13.4 – Tests for Equal Variances

Introduction
More about the F-test
R code
Levene’s test of equal variances
Cahoy’s bootstrap method
Questions
Quiz
R code reminders
Chapter 13 contents

Introduction

In order to carry out statistical tests correctly, we must test our data first to see if our sample conforms to the assumptions of the statistical test. If our data do not meet these assumptions, then inferences drawn may be incorrect. How far off our inferences may be depends on a number of factors, but mostly it depends on how far from the expectations our data are.

One assumption we make with parametric tests involving ratio scale data is that the data could be from a normally distributed population. The other key assumption introduced, but not described in detail for the two-sample t-test, was that the variability in the two groups must be the same, ie, homoscedasticity. Thus, in order to carry out the independent sample t-test, we must assume that the variances are equal.

There are two general reasons we may want to concern ourselves with a test for the differences between two variances

The t-test (and other tests like one-way ANOVA) requires that the two samples compared have the same variances. If the Variances are Not Equal we need to perform a modified t-test (see Welch’s formula).
We may also be interested in the differences between the variances in two populations.

Example 1: In genetics we might be interested in the difference between the variability of response of inbred lines (little genetic variation but environmental variation) versus an outbred populations (lots of genetic and environmental variation).

Example 2: Environmental stress can cause organisms to have developmental instability. This might cause organisms to be more variable in morphology or the two sides (right & left) of an organism may develop non-symmetrically. Therefore, polluted environments might cause organisms to have greater variability compared to non-polluted environments.

The first way to test the variances is to use the F test. This works for two groups.

$\begin{align*} F = \frac{s_{1}^{2}}{s_{2}^{2}} \end{align*}$

For more than two groups, we’ll use different tests (eg, Bartlett’s test, Levene’s test).

Remember that the formula for the sample variance is

The Null Hypothesis is that the two samples have the same variances:

$\begin{align*} H_{0}: s_{1}^{2}=s_{1}^{2} \end{align*}$

The Alternative Hypothesis is that the two samples do not have the same variances:

$\begin{align*} H_{0}: s_{1}^{2}\ne s_{1}^{2} \end{align*}$

Note: I prefer to evaluate this as a one-tailed test: identify the larger of the two variances and take that as the numerator Then, the null hypothesis is that

$\begin{align*} H_{0}: s_{1}^{2}\leq s_{1}^{2} \end{align*}$

and therefore, the alternative hypothesis is that (ie, a one-tailed test).

$\begin{align*} H_{0}: s_{1}^{2} > s_{1}^{2} \end{align*}$

Another way to state equal variance test is that we are testing for homogeneity of variances. You may run across the term homoscedasticity; it is the same thing, just a different three dollar word for “equal variances.”

Stated yet another way, if we reject the null hypothesis, then the variances are unequal or show heterogeneity. An additional and equivalent $3 word for inequality of variances is called heteroscedasticity.

More about the F-test

For the F-test, the null hypothesis is that the variances are equal. This means that the “expected” F value will be one: F = 1.0. (The F-distribution differs from t-distribution because it requires 2 values for DF, and ranges from 1 to infinity for every possible combination of v₁ and v₂).

To evaluate the null hypothesis we need the degrees of freedom. For the F test we need two different degrees of freedom, one set for each group): from Table 2, Appendix – F distribution, look up 5% Type I error line in this table because we make it one tailed.

I need the F-test statistic at $F_{0.05_{1},v_{1},v_{2}}$

Examples of difference between two variances, Table 1.

Sample 1: Aggressiveness of Inbred Mice (number of bites in 30 minutes)

Sample 2: Aggressiveness of Outbred Mice (number of bites in 30 minutes)

Table 1. Aggression by inbred and outbred mice.

