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| '''Step 3''' is to calculate the between-groups sum of squares(''SS,,B,,''):<<BR>> | '''Step 3''' is to calculate the between-groups sum-of-squares (''SS,,B,,''):<<BR>> |
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| where ''n'' is the number of subjects in each group, {{attachment:xbar.gif|X-bar}}'',,g,,'' is the mean for group ''g'', and {{attachment:xbar.gif|X-bar}} is the overall mean (across all groups). | where ''n'' is the number of subjects in each group, {{attachment:xbar.gif|X-bar}}'',,g,,'' is the mean for group ''g'', and {{attachment:xbar.gif|X-bar}} is the grand mean (across all groups). |
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| Now, '''Step 4''' , we'll calculate the sum-of-squares within groups (''SS,,W,,''). For a group ''g'', this is | Now, '''Step 4''' , we'll calculate the sum-of-squares, degrees of freedom, and mean square error ''within groups'' as a way of determining how much variability there is in the sample observation values. First, compute the within-groups sum-of-squares (''SS,,W,,'') by summing the sum-of-squares for each cell group. More formally, we can write this as: |
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| where ''SS,,g,,'' is the sum-of-squares for group ''g''. So: <<BR>> | where ''SS,,g,,'' is the sum-of-squares for cell group ''g''. So: <<BR>> |
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| To calculate the degrees of freedom within groups, we use the following: | Next, to calculate the ''degrees of freedom'' within groups (''df,,W,,''), we use the following: |
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| where ''N''is the total number of subjects (or observations) in the experiment, ''r'' is the number of rows, and ''cc'' is the number of columns. So:<<BR>> '''''df,,W,,''' = N - rc''<<BR>> |
where ''N''is the total number of observations in the experiment, ''r'' is the number of rows, and ''c'' is the number of columns. So: <<BR>> '''''df,,W,,''' = N - rc'' <<BR>> |
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And finally, we compute the ''mean square error'' within groups ('''''s,,W,,^2^''''') by dividing ''SS,,W,,''by ''df,,W,,'':<<BR>> |
|
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| where ''r'' ranges over rows, meaning that {{attachment:xbar.gif|X-bar}}'',,r,,'' is the mean of row ''r'', and {{attachment:xbar.gif|X-bar}} is the overall mean (across all groups). | where ''r'' ranges over rows, meaning that {{attachment:xbar.gif|X-bar}}'',,r,,'' is the mean of row ''r'', and {{attachment:xbar.gif|X-bar}} is the grand mean. |
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| where ''c'' ranges over columns, meaning that {{attachment:xbar.gif|X-bar}}'',,c,,'' is the mean of column ''c'', and {{attachment:xbar.gif|X-bar}} is the overall mean (across all groups). | where ''c'' ranges over columns, meaning that {{attachment:xbar.gif|X-bar}}'',,c,,'' is the mean of column ''c'', and {{attachment:xbar.gif|X-bar}} is the grand mean. |
Situations with Two or More Variables of Interest
When considering the relationship among three or more variables, an interaction may arise. Interactions describe a situation in which the simultaneous influence of two variables on a third is not additive. Most commonly, interactions are considered in the context of multiple regression analyses, but they may also be evaluated using two-way ANOVA.
An Example Problem
To determine if prenatal exposure to cocaine alters dendritic spine density within prefrontal cortex, 20 rats were equally divided between treatment groups that were prenatally exposed to either cocaine or placebo. Further, because any effect of prenatal drug exposure might be evident at one age but not another, animals within each treatment group were further divided intow two groups. One sub-group was studied at 4 weeks of age and the other was studied at 12 weeks of age. Thus, our independent variables are treatment (prenatal drug/placebo exposure) and age (lets say 4 and 12 weeks of age), and our dependent variable is spine density. The following table shows one possible outcome of such a study:
DENDRITIC SPINE DENSITY |
|||
4-Week Control |
4-Week Cocaine |
12-Week Control |
12-Week Cocaine |
7.5 |
5.5 |
8.0 |
5.0 |
8.0 |
3.5 |
10.0 |
4.5 |
6.0 |
4.5 |
13.0 |
4.0 |
7.0 |
6.0 |
9.0 |
6.0 |
6.5 |
5.0 |
8.5 |
4.0 |
There are three null hypotheses we may want to test. The first two test the effects of each independent variable (or factor) under investigation:
H01: Treatment substance (control or cocaine) has no effect on dendritic spine density.
