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You can use the Statistics and Machine Learning Toolbox™ function `anova1`

to perform one-way analysis of variance
(ANOVA). The purpose of one-way ANOVA is to determine whether data
from several groups (levels) of a factor have a common mean. That
is, one-way ANOVA enables you to find out whether different groups
of an independent variable have different effects on the response
variable *y*. Suppose, a hospital wants to determine
if the two new proposed scheduling methods reduce patient wait times
more than the old way of scheduling appointments. In this case, the
independent variable is the scheduling method, and the response variable
is the waiting time of the patients.

One-way ANOVA is a simple special case of the linear model. The one-way ANOVA form of the model is

$${y}_{ij}={\alpha}_{.j}+{\epsilon}_{ij}$$

with the following assumptions:

*y*_{ij}is an observation, in which*i*represents the observation number, and*j*represents a different group (level) of the predictor variable*y*. All*y*_{ij}are independent.*α*_{j}represents the population mean for the*j*th group (level or treatment).*ε*_{ij}is the random error, independent and normally distributed, with zero mean and constant variance, i.e.,*ε*_{ij}~ N(0,*σ*^{2}).

This model is also called the *means model*.
The model assumes that the columns of *y* are the
constant *α*_{j} plus
the error component *ε*_{ij}.
ANOVA helps determine if the constants are all the same.

ANOVA tests the hypothesis that all group means are equal versus the alternative hypothesis that at least one group is different from the others.

$$\begin{array}{l}{H}_{0}:{\alpha}_{1}={\alpha}_{2}=\mathrm{...}={\alpha}_{k}\\ {H}_{1}:\text{notallgroupmeansareequal}\end{array}$$

`anova1(y)`

tests
the equality of column means for the data in matrix `y`

,
where each column is a different group and has the same number of
observations (i.e., a balanced design). `anova1(y,group)`

tests
the equality of group means, specified in `group`

,
for the data in vector or matrix `y`

. In this case,
each group or column can have a different number of observations (i.e.,
an unbalanced design).

ANOVA is based on the assumption that all sample populations
are normally distributed. It is known to be robust to modest violations
of this assumption. You can check the normality assumption visually
by using a normality plot (`normplot`

).
Alternatively, you can use one of the Statistics and Machine Learning Toolbox functions
that checks for normality: the Anderson-Darling test (`adtest`

), the chi-squared goodness of fit
test (`chi2gof`

), the Jarque-Bera
test (`jbtest`

), or the Lilliefors
test (`lillietest`

).

You can provide sample data as a vector or a matrix.

If the sample data is in a vector,

`y`

, then you must provide grouping information using the`group`

input variable:`anova1(y,group)`

.`group`

must be a categorical variable, numeric vector, logical vector, character array, string array, or cell array of character vectors, with one name for each element of`y`

. The`anova1`

function treats the`y`

values corresponding to the same value of`group`

as part of the same group. For example,Use this design when groups have different numbers of elements (unbalanced ANOVA).

If the sample data is in a matrix,

`y`

, providing the group information is optional.If you do not specify the input variable

`group`

, then`anova1`

treats each column of`y`

as a separate group, and evaluates whether the population means of the columns are equal. For example,Use this form of design when each group has the same number of elements (balanced ANOVA).

If you specify the input variable

`group`

, then`group`

must be a character array, string array, or cell array of character vectors, with one name for each column of`y`

. The`anova1`

function treats the columns with the same group name as part of the same group. For example,

If `group`

contains empty or `NaN`

valued elements,
`anova1`

disregards the corresponding observations in
`y`

.

This example shows how to perform one-way ANOVA to determine whether data from several groups have a common mean.

Load and display the sample data.

```
load hogg
hogg
```

`hogg = `*6×5*
24 14 11 7 19
15 7 9 7 24
21 12 7 4 19
27 17 13 7 15
33 14 12 12 10
23 16 18 18 20

The data comes from a Hogg and Ledolter (1987) study on bacteria counts in shipments of milk. The columns of the matrix `hogg`

represent different shipments. The rows are bacteria counts from cartons of milk chosen randomly from each shipment.