Sample 1 Aggressiveness of Inbred Mice (number of bites in 30 minutes)	Sample 2 Aggressiveness of Outbred Mice (number of bites in 30 minutes)
3	4
5	10
4	4
3	7
4	7
5	10
4 = mean	7 = mean

Identify the null and alternative hypotheses
Calculate Variances
Calculate F-test
The “test statistic” for this hypothesis test was F = 7.2 / 0.8 = 9.0
Determine Critical Value of the F table (Table 2, Appendix – F distribution)

Example of how to find the critical values of the F distribution for $F_{0.05_{2}}$ and numerator $DF_{v_{1}}=5$ and denominator $DF_{v_{2}}=5$ .

Table 2. Portion of F distribution, see Appendix – F distribution.

α = 0.05		v1
α = 0.05		1	2	3	4	5	6	7	8
v2	1					230
	2					19.3
	3					9.01
	4					6.26
	5	6.61	5.79	5.41	5.19	5.05	4.95	4.88	4.82
	6

Or, instead of using tables, use R

Rcmdr: Distributions → F distribution → F probabilities

and enter the numbers as shown below (Fig 1).

Screenshot Rcmdr menu F distribution

Figure 1. Screenshot R Commander F distribution probabilities.

This will return the p-value, and you would interpret this against your Type I error rate of 5% as you do for other tests.

From Table 2, Appendix – F distribution we find $F_{0.05\left ( 1 \right ),5,5}=5.05$

And the p-value = 0.015

pf(c(9), df1=5, df2=5, lower.tail=FALSE)
[1] 0.01537472

Question: Reject or Accept null hypothesis?

Question: What is the biological interpretation or conclusion from this result?

R code

Rather than play around with the tables of critical values, which are awkward and old-school (and I am showing you these stats tables so that you get a feel for the process, not so you’d actually use them in real practice), use Rcmdr to generate the F test and therefore return the F distribution probability value. As you may expect, R provides a number of options for testing the equal variances assumption, including the F test. The F test is limited to only two groups and, because it is a parametric test, it also makes the assumption of normality, so the F test should not be viewed as necessarily the best test for the equal variances assumption among groups. We present it here because it is a logical test to understand and because of its relevance to the Mean Square ratios in the ANOVA procedures.

So, without further justification, here is the presentation on how to get the F test in Rcmdr. At the end of this section I present a better procedure than the F test for evaluating the equal variance assumption called the Levene test.

Return to the bite data in the table above and enter the data into an R data frame. Note that the data in the table above are unstacked; R expects the data to be stacked, so either create a stacked worksheet and transcribe the data appropriately into the cells of the worksheet, or, go ahead and enter the values into two separate columns then use the Stack variables in active data set… command from the Data menu in Rcmdr.

Then, proceed to perform the F test.

Rcmdr: Statistics → Variances → Two variances F-test…

The first context menu popup is where you enter the variables (Fig 2).

Screenshot Rcmdr variances F-test menu

Figure 2. Screenshot data options R Commander F test

Because there are only two variables in the data set and because Strain contains the text labels of inbred or outbred whereas the other variable is numeric data type, R will correctly select the variables for you by default. Select the “Options” tab to set the parameters of the F test (Fig 3).

Screenshot Rcmdr F-test menu

Figure 3. Screenshot menu options R Commander F test.

When you are finished setting the alternative hypothesis and confidence levels, proceed with the F test by clicking the OK button.

    F test to compare two variances

data:  mice.aggression
F = 0.1111, num df = 5, denom df = 5, p-value = 0.03075
alternative hypothesis: true ratio of variances is not equal to 1
95 percent confidence interval:
 0.01554788 0.79404243
sample estimates:
ratio of variances 
         0.1111111

End of R output.

Levene’s test of equal variances

We will discuss this test in more detail following our presentation on ANOVA. For now, we note that the test works on two or more groups and is a conservative test of the equal variance assumption. Nonparametric tests in general make fewer assumptions about the data and in particular make no assumption of normality like the F test. It is in this context that the Levene’s test would be preferable over the F test. Below we present only how to calculate the statistic in R and Rcmdr and provide the output for the same mouse data set.

Assuming the data are stacked, obtain the Levene’s test in Rcmdr by clicking on (Fig 4).

