats.
Exercise 13.8.4.
Refer to Exercise 13.8.2 above. Determine whether or not the variance in mileage driven is statistically the same among the working class and professional (middle income) groups.
For the next two problems, refer to the data from Terri Vogel’s Log Book [link pending].
Exercise 13.8.5. (Go to Solution)
Examine the 7 practice laps. Determine whether the average lap time is statistically the same for the 7 practice laps, or if there is at least one lap that has a different average time from the others.
Exercise 13.8.6.
Examine practice laps 3 and 4. Determine whether or not the variance in lap time is statistically the same for those practice laps.
For the next four problems, refer to the following data.
The following table lists the number of pages in four different types of magazines.
home decorating  news  health  computer 

172  87  82  104 
286  94  153  136 
163  123  87  98 
205  106  103  207 
197  101  96  146 
Exercise 13.8.7. (Go to Solution)
Using a significance level of 5%, test the hypothesis that the four magazine types have the same average length.
Exercise 13.8.8.
Eliminate one magazine type that you now feel has an average length different than the others. Redo the hypothesis test, testing that the remaining three averages are statistically the same. Use a new solution sheet. Based on this test, are the average lengths for the remaining three magazines statistically the same?
Exercise 13.8.9.
Which two magazine types do you think have the same variance in length?
Exercise 13.8.10.
Which two magazine types do you think have different variances in length?
Solution to Exercise 13.8.1. (Return to Exercise)
a. H _{ o } : μ _{ L } = μ _{ T } = μ _{ J } 
c. ; 
e. 0.67 
f. 0.5305 
h. Decision: Do not reject null; Conclusion: Means are same 
Solution to Exercise 13.8.3. (Return to Exercise)
c. ; 
e. 3.00 
f. 2 ( 0 . 1563 ) = 0 . 3126 
h. Decision: Do not reject null; Conclusion: Variances are same 
Solution to Exercise 13.8.5. (Return to Exercise)
c. ; 
e. 1.69 
f. 0.1319 
h. Decision: Do not reject null; Conclusion: Average lap times are the same 
Solution to Exercise 13.8.7. (Return to Exercise)
a. H _{ o } : μ _{ d } = μ _{ n } = μ _{ h } = μ _{ c } 
b. At least one average is different 
c. ; 
e. 8.69 
f. 0.0012 
h. Decision: Reject null; Conclusion: At least one average is different 
The next two questions refer to the following situation:
Suppose that the probability of a drought in any independent year is 20%. Out of those years in which a drought occurs, the probability of water rationing is 10%. However, in any year, the probability of water rationing is 5%.
Exercise 13.9.1. (Go to Solution)
What is the probability of both a drought and water rationing occurring?
Exercise 13.9.2. (Go to Solution)
Out of the years with water rationing, find the probability that there is a drought.
The next three questions refer to the following survey:
apple  pumpkin  pecan  