Test if some shipments have higher counts than others. By default, `anova1`

returns two figures. One is the standard ANOVA table, and the other one is the box plots of data by group.

[p,tbl,stats] = anova1(hogg);

p

p = 1.1971e-04

The small *p*-value of about 0.0001 indicates that the bacteria counts from the different shipments are not the same.

You can get some graphical assurance that the means are different by looking at the box plots. The notches, however, compare the medians, not the means. For more information on this display, see `boxplot`

.

View the standard ANOVA table. `anova1`

saves the standard ANOVA table as a cell array in the output argument `tbl`

.

tbl

`tbl = `*4x6 cell array*
Columns 1 through 5
{'Source' } {'SS' } {'df'} {'MS' } {'F' }
{'Columns'} {[ 803.0000]} {[ 4]} {[200.7500]} {[ 9.0076]}
{'Error' } {[ 557.1667]} {[25]} {[ 22.2867]} {0x0 double}
{'Total' } {[1.3602e+03]} {[29]} {0x0 double} {0x0 double}
Column 6
{'Prob>F' }
{[1.1971e-04]}
{0x0 double }
{0x0 double }

Save the *F*-statistic value in the variable `Fstat`

.

Fstat = tbl{2,5}

Fstat = 9.0076

View the statistics necessary to make a multiple pairwise comparison of group means. `anova1`

saves these statistics in the structure `stats`

.

stats

`stats = `*struct with fields:*
gnames: [5x1 char]
n: [6 6 6 6 6]
source: 'anova1'
means: [23.8333 13.3333 11.6667 9.1667 17.8333]
df: 25
s: 4.7209

ANOVA rejects the null hypothesis that all group means are equal, so you can use the multiple comparisons to determine which group means are different from others. To conduct multiple comparison tests, use the function `multcompare`

, which accepts `stats`

as an input argument. In this example, `anova1`

rejects the null hypothesis that the mean bacteria counts from all four shipments are equal to each other, i.e., $${H}_{0}:{\mu}_{1}={\mu}_{2}={\mu}_{3}={\mu}_{4}$$.

Perform a multiple comparison test to determine which shipments are different than the others in terms of mean bacteria counts.

multcompare(stats)

`ans = `*10×6*
1.0000 2.0000 2.4953 10.5000 18.5047 0.0059
1.0000 3.0000 4.1619 12.1667 20.1714 0.0013
1.0000 4.0000 6.6619 14.6667 22.6714 0.0001
1.0000 5.0000 -2.0047 6.0000 14.0047 0.2119
2.0000 3.0000 -6.3381 1.6667 9.6714 0.9719
2.0000 4.0000 -3.8381 4.1667 12.1714 0.5544
2.0000 5.0000 -12.5047 -4.5000 3.5047 0.4806
3.0000 4.0000 -5.5047 2.5000 10.5047 0.8876
3.0000 5.0000 -14.1714 -6.1667 1.8381 0.1905
4.0000 5.0000 -16.6714 -8.6667 -0.6619 0.0292

The first two columns show which group means are compared with each other. For example, the first row compares the means for groups 1 and 2. The last column shows the *p*-values for the tests. The *p*-values 0.0059, 0.0013, and 0.0001 indicate that the mean bacteria counts in the milk from the first shipment is different from the ones from the second, third, and fourth shipments. The *p*-value of 0.0292 indicates that the mean bacteria counts in the milk from the fourth shipment is different from the ones from the fifth. The procedure fails to reject the hypotheses that the other group means are different from each other.

The figure also illustrates the same result. The blue bar shows the comparison interval for the first group mean, which does not overlap with the comparison intervals for the second, third, and fourth group means, shown in red. The comparison interval for the mean of fifth group, shown in gray, overlaps with the comparison interval for the first group mean. Hence, the group means for the first and fifth groups are not significantly different from each other.