Rcmdr: Statistics → Variances → Levene’s test…

Screenshot menu options R Commander Levene's test

Figure 4. Screenshot menu options R Commander Levene’s test.

Select the median and the factor variable (in our case “Strain”) and the numeric outcome variable (“Bites”), then click OK button.

Tapply(bites ~ strain, var, na.action=na.omit, data=mice.aggression) # variances by group
inbred outbred
   0.8     7.2

leveneTest(bites ~ strain, data=mice.aggression, center="median")
Levene's Test for Homogeneity of Variance (center = median)
      Df F value  Pr(>F)
group  1       4 0.07339 .
      10
---
Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1

End of R output

Compare the p-values. In both cases — F test and Levene’s test — we’re testing the same hypothesis (equal variances), but the p-values do not agree between the parametric F test and the nonparametric Levene’s test! If we were to go by the results of the F test, p-value was 0.031, less than out Type I error of 5% and we would tentatively conclude that the assumption of equal variances may not apply. On the other hand, if we go with the Levene’s test the p-value was 0.074, which is greater than our Type I error rate of 5% and we would therefore conclude the opposite, that the assumption of equal variances might apply! Both conclusions can’t hold, so which test result of equal variances do we prefer, the parametric F test or the nonparametric Levene’s test? Answer — we’d go with Levene’s because the F test is parametric and, therefore, also assumes normality.

Cahoy’s bootstrap method

Draft. Cahoy (2010). Variance-based statistic bootstrap test of heterogeneity of variances. we discuss bootstrap methods in Chapter 19.2; in brief, bootstrapping involves resampling the dataset and computing a statistic on each sample. This method may be more powerful, that is, more likely to correctly reject the null hypothesis when warranted, compared to Levene’s test.

for now, install package testequavar.

Function for testing two samples.

equa2vartest(inbred, outbred, 0.05, 999)

R output

[[1]]
[1] "Decision: Reject the Null"
$Alpha
[1] 0.05
$NumberOfBootSamples
[1] 999
$BootPvalue
[1] 0.006

The output “Decision: Reject the Null” reflects output from a box-type acceptance region.

Compare results from Levene’s test: p-value 0.07339 suggests accept hypothesis of equal variances, whereas bootstrap method indicates variances heterogenous, ie, reject equal variance hypothesis. However, re-running the test without setting the seed for R’s pseudorandom number generator will result in different p-values. For example, I re-ran Cahoy’s test five times with the following results:

Questions

Test assumption of equal variances by Bartlett’s method and by Levene’s test on OliveMoment variable from the Comet tea data set introduced in Chapter 12.1. Do the methods agree? If not, which test result would you choose?
- BONUS. Retest homogeneous variance hypothesis by equa3vartest(Copper.Hazel, Copper, Hazel, 0.05, 999). Reject or fail to reject the null hypothesis by bootstrap method?
Test assumption of equal variances by Bartlett’s method and by Levene’s test on Height from the O’hia data set introduced in Chapter 12.7. Do the methods agree? If not, which result would you choose?
- BONUS. Retest homogeneous variance hypothesis by equa3vartest(M.1, M.2, M.3, 0.05, 999). Reject or fail to reject the null hypothesis by bootstrap method?

Quiz Chapter 13.4

Tests for Equal Variances

R code reminders

The bootstrap method expects the variables in unstacked format. A simple method to extract the variables from the stacked data is to use a command like the following. For example, extract OliveMoment values for Copper-Hazel treatment.

Copper.Hazel <- cometTea$OliveMoment[1:10]

For Copper, Hazel, replace above with [11:20], and [21-30] respectively. Note: changed variable name from Copper-Hazel to Copper.Hazel. The hyphen is a reserved character in R.

Chapter 13 contents

12.2 – One way ANOVA

Introduction
Three important terms you’ll need for one-way ANOVA
Note on experimental design
ANOVA sources of variation
Mean squares
Worked example with R
Questions
Quiz
Data set
Chapter 12 contents

Introduction

In ANalysis Of VAriance, or ANOVA for short, we likewise have tools to test the null hypothesis of no difference between between categorical independent variables — often called factors when there’s just a few levels to keep track of — and a single, dependent response variable. But now, the response variable is quantitative, not qualitative like the $\chi^2$ tests.