H02: Age has no effect on dendritic spine density.
And the third tests for an interaction between these two factors:
H03: The two factors (treatment and age) are independent; i.e., there is no interaction effect.
Two-Way ANOVA
A two-way ANOVA is an analysis technique that quantifies how much of the variance in a sample can be accounted for by each of two categorical variables and their interactions. Note that different rats are used in each of the four experimental groups, so a standard two-way ANOVA should be used. If the same rats were used in each experimental condition, one would want to use a two-way repeated-measures ANOVA.
Step 1 is to compute the group means (for each cell, row, and column) and the grand mean (for all observations in the whole experiment):
GROUP MEANS |
|||
|
4-Week |
12-Week |
All Ages |
Control |
7 |
9.7 |
8.35 |
Cocaine |
4.9 |
4.7 |
4.8 |
All Prenatal Exposures |
5.95 |
7.2 |
6.575 |
Take a moment to define your variables
Before we go any further, we should define some variables that we'll use to complete the computations needed for the rest of the analysis:
n = number of observations in each group (here, it's 5 since there are 5 subjects per group)
N = total number of observations in the whole experiment (here, 20)
r = number of rows (here 2, one for each age group)
c = number of columns (here 2, one for each treatment substance condition
g = mean of a particular cell group g (for example, 12-Week, Control = 9.7).
R = mean of a particular row R (for example, 4-Week = 5.95)
C = mean of a particular column C (for example, Cocaine = 4.8)
And finally, to refer to the grand mean (the overall mean of all observations in the experiment), we'll simply use the notation . Here, the grand mean, , is 6.575.
Now that we've calculated the means, it's also helpful to also plot the mean data. (It's easier to understand it that way!)
Step 2 is to calculate the sum of squares (SS) for each group (cell) using the following formula:
![\[
SS = \sum_{i=1} (x_{i,g} - \overline{X}_g)^2
\]](/StatsWiki/MoreThanTwoVariables?action=AttachFile&do=get&target=latex_f69f0fdaf675e595f329938fb5a945e50c72d93d_p1.png "\[
SS = \sum_{i=1} (x_{i,g} - \overline{X}_g)^2
\]")
where xi,g is the ith measurement for group g, and g is the mean of group g.
For each group, this formula is implemented as follows:
4-Week Control:
{7.5, 8, 6, 7, 6.5},
4-week, control = 7
4-week, control = (7.5-7)2 + (8-7)2 + (6-7)2 + (7-7)2 + (6.5-7)2 = 2.5
4-Week Cocaine:
{5.5, 3.5, 4.5, 6, 5},
4-week, cocaine = 4.9
4-week, cocaine = (5.5-4.9)2 + (3.5-4.9)2 + (4.5-4.9)2 + (6-4.9)2 + (5-4.9)2 = 3.7
12-Week Control:
{8, 10, 13, 9, 8.5},
12-week, control = 9.7
12-week, control = (8-9.7)2 + (10-9.7)2 + (13-9.7)2 + (9-9.7)2 + (8.5-9.7)2 = 15.8
12-Week Cocaine:
{5, 4.5, 4, 6, 4},
12-week, cocaine = 4.7
12-week, cocaine = (5-4.7)2 + (4.5-4.7)2 + (4-4.7)2 + (6-4.7)2 + (4-4.7)2 = 2.8
Step 3 is to calculate the between-groups sum-of-squares (SSB):
![\[
SS_B = n \cdot \sum_{g} (\overline{X}_{g} - \overline{X})^2
\]](/StatsWiki/MoreThanTwoVariables?action=AttachFile&do=get&target=latex_fbe398bc5e0bcdd916b5a33ab0662fac1026241a_p1.png "\[
SS_B = n \cdot \sum_{g} (\overline{X}_{g} - \overline{X})^2
\]")
where n is the number of subjects in each group, g is the mean for group g, and is the grand mean (across all groups).
