OneWay Analysis of Variance: Example
In this lesson, we apply oneway analysis of variance to some fictitious data, and we show how to interpret the results of our analysis.
Note: Computations for analysis of variance are usually handled by a software package.
For this example, however, we will do the computations "manually", since the gory details have educational value.
Problem Statement
A pharmaceutical company conducts an experiment to test the effect of a new cholesterol medication.
The company selects 15 subjects randomly from a larger population. Each subject is randomly assigned
to one of three treatment groups. Within each treament group, subjects receive a different dose of
the new medication. In Group 1, subjects receive 0 mg/day; in Group 2, 50 mg/day; and in Group 3, 100 mg/day.
The treatment levels represent all the levels of interest to the experimenter, so this experiment used a
fixedeffects model to select
treatment levels for study.
After 30 days, doctors measure the cholesterol level of each subject. The results for all 15 subjects appear in the table below:
Dosage 
Group 1, 0 mg 
Group 2, 50 mg 
Group 3, 100 mg 
210 
210 
180 
240 
240 
210 
270 
240 
210 
270 
270 
210 
300 
270 
240 
In conducting this experiment, the experimenter had two research questions:
 Does dosage level have a significant effect on cholesterol level?
 How strong is the effect of dosage level on cholesterol level?
To answer these questions, the experimenter intends to use oneway analysis of variance.
Is OneWay ANOVA the Right Technique?
Before you crunch the first number in oneway analysis of variance, you must be sure that oneway analysis of variance
is the correct technique. That means you need to ask two questions:
 Is the experimental design compatible with oneway analysis of variance?
 Does the dataset satisfy the critical assumptions required for oneway analysis of variance?
Let's address both of those questions.
Experimental Design
As we discussed in the previous lesson (see OneWay Analysis of Variance: Fixed Effects),
oneway analysis of variance is only appropriate with one experimental design  a completely randomized design. That is
exactly the design used in our cholesterol study, so we can check the experimental design box.
Critical Assumptions
We also learned in the previous lesson that oneway analysis of variance makes three critical assumptions:
 Independence. The dependent variable score for each experimental unit is independent of the score for any other unit.
 Normality. In the population, dependent variable scores are normally distributed within treatment groups.
 Equality of variance. In the population, the variance of dependent variable scores in each treatment group is equal.
(Equality of variance is also known as homogeneity of variance or homoscedasticity.)
Therefore, for the cholesterol study, we need to make sure our dataset is consistent with the critical assumptions.
Independence of Scores
The assumption of independence is the most important assumption. When that assumption is violated, the resulting
statistical tests can be misleading.
The independence assumption is satisfied by the design of the study, which features random selection of subjects and random
assignment to treatment groups. Randomization tends to distribute effects of extraneous variables evenly across groups.
Normal Distributions in Groups
Violations of normality can be a problem when sample size is small, as it is in this cholesterol study. Therefore, it
is important to be on the lookout for any indication of nonnormality.
There are many different ways to check for normality. On this website, we describe three at:
How to Test for Normality: Three Simple Tests.
Given the small sample size, our best option for testing normality is to look at the following
descriptive statistics:
 Central tendency. The
mean and the
median are summary measures used
to describe central tendency  the most "typical" value in a set of values. With a normal
distribution, the mean is equal to the median.
 Skewness. Skewness is a measure of the assymmetry of a probability distribution.
If observations are equally distributed around the mean, the skewness value is zero; otherwise, the skewness
value is positive or negative. As a rule of thumb, skewness between 2 and +2 is consistent with
a normal distribution.
 Kurtosis. Kurtosis is a measure of whether observations cluster around the
mean of the distribution or in the tails of the distribution. The normal distribution has a
kurtosis value of zero. As a rule of thumb, kurtosis between 2 and +2 is consistent with a normal
distribution.
The table below shows the mean, median, skewness, and kurtosis for each group from our study.

Group 1, 0 mg 
Group 2, 50 mg 
Group 3, 100 mg 
Mean 
258 
246 
210 
Median 
270 
240 
210 
Range 
90 
60 
60 
Skewness 
0.40 
0.51 
0.00 
Kurtosis 
0.18 
0.61 
2.00 
In all three groups, the difference between the mean and median looks small (relative to the
range). And skewness and kurtosis measures are
consistent with a normal distribution (i.e., between 2 and +2). These are crude tests,
but they provide some confidence for the assumption of normality in each group.
Note: With Excel, you can easily compute the descriptive statistics in Table 1. To see how, go to:
How to Test for Normality: Example 1.
Homogeneity of Variance
When the normality of variance assumption is satisfied,
you can use Hartley's Fmax test to test for homogeneity of variance.
Here's how to implement the test:
 Step 1. Compute the sample variance ( s^{2}_{j} ) for each group.