Analysis of variance, ANOVA, is such an important statistical test in biology that we will take the time to “build it” from scratch. We begin by reminding you of where you’ve been with the material.

We already saw an example of this null hypothesis. When there’s only one factor (but with two or more levels), we call the analysis of means and “one-way ANOVA.” In the independent sample t-test, we tested whether two groups had the same mean. We made the connection between the confidence interval of the difference between the sample means and whether or not it includes zero (ie, no difference between the means). In ANOVA, we extend this idea to a test of whether two or more groups have the same mean. In fact, if you perform an ANOVA on only two groups, you will get the exact same answer as the independent two-sample t-test, although they use different distributions of critical values (t for the t-test, F for the ANOVA — a nice little fact for you square the t-test statistic, you’ll get the F-test statistic: $t^{2} = F$ ).

Let’s say we have an experiment where we’ve looked at the effect of different three different calorie levels on weight change in middle-aged men.

I’ve created a simulated dataset which we will use in our ANOVA discussions. The data set is available at the end of this page (scroll down or click here).

We might graph the mean weight changes ( $\pm$ SEM); Below are two possible outcomes of our experiment (Fig 1).

Figure 1. Hypothetical results, box plots.

Figure 1. Hypothetical results of an experiment, box plots. Left, no difference among groups; Right, large differences among groups.

As statisticians and scientists, we would first calculate an overall or grand mean for the entire sample of observations; we know this as the sample mean whose symbol is $\bar{X}$ . But this overall mean is made up of the average of the sample means. If the null hypothesis is true, then the all of the sample means all estimate the overall mean. Put another way, being a member of a treatment group doesn’t matter, ie, there is no systematic effect, all differences among subjects are due to random chance.

$\begin{align*} H_{0} : \bar{X} = \bar{X}_{A} = \bar{X}_{B} = \bar{X}_{C} \end{align*}$

The hypotheses among are three groups or treatment levels then are:

$\begin{align*} H_{0} : \bar{X}_{A} = \bar{X}_{B} = \bar{X}_{C}&& \end{align*}$

Null that there are no differences among the group means. And the alternative hypotheses include any (or all) of the following possibilities

$\begin{align*} H_{0} : \bar{X}_{A} \ne \bar{X}_{B} = \bar{X}_{C} \end{align*}$

or maybe

$\begin{align*} H_{0} : \bar{X}_{A} = \bar{X}_{B} \ne \bar{X}_{C} \end{align*}$

or… have we covered all possible alternate outcomes among three groups?

In either case, we could use one-way ANOVA to test for “statistically significant differences.”

Three important terms you’ll need for one-way ANOVA

FACTOR: We have one factor of interest. For example, a factor of interest might be

Diet fed to hypertensive subjects (men and women)
Distribution of coral reef sea cucumber species in archipelagos
Antibiotic drug therapy for adolescents with Acne vulgaris (click see Webster 2002 for review).

Note 1: Character vs factor. In R programming, while it is true that a treatment variable can be a character vector like Gender*, such vectors are handled by R as string. In contrast, by claiming Gender as a factor, R handles the variable as a nominal data type. Hence, ANOVA and other GLM functions expect any treatment variable to be declared as a factor:

Gender <- c("Gender non-conforming", "Men", "Non-binary", "Transgender", "Women", "Declined"); Gender
[1] "Gender non-conforming" "Men" "Non-binary" "Transgender" 
[5] "Women" "Declined"

# it’s a string vector, not a factor

is.factor(Gender)
[1] FALSE

Change string vector to a factor.

Gender <- as.factor(Gender); is.factor(Gender) [1] TRUE

* For discussions of ethics of language inclusivity and whether or not such information should be gathered, see Henrickson et al 2020.