female  40  10  30 
male  20  30  10 
Exercise 13.9.3. (Go to Solution)
Suppose that one individual is randomly chosen. Find the probability that the person’s favorite pie is apple or the person is male.
Exercise 13.9.4. (Go to Solution)
Suppose that one male is randomly chosen. Find the probability his favorite pie is pecan.
Exercise 13.9.5. (Go to Solution)
Conduct a hypothesis test to determine if favorite pie type and gender are independent.
The next two questions refer to the following situation:
Let’s say that the probability that an adult watches the news at least once per week is 0.60.
Exercise 13.9.6. (Go to Solution)
We randomly survey 14 people. On average, how many people do we expect to watch the news at least once per week?
Exercise 13.9.7. (Go to Solution)
We randomly survey 14 people. Of interest is the number that watch the news at least once per week. State the distribution of X . X ~
Exercise 13.9.8. (Go to Solution)
The following histogram is most likely to be a result of sampling from which distribution?
Figure 13.4.
A. ChiSquare 
B. Geometric 
C. Uniform 
D. Binomial 
Exercise 13.9.9.
The ages of De Anza evening students is known to be normally distributed. A sample of 6 De Anza evening students reported their ages (in years) as: 28; 35; 47; 45; 30; 50. Find the probability that the average of 6 ages of randomly chosen students is less than 35 years.
The next three questions refer to the following situation:
The amount of money a customer spends in one trip to the supermarket is known to have an exponential distribution. Suppose the average amount of money a customer spends in one trip to the supermarket is $72.
Exercise 13.9.10. (Go to Solution)
Find the probability that one customer spends less than $72 in one trip to the supermarket?
Exercise 13.9.11. (Go to Solution)
Suppose 5 customers pool their money. (They are poor college students.) How much money altogether would you expect the 5 customers to spend in one trip to the supermarket (in dollars)?
Exercise 13.9.12. (Go to Solution)
State the distribution to use is if you want to find the probability that the average amount spent by 5 customers in one trip to the supermarket is less than $60.
Exercise 13.9.13. (Go to Solution)
A math exam was given to all the fifth grade children attending Country School. Two random samples of scores were taken. The null hypothesis is that the average math scores for boys and girls in fifth grade are the same. Conduct a hypothesis test.
n  s ^{ 2 }  

Boys  55  82  29 
Girls  60  86  46 
Exercise 13.9.14. (Go to Solution)
In a survey of 80 males, 55 had played an organized sport growing up. Of the 70 females surveyed, 25 had played an organized sport growing up. We are interested in whether the proportion for males is higher than the proportion for females. Conduct a hypothesis test.
Exercise 13.9.15. (Go to Solution)
Which of the following is preferable when designing a hypothesis test?
A. Maximize α and minimize β 
B. Minimize α and maximize β 
C. Maximize α and β 
D. Minimize α and β 
The next three questions refer to the following situation:
120 people were surveyed as to their favorite beverage (nonalcoholic). The results are below.
0 – 9  10 – 19  20 – 29  30 +  Totals  

Milk  14  10  6  0  30  
Soda  3  8  26  15  52  
Juice  7  12  12  7  38  
Totals  24  30  44  22  120 
Exercise 13.9.16. (Go to Solution)
Are the events of milk and 30+:
a. Independent events? Justify your answer. 
b. Mutually exclusive events? Justify your answer. 
Exercise 13.9.17. (Go to Solution)
Suppose that one person is randomly chosen. Find the probability that person is 10 – 19 given that he/she prefers juice.
Exercise 13.9.18. (Go to Solution)
Are Preferred Beverage and Age independent events? Conduct a hypothesis test.
Exercise 13.9.19. (Go to Solution)
Given the following histogram, which distribution is the data most likely to come from?
Figure 13.5.
A. uniform 
B. exponential 
C. normal 
D. chisquare 
Solution to Exercise 13.9.5. (Return to Exercise)
pvalue = 0; Reject null; Conclude dependent events
Solution to Exercise 13.9.13. (Return to Exercise)
pvalue = 0.0006; Reject null; Conclude averages are not equal
Solution to Exercise 13.9.14. (Return to Exercise)
pvalue = 0; Reject null; Conclude proportion of males is higher
Class Time:
Names:
The student will conduct a simple ANOVA test involving three variables.
Record the price per pound of 8 fruits, 8 vegetables, and 8 breads in your local supermarket.
Fruits  Vegetables  Breads 

Explain how you could try to collect the data randomly.
Compute the following:
a. Fruit:
 
a. Vegetables:
 
a. Bread:

Find the following:
a. df(num) = 
b. df(denom) = 
State the approximate distribution for the test.
Test statistic: F =
Sketch a graph of this situation. CLEARLY, label and scale the horizontal axis and shade the region(s) corresponding to the pvalue.
pvalue =
Test at α = 0.05. State your decision and conclusion.
a. Decision: Why did you make this decision? 
b. Conclusion (write a complete sentence). 
c. Based on the results of your study, is there a need to further investigate any of the food groups’ prices? Why or why not? 