kΣj=1
( X_{ i, j}  X_{ j} )^{ 2}

s^{2}_{j} = 


( n_{ j}  1 ) 
where X_{ i, j} is the score for observation i in Group j ,
X_{ j} is the mean of Group j, and
n_{ j} is the number of observations in Group j.
Here is the variance ( s^{2}_{j} ) for each group in the cholesterol study.
Group 1, 0 mg 
Group 2, 50 mg 
Group 3, 100 mg 
1170 
630 
450 
 Step 2. Compute an F ratio from the following formula:
F_{RATIO} = s^{2}_{MAX} / s^{2}_{MIN}
F_{RATIO} = 1170 / 450
F_{RATIO} = 2.6
where s^{2}_{MAX} is the largest group variance,
and s^{2}_{MIN} is the smallest group variance.
 Step 3. Compute degrees of freedom ( df ).
df = n  1
df = 5  1
df = 4
where n is the largest sample size in any group.
 Step 4. Based on the degrees of freedom ( 4 ) and the number of groups ( 3 ),
Find the critical F value from the Table of Critical F Values for Hartley's Fmax Test.
From the table, we see that the critical Fmax value is 15.5.
Note: The critical F values in the table are based on a
significance level of 0.05.
 Step 5. Compare the observed F ratio computed in Step 2 to the critical
F value recovered from the Fmax table in Step 4. If the F ratio is smaller than the Fmax table value,
the variances are homogeneous. Otherwise, the variances are heterogeneous.
Here, the F ratio (2.6) is smaller than the Fmax value (15.5), so we conclude that the variances are homogeneous.
Note: Other tests, such as
Bartlett's test, can also test for homogeneity of variance.
For the record, Bartlett's test yields the same conclusion for the cholesterol study; namely, the
variances are homogeneous.
Analysis of Variance
Having confirmed that the critical assumptions are tenable, we can proceed with a oneway analysis of variance.
That means taking the following steps:
 Specify a mathematical model to describe the causal factors that affect the dependent variable.
 Write statistical hypotheses to be tested by experimental data.
 Specify a significance level for a hypothesis test.
 Compute the grand mean and the mean scores for each group.
 Compute sums of squares for each effect in the model.
 Find the degrees of freedom associated with each effect in the model.
 Based on sums of squares and degrees of freedom, compute mean squares for each effect in the model.
 Compute a test statistic, based on observed mean squares and their expected values.
 Find the P value for the test statistic.
 Accept or reject the null hypothesis, based on the P value and the significance level.
 Assess the magnitude of the effect of the independent variable, based on sums of squares.
Now, let's execute each step, onebyone, with our cholesterol medication experiment.
Mathematical Model
For every experimental design, there is a mathematical model that accounts for all of the
independent and extraneous variables that affect the dependent variable. In our experiment,
the dependent variable ( X ) is the cholesterol level of a subject, and the
independent variable ( β ) is the dosage level administered to a subject.
For example,
here is the fixedeffects model for a completely randomized design:
X_{ i j} = μ + β_{ j} + ε_{ i ( j )}
where X_{ i j} is the cholesterol level for subject i in treatment group j,
μ is the population mean,
β_{ j} is the effect of the dosage level administered to subjects in group j;
and ε_{ i ( j )} is the effect of all other extraneous variables on subject i
in treatment j.
Statistical Hypotheses
For fixedeffects models, it is common practice to write statistical hypotheses in terms of the treatment effect β_{ j}.
With that in mind, here is the null hypothesis and the alternative hypothesis for a oneway analysis of variance:
If the null hypothesis is true, the mean score (i.e., mean cholesterol level) in each treatment group should equal the population mean. Thus,