LEVELS: We can have multiple levels (2 or more) within the single factor. Some examples of levels for the Factors listed:

Three diets (DASH), diet rich in fruits & vegetables, control diet)
Five archipelagos (Hawaiian Islands, Line Islands, Marshal Islands, Bonin Islands, and Ryukyu Islands)
Five antibiotics (ciprofloxacin, cotrimoxazole, erythromycin, doxycycline, minocycline).

RESPONSE: There is one outcome or measurement variable. This variable must be quantitative (ie, on the ratio or interval scale). Continuing our examples then

Reduction in systolic pressure
Numbers of individual sea cucumbers in a plot
number of microcomedo.

Note 2: A comedo is a clogged pore in the skin; a microcomedo refers to the small plug. Yes, I had to look that up, too.

The response variable can be just about anything we can measure, but because ANOVA is a parametric test, the response variable must be normally distributed!

Note on experimental design

As we discuss ANOVA, keep in mind we are talking about analyzing results from an experiment. Understanding statistics, and in particular ANOVA, informs how to plan an experiment. The basic experimental design is called a completely randomized experimental design, or CDR, where treatments are assigned to experimental units at random.

In this experimental design, subjects (experimental units) must be randomly assigned to each of these levels of the factor. That is, each individual should have had the same probability of being found in any one of the levels of the factor. The design is complete because randomization is conducted for all levels, all factors.

Thinking about how you would describe an experiment with three levels of some treatment, we would have

Table 1. Summary statistics of three levels for some ratio scale response variable.

Level 1

$\bar_{X}_{1}$

sample mean₁

$s_{1}$

sample standard deviation₁

Level 2

$\bar_{X}_{2}$

sample mean₂

$s_{2}$

sample standard deviation₂

Level 3

$\bar_{X}_{3}$

sample mean₃

$s_{3}$

sample standard deviation₃

Note 3: Table 1 is the basis for creating the box plots in Figure 1.

ANOVA sources of variation

ANOVA works by partitioning total variability about the means (the grand mean, the group means). We will discuss the multiple samples and how the ANOVA works in terms of the sources of variation. There are two “sources” of variation that can occur:

Within Group Variation
Among Groups Variation

So let’s look first at the variability within groups, also called the Within Group Variation.

Consider and experiment to see if DASH diet reduces systolic blood pressure in USA middle-aged men and women with hypertension (Moore et al 2001). After eight weeks we have

We get the corrected sum of squares, SS, for within group

$\begin{align*} within-groups \ SS = \sum_{i=1}^{k}\left [ \sum_{j=1}^{n_{i}}\left ( X_{ij}-\bar{X}_{i} \right )^2 \right ] \end{align*}$

and the degrees of freedom, DF, for within groups

$\begin{align*} within-groups \ DF = \sum_{i=1}^{k}\left ( n_{i}-1 \right ) = N - k \end{align*}$

where i is the identity of the groups, $X_{ij}$ is the individual observations within group i, $\bar{X}_{i}$ is the group i mean, n_i is the sample size within each group, N is the total sample size of all subjects for all groups, and k is the number of groups.

Importantly, this value is also referred to as “error sums of squares” in ANOVA. It’s importance is as follows — In our example, the within group variability would be zero if and only if all subjects within the same diet lost exactly the same amount of weight. This is hardly ever the case of course in a real experiment. Because there are almost always some other unknown factors or measurement error that affect the response variable, there will be some unknown variation among individuals who received the same treatment (within the same group). Thus, the error variance will generally be larger than zero.

The first point to consider: your ANOVA will never result in statistical differences among the groups if the error variance is greater than the second type of variability, the variability between groups.