ANOVA tests for the difference in the group means by partitioning the total variation in the data into two components:

Variation of group means from the overall mean, i.e., $${\overline{y}}_{.j}-{\overline{y}}_{\mathrm{..}}$$ (variation between groups), where $${\overline{y}}_{.j}$$ is the sample mean of group

*j*, and $${\overline{y}}_{\mathrm{..}}$$ is the overall sample mean.Variation of observations in each group from their group mean estimates, $${y}_{ij}-{\overline{y}}_{.j}$$ (variation within group).

In other words, ANOVA partitions the total sum of squares (SST) into sum of squares due to between-groups effect (SSR) and sum of squared errors(SSE).

$$\underset{SST}{\underbrace{{\displaystyle \sum _{i}{\displaystyle \sum _{j}{\left({y}_{ij}-{\overline{y}}_{\mathrm{..}}\right)}^{2}}}}}=\underset{SSR}{\underbrace{{\displaystyle \sum _{j}{n}_{j}{\left({\overline{y}}_{.j}-{\overline{y}}_{\mathrm{..}}\right)}^{2}}}}+\underset{SSE}{\underbrace{{\displaystyle \sum _{i}{\displaystyle \sum _{j}{\left({y}_{ij}-{\overline{y}}_{.j}\right)}^{2}}}}},$$

where *n _{j}* is the sample
size for the

Then ANOVA compares the variation between groups to the variation
within groups. If the ratio of within-group variation to between-group
variation is significantly high, then you can conclude that the group
means are significantly different from each other. You can measure
this using a test statistic that has an *F*-distribution
with (*k* – 1, *N* – *k*)
degrees of freedom:

$$F=\frac{\raisebox{1ex}{$SSR$}\!\left/ \!\raisebox{-1ex}{$k-1$}\right.}{\raisebox{1ex}{$SSE$}\!\left/ \!\raisebox{-1ex}{$N-k$}\right.}=\frac{MSR}{MSE}~{F}_{k-1,N-k},$$

where *MSR* is the mean squared treatment, *MSE* is
the mean squared error, *k* is the number of groups,
and *N* is the total number of observations. If the *p*-value
for the *F*-statistic is smaller than the significance
level, then the test rejects the null hypothesis that all group means
are equal and concludes that at least one of the group means is different
from the others. The most common significance levels are 0.05 and
0.01.

The ANOVA table captures the variability in the model by source,
the *F*-statistic for testing the significance of
this variability, and the *p*-value for deciding
on the significance of this variability. The *p*-value
returned by `anova1`

depends on assumptions about
the random disturbances *ε*_{ij} in
the model equation. For the *p*-value to be correct,
these disturbances need to be independent, normally distributed, and
have constant variance. The standard ANOVA table has this form:

`anova1`

returns the standard ANOVA table as
a cell array with six columns.

Column | Definition |
---|---|

`Source` | Source of the variability. |

`SS` | Sum of squares due to each source. |

`df` | Degrees of freedom associated with each source. Suppose N is
the total number of observations and k is the number
of groups. Then, N – k is
the within-groups degrees of freedom (`Error` ), k –
1 is the between-groups degrees of freedom (`Columns` ),
and N – 1 is the total degrees of freedom: N –
1 = (N – k) + (k –
1). |

`MS` | Mean squares for each source, which is the ratio `SS/df` . |

`F` | F-statistic, which is the ratio of the mean
squares. |

`Prob>F` | p-value, which is the probability that the F-statistic
can take a value larger than the computed test-statistic value. `anova1` derives
this probability from the cdf of the F-distribution. |

The rows of the ANOVA table show the variability in the data, divided by the source.

Row (Source) | Definition |
---|---|

`Groups` or `Columns` | Variability due to the differences among the group means (variability between groups) |

`Error` | Variability due to the differences between the data in each
group and the group mean (variability within groups) |

`Total` | Total variability |

[1] Wu, C. F. J., and M. Hamada. *Experiments:
Planning, Analysis, and Parameter Design Optimization*,
2000.

[2] Neter, J., M. H. Kutner, C. J. Nachtsheim,
and W. Wasserman. 4th ed. *Applied Linear Statistical Models*.
Irwin Press, 1996.

`anova1`

| `kruskalwallis`

| `multcompare`