if the null hypothesis is true, mean scores in the k
treatment groups should be equal. If the null hypothesis is false, at least one pair of mean scores should be unequal.
Significance Level
The significance level (also known as alpha or α) is the probability of rejecting the null hypothesis when it
is actually true. The significance level for an experiment is specified by the experimenter, before data collection
begins.
Experimenters often choose significance levels of 0.05 or 0.01. For this experiment, let's use a significance level of 0.05.
Mean Scores
Analysis of variance begins by computing a grand mean and group means:
X_{ j} = ( 1 / n
_{ j} )
n_{ j}Σi=1
( X
_{ i j} )
X_{ 1} = 258
X_{ 2} = 246
X_{ 3} = 210
In the equations above, n is the total sample size across all groups; and
n_{ j} is the sample size in Group j .
Sums of Squares
A sum of squares is the sum of squared deviations from a mean score. Oneway analysis of variance makes use of three sums of squares:
SST = 784 + 4 + 1084 + ... + 784 + 784 + 4
SST = 15,240
It turns out that the total sum of squares is equal to the betweengroups sum of squares plus the withingroups sum of squares, as shown below:
SST = SSB + SSW
15,240 = 6240 + 9000
Degrees of Freedom
The term degrees of freedom (df) refers to the number of independent sample points used to compute a
statistic minus the number of
parameters estimated from the sample points.
To illustrate what is going on, let's find the degrees of freedom associated with the various sum of squares computations:
 Betweengroups degrees of freedom. The betweengroups sum of squares formula appears below:
SSB =
kΣj=1
n
_{j} (
X_{ j} 
X )
^{2}
Here, the formula uses k independent sample points, the sample means X_{ j} .
And it uses one parameter estimate, the grand mean X, which was estimated from the sample points. So, the
betweengroups sum of squares has k  1 degrees of freedom ( df_{BG} ).
df_{BG} = k  1 = 5  1 = 4
 Withingroups degrees of freedom. The withingroups sum of squares formula appears below:
SSW =
kΣj=1
n_{ j}Σi=1
( X
_{ i j} 
X _{j} )
^{2}
Here, the formula uses n independent sample points, the individual subject scores X_{ i j} .
And it uses k parameter estimates, the group means X _{j} ,
which were estimated from the sample points. So, the
withingroups sum of squares has n  k degrees of freedom ( df_{WG} ).
n = Σ n_{ i} = 5 + 5 + 5 = 15
df_{WG} = n  k = 15  3 = 12
 Total degrees of freedom. The total sum of squares formula appears below:
SST =
kΣj=1
n_{ j}Σi=1
( X
_{ i j} 
X )
^{2}
Here, the formula uses n independent sample points, the individual subject scores X_{ i j} .
And it uses one parameter estimate, the grand mean X, which was estimated from the sample points. So, the
total sum of squares has n  1 degrees of freedom ( df_{TOT} ).
df_{TOT} = n  1 = 15  1 = 14
The degrees of freedom for each sum of squares are summarized in the table below:
Sum of squares 
Degrees of freedom 
Betweengroups 
k  1 = 2 
Withingroups 
n  k =12 
Total 
n  1 = 14 
Mean Squares
A mean square is an estimate of population variance. It is computed by dividing
a sum of squares (SS) by its corresponding degrees of freedom (df), as shown below:
MS = SS / df
To conduct a oneway analysis of variance, we are interested in two mean squares:
Expected Value
The expected value
of a mean square is the average value of the mean square over a large number of experiments.
Statisticians have derived formulas for the expected value of
the withingroups mean square ( MS_{WG} ) and for the expected value of the betweengroups mean square
( MS_{BG} ). For oneway analysis of variance, the
expected value formulas are:
Fixed and RandomEffects:
E( MS_{WG} ) = σ_{ε}^{2}
FixedEffects:


E( MS_{BG} ) = σ_{ε}^{2} + 


( k  1 ) 
RandomEffects:
E( MS_{BG} ) = σ_{ε}^{2} + nσ_{β}^{2}
In the equations above, E( MS_{WG} ) is the expected value of the withingroups mean square;
E( MS_{BG} ) is the expected value of the betweengroups mean square;
n is total sample size; k is the number of treatment groups;
β_{ j} is the treatment effect in Group j; σ_{ε}^{2}
is the variance attributable to everything except the treatment effect (i.e., all the extraneous variables); and
σ_{β}^{2} is the variance due to random selection of treatment levels.
Notice that MS_{BG} should equal MS_{WG}
when the variation due to treatment effects
( β_{ j} for fixed effects and σ_{β}^{2} for random effects)
is zero (i.e., when the independent variable does not affect the
dependent variable). And MS_{BG} should be bigger than the MS_{WG}
when the variation due to treatment effects is not zero (i.e., when the independent variable does affect the
dependent variable)
Conclusion: By examining the relative size of the mean squares, we can make a judgment about whether an
independent variable affects a dependent variable.
Test Statistic
Suppose we use the mean squares to
define a test statistic F as follows:
F(v_{1}, v_{2}) = MS_{BG} / MS_{WG}
F(2, 12) = 3120 / 750 = 4.16
where MS_{BG} is the betweengroups mean square,
MS_{WG} is the withingroups mean square, v_{1} is the degrees of freedom for MS_{BG},
and v_{2} is the degrees of freedom for MS_{WG}.
Defined in this way, the F ratio measures the size of MS_{BG} relative to MS_{WG}.
The F ratio is a convenient measure that we can use to test the null hypothesis. Here's how:
 When the F ratio is close to one, MS_{BG} is approximately equal to MS_{WG}.
This indicates that the independent variable did not affect the dependent variable, so we cannot
reject the null hypothesis.
 When the F ratio is significantly greater than one, MS_{BG} is bigger than MS_{WG}.
This indicates that the independent variable did affect the dependent variable, so we must reject the null hypothesis.
What does it mean for the F ratio to be significantly greater than one?
To answer that question, we need to talk about the Pvalue.
PValue
In an experiment, a Pvalue is the probability of obtaining a result more extreme than the observed experimental outcome,
assuming the null hypothesis is true.
With analysis of variance, the F ratio is the observed experimental outcome that we are interested in.
So, the Pvalue would be the probability that an F statistic would be more extreme (i.e., bigger) than the
actual F ratio computed from experimental data.
We can use Stat Trek's
F Distribution Calculator
to find the probability that an F statistic will be bigger than the actual F ratio observed in the experiment.
Enter the betweengroups degrees of freedom (2), the withingroups degrees of freedom (12),
and the observed F ratio (4.16) into the calculator; then, click the Calculate button.
From the calculator, we see that P( F < 4.16 ) equals 0.96; so the P ( F > 4.16 ) equals 1 minus 0.96 or 0.04.
Therefore, the PValue is 0.04.
Hypothesis Test
Recall that we specified a significance level 0.05 for this experiment.
Once you know the significance level and the Pvalue, the hypothesis test is routine.
Here's the decision rule for accepting or rejecting the null hypothesis:
 If the Pvalue is bigger than the significance level, accept the null hypothesis.
 If the Pvalue is equal to or smaller than the significance level, reject the null hypothesis.
Since the Pvalue (0.04) in our experiment is smaller than the significance level (0.05), we
reject the null hypothesis that drug dosage had no effect on cholesterol level. And we
conclude that the mean cholesterol level in at least one treatment group differed
significantly from the mean cholesterol level in another group.
Magnitude of Effect
The hypothesis test tells us whether the independent variable in our experiment has a statistically
significant effect on the dependent variable, but it does not address the magnitude
of the effect. Here's the issue:
 When the sample size is large, you may find that even small differences in treatment means are
statistically significant.
 When the sample size is small, you may find that even big differences in treatment means are
not statistically significant.
With this in mind, it is customary to supplement analysis of variance with an appropriate measure
of effect size. Eta squared (η^{2}) is one such measure. Eta squared is the proportion of variance in the
dependent variable that is explained by a treatment effect. The eta squared formula
for oneway analysis of variance is:
η^{2} = SSB / SST
where SSB is the betweengroups sum of squares and SST is the total sum of squares.
Given this formula, we can compute eta squared for this drug dosage experiment, as shown below:
η^{2} = SSB / SST = 6240 / 15240 = 0.41
Thus, 41 percent of the variance in our dependent variable (cholesterol level) can be explained by
variation in our independent variable (dosage level). It appears that the relationship between
dosage level and cholesterol level is significant not only in a statistical sense; it is
significant in a practical sense as well.
ANOVA Summary Table
It is traditional to summarize ANOVA results in an analysis of variance table. The analysis
that we just conducted provides all of the information that we need to produce the following
ANOVA summary table:
Analysis of Variance Table
Source 
SS 
df 
MS 
F 
P 
BG 
6,240 
4 
750 
4.16 
0.04 
WG 
9,000 
12 
3,120 


Total 
15,240 
14 



This ANOVA table allows any researcher to interpret the results of the experiment, at a glance.
The Pvalue (shown in the last column of the ANOVA table) is the probability that an F statistic would be more extreme (bigger) than the
F ratio shown in the table, assuming the null hypothesis is true. When the Pvalue is bigger
than the significance level, we accept the null hypothesis; when it is smaller, we reject it.
Here, the Pvalue (0.04) is smaller than the significance level (0.05), so
we reject the null hypothesis.
To assess the strength of the treatment effect, an experimenter might compute eta squared (η^{2}). The
computation is easy, using sum of squares entries from the ANOVA table, as shown below:
η^{2} = SSB / SST = 6,240 / 15,240 = 0.41
where SSB is the betweengroups sum of squares and SST is the total sum of squares.
For this experiment, an eta squared of 0.41 means that 41% of the variance in the dependent variable
can be explained by the effect of the independent variable.
An Easier Option
In this lesson, we showed all of the hand calculations for a oneway analysis of variance. In the real world,
researchers seldom conduct analysis of variance by hand. They use statistical software. In the next lesson,
we'll analyze data from this problem with Excel. Hopefully, we'll get the same result.