The second type of variability in ANOVA is that due to the groups or treatments. Individuals given a calorie-restricted diet will lose some weight; individuals allowed to eat a calorie-rich diet likely will gain weight, therefore there will be variability (a difference) due to the treatment. So we can calculate the variability among groups. We get the corrected sum of squares for among groups

$\begin{align*} among-groups \ SS = \sum_{i=1}^{k}n_{i}\left ( X_{i}-\bar{X}\right )^2 \end{align*}$

and the degrees of freedom for among groups

$\begin{align*} among-groups \ DF = k - 1 \end{align*}$

where i is the identity of the groups, $\bar{X}$ is the grand mean as defined in Measures of Central Tendency (Chapter 3.1), $\bar{X}_{i}$ is the group i mean, n_i is the sample size within each group, N is the total sample size of all subjects for all groups, and k is the number of groups.

The sums of squares here is simply subtracting the mean of each population from the overall mean.

If the Factor is not important in explaining the variation among individuals then all the population means will be similar and the sums of squares among populations would be small.
If the Factor is important in explaining some of the variation among the individuals then all the population means will NOT be the same and the sums of squares among populations would be large.

Finally, we can identify the total variation in the entire experiment. We have the total sum of squares.

$\begin{align*} Total \ SS = among-groups \ SS + within-groups \ SS \end{align*}$

Thus, the insight of ANOVA is that variation in the dataset may be attributed to a portion explained by differences among the groups and differences among individual observations within each group. The inference comes from recognizing that if the among group effect is greater than the within group effect, then there will be a difference due to the treatment levels.

Mean squares

To decide whether the variation associated with the among group differences are greater than the within group variation, we calculate ratios of the sums of squares. These are called Mean Squares or MS for short. The ratio of the Mean Squares is called F, the test statistic for ANOVA.

For the one-way ANOVA we will have two Mean Squares and one F, tested with degrees of freedom for both the numerator $MS_{groups}$ , and denominator $MS_{error}$ ).

The Mean Square for (among) groups is

$\begin{align*} MS_{group} = \frac{among-groups \ SS}{among-groups \ DF} \end{align*}$

The Mean Square for error is

$\begin{align*} MS_{error} = \frac{within-groups \ SS}{within-groups \ DF} \end{align*}$

And finally, the value for F, the test statistic for ANOVA, is

$\begin{align*} F = \frac{MS_{group}}{MS_{error}} \end{align*}$

Worked example with R

A factor with three levels, A, B, and C

group <- c("A", "A", "A", "B", "B", "B", "C", "C", "C")

and their responses, simulated

response <- c(10.8, 11.8, 12.8, 6.5, 7, 8, 3.8, 2.8, 3)

We create a data frame

all <- data.frame(group, response)

Of course, you could place the data into a worksheet and then import the worksheet into R. Regardless, we now have our dataset.

Now, call the ANOVA function, aov, and assign the results to an object (eg, Model.1)

Model.1 <- aov(response ~ group, data=all)

Now, visualize the ANOVA table

summary(Model.1)

and the output from R, the ANOVA table, is shown below

Note 4: Rcmdr: Statistics → Means → One-way ANOVA… see figures 2 and 3 below

Table 2. Output from aov() command, the ANOVA table, for the “Difference” outcome variable.

           Df   Sum Sq    Mean Sq   F value       Pr(>F) 
group       2   111.16      55.58     89.49    0.0000341 ***
Residuals   6     3.73       0.62
---
Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1

Let’s take the ANOVA table one row at a time

The first row has subject headers defining the columns.
The second row of the table “groups” contains statistics due to group, and provides the comparisons among groups.
The third row, “Residuals” is the error, or the differences within groups.

Moving across the columns, then, for each row, we have in turn,

the degrees of freedom (there were 3 groups, therefore 2 DF for group),
the Sums of Squares, the Mean Squares,
the value of F, and finally,
the P-value.

R provides a helpful guide on the last line of the ANOVA summary table, the “Signif[icance] codes,” which highlights the magnitude of the P-value.

Note 5: On the use of “error” and “residuals.” Error refers to the difference between an observed value and the true population value. Residual is the difference between an observed value and a value predicted by a model. In ANOVA, the “error variance” or “residual variance” is the variation within groups unexplained by the tested factor. It is what remains after the effects of the model are accounted for.

What to report? ANOVA problems can be much more complicated than the simple one-way ANOVA introduced here. For complex ANOVA problems, report the ANOVA table itself! But for the one-way ANOVA it would be sufficient to report the test statistic, the degrees of freedom, and the p-value, as we have in previous chapters (eg, t-test, chi-square, etc.). Thus, we would report:

F = 89.49, df = 2 and 6, p = 0.0000341

where F = 89.49 is the test statistic, df = 2 (degrees of freedom for the among group mean square) and 6 (degrees of freedom for the within group mean square), and p = 0.0000341 is the p-value.

In Rcmdr, the appropriate command for the one-way ANOVA is simply

Rcmdr: Statistics → Means

Screenshot Rcmdr select one-way ANOVA

Figure 2. Screenshot Rcmdr select one-way ANOVA.

which brings up a simple dialog. R Commander anticipates factor (Groups) and Response variable. Optional, choose Pairwise comparisons of means for post-hoc test (Tukey’s) and, if you do not want to assume equal variances (see Chapter 13), select Welch F-test.

Screenshot Rcmdr select one-way ANOVA options

Figure 3. Screenshot Rcmdr select one-way ANOVA options.

Questions

1. Review the example ANOVA Table and confirm the following

How many levels of the treatment were there?
How many sampling units were there?
Confirm the calculation of $MS_{group}$ and $MS_{error}$ using the formulas contained in the text.
Confirm the calculation of F using the formula contained in the text.
The degrees of freedom for the F statistic in this example were 2 and 6 (F_2,6). Assuming a two-tailed test with Type I error rate of 5%, What is the critical value of the F distribution (see Appendix 20.4)

2. Repeat the one-way ANOVA using the simulated data, but this time, calculate the ANOVA problem for the “No.difference” response variable.

3. Leaf lengths from three strains of Arabidopsis thaliana plants grown in common garden are shown in Fig 2. Data are provided for you in the following R script.

arabid <- c("wt","wt","wt","AS1","AS1","AS1","AS2","AS2","AS2")
leaf <- c(4.909,5.736,5.108,6.956,5.809,6.888,4.768,4.209,4.065)
leaves <- data.frame(arabid,leaf)

Write out a null and alternative hypotheses
Conduct a test of the null hypothesis by one-way ANOVA
Report the value of the test statistic, the degrees of freedom, and the P-value
Do you accept or reject the null hypothesis? Explain your choice.

Box plot of length of leaves from three strains of Arabidopsis thaliana

Figure 4. Box plot of lengths of leaves of a 10-day old plant from on of three strains of Arabidopsis thaliana.

4. Return to your answer to question 7 from Chapter 12.1 and review your answer and modify as appropriate to correct your language to that presented here about factors and levels.

5. Conduct the one-way ANOVA test on the Comet assay data presented question 7 from Chapter 12.1. Data are copied for your convenience to end of this page. Obtain the ANOVA table and report the value of the test statistics, degrees of freedom, and p-value.
a. Based on the ANOVA results, do you accept or reject the null hypothesis? Explain your choice.

Quiz Chapter 12.2

One-way ANOVA

Data used in this page

Difference, no difference

Table 12.2 Difference or No Difference

Group	No.difference	Difference
A	12.04822	11.336161
A	12.67584	13.476142
A	12.99568	12.96121
A	12.01745	11.746712
A	12.34854	11.275492
B	12.17643	7.167262
B	12.77201	5.136788
B	12.07137	6.820242
B	12.94258	5.318743
B	12.0767	7.153992
C	12.58212	3.344218
C	12.69263	3.792337
C	12.60226	2.444438
C	12.02534	2.576014
C	12.6042	4.575672

Simulated data

Comet assay, antioxidant properties of tea

Data presented in Chapter 12.1

Chapter 12 contents

6.11 – F distribution

Introduction
χ2, t and F distributions are related
Questions
Quiz
Chapter 6 contents

Introduction

The F distribution is the probability distribution associated with the F statistic and named in honor of R. A. Fisher. The F distribution is used as the null distribution of the ANOVA test statistic. The F distribution is the ratio of two chi-square distributions, with degrees of freedom v₁ and v₂ for numerator and denominator, respectively.

Note 1: George W. Snedecor named the F-distribution after Sir Ronald Fisher in honor of his work on variance ratio. Fisher first developed the concept of the variance ratio in the 1920s, calling it the “general z-distribution”.

We can for illustration purposes define the F statistic as a ratio of two variances,

$\begin{align*} F = \frac{s_{2}^2}{s_{1}^2} \end{align*}$

Note 2: We test the ratio of variances to infer the difference between means because a large ratio suggests the variation between groups is much greater than the variation within groups. This comparison, key to Analysis of Variance (ANOVA), allows us to decide if the group means are different beyond random chance.

The F statistic has two sets of degrees of freedom, one for the numerator and one for the denominator. The actual formula for the F distribution is quite complicated and in general we don’t use the F distribution in a way that involves parameter estimation. Rather, it is used in evaluating the statistical significance of the F statistic. Therefore, we produce but a few graphs and a table of critical values to illustrate the distribution.

We call the result of this calculation the F test statistic. We evaluate how often that value or greater of a test statistic will occur by applying the F distribution function. A few graphs to get a sense of what the distribution looks like for varying v₁, v₂ held to ten degrees of freedom (Fig 1).
animated GIF, F distribution

Figure 1. Animated GIF plot of F distribution value for range of degrees of freedom.

By convention in the Null Hypothesis Significance Testing protocol (NHST), we compare the test statistic to a critical value. The critical value is defined as the value of the test statistic that occurs at the Type I error rate, which is typically set to 5%., per our presentations in Chapter 6.7, 6.9, and 6.10. The justification for NHST approach to testing of statistical significance is developed in Chapter 8.

Table of Critical values of the F distribution, one tail (upper)

Degrees of freedom v1 = 1 – 4, v2 = 10

Fv1, 10	α = 0.05	α = 0.025	α = 0.01
1	4.964	6.937	10.044
2	4.103	5.456	7.559
3	3.708	4.826	6.552
4	3.478	4.468	5.994

For the complete F table see Appendix 20.4

χ², t and F distributions are related

χ², t and F distributions are all distributions indexed by their degrees of freedom. With some algebra, these three distributions can be shown to be related to each other. The probabilities tabled in the chi-squared are part of the F-distribution.

Some interesting relationships between the F distribution and other distributions can be shown. By definition we claimed that the F distribution is built on ratio of chi-square distributions, so that should indicate to you the relationship between the two kinds of continuous probability distributions. However, one can also show relationships to other distributions for the F distribution. For example, for the case of v₁ = 1 and v₂ = any value, then F_1,v2 = t², where t refers to the t distribution.

Questions

What happens to the shape of the F distribution as df1 is held constant, but df2 degrees of freedom are increased from 1 to 5 to 20 to 100?
What happens to the shape of the F distribution as df1 degrees of freedom are increased from 1 to 5 to 20 to 100 while df2 is held constant?

Be able to answer these questions using the F table, Appendix 20.4, or using R/Rcmdr

For probability α = 5%, and numerator degrees of freedom equal to 1, what is the critical value of the F distribution (upper tail) for 5 degree of freedom? For 20 df? For 30 df?

Quiz Chapter 6.11

F distribution

13.4 – Tests for Equal Variances

Introduction

More about the F-test

R code

Levene’s test of equal variances

Cahoy’s bootstrap method

Questions

R code reminders

Chapter 13 contents

12.2 – One way ANOVA

Introduction

Three important terms you’ll need for one-way ANOVA

Note on experimental design

ANOVA sources of variation

Mean squares

Worked example with R

Questions

Data used in this page

Table 12.2 Difference or No Difference

Comet assay, antioxidant properties of tea

Chapter 12 contents

6.11 – F distribution

Introduction

Table of Critical values of the F distribution, one tail (upper)

χ2, t and F distributions are related

Questions

Chapter 6 contents

χ², t and F distributions are related