SSB = n [(
4-week, control - )2 + ( 4-week, cocaine - )2 + ( 12-week, control - )2 + ( 12-week, cocaine - )2]
= 5 [(7 - 6.575 )2 + (4.9 - 6.575)2 + (9.7 - 6.575)2 + (4.7 - 6.575)2]
= 5 [0.180625 + 2.805625 + 9.765625 + 3.515625]
= 5 [16.2675]
= 81.3375
Now, Step 4 , we'll calculate the sum-of-squares, degrees of freedom, and mean square error within groups as a way of determining how much variability there is in the sample observation values. First, compute the within-groups sum-of-squares (SSW) by summing the sum-of-squares for each cell group. More formally, we can write this as:
![\[
SS_W = \sum_{g} SS_g
\]](/StatsWiki/MoreThanTwoVariables?action=AttachFile&do=get&target=latex_4661d70fdac9a975e5dfe848b22a69a682186df1_p1.png "\[
SS_W = \sum_{g} SS_g
\]")
where SSg is the sum-of-squares for cell group g. So:
SSW = SS4-week, control + SS4-week, cocaine + SS12-week, control + SS12-week, cocaine
= 2.5 + 3.7 + 15.8 + 2.8
= 24.8
Next, to calculate the degrees of freedom within groups (dfW), we use the following:
dfW = N - rc
where Nis the total number of observations in the experiment, r is the number of rows, and c is the number of columns. So:
dfW = N - rc
= 20 - (2 * 2)
= 16
And finally, we compute the mean square error within groups (sW2) by dividing SSWby dfW:
sW2 = SSW / dfW
= 24.8 / 16
= 1.55
Note that SSW is also known as the "residual" or "error" since it quantifies the amount of variability after the condition means are taken into account. The degrees of freedom here are N - rc because there are N data points, but the number of means fit is r*c, giving a total of N - rc variables that are free to vary.
For Step 5, we'll calculate the sum-of-squares for the rows (SSR):
![\[
SS_r = \sum_{r} (\overline{X}_r - \overline{X})^2
\]](/StatsWiki/MoreThanTwoVariables?action=AttachFile&do=get&target=latex_a0f574792aac4e4b8bf3bf3c6832a6f7adf06754_p1.png "\[
SS_r = \sum_{r} (\overline{X}_r - \overline{X})^2
\]")
where r ranges over rows, meaning that r is the mean of row r, and is the grand mean.
SSR = n [(
control - )2 + ( cocaine - )2]
= 10 [(8.35 - 6.575)2 + (4.8 - 6.575)2]
= 10 [3.150625 + 3.150625]
= 10 [6.30125]
= 63.0125
dfR = r - 1
= 2-1
= 1
sR2 = SSR / dfR
= 63.0125 / 1
= 63.0125
In Step 6, we calculate the sum-of-squares for the columns (SSc):
![\[
SS_c = \sum_{c} (\overline{X}_c - \overline{X})^2
\]](/StatsWiki/MoreThanTwoVariables?action=AttachFile&do=get&target=latex_f6c3dc8465d80e5fbe0ef5018b5dacef5e32d0cb_p1.png "\[
SS_c = \sum_{c} (\overline{X}_c - \overline{X})^2
\]")
where c ranges over columns, meaning that c is the mean of column c, and is the grand mean.
SSc =
( 4-week - )2 + ( 12-week - )2]
= 10 [(5.95 - 6.575)2 + (7.2 - 6.575)2]
= 10 [0.390625 + 0.390625]
= 10 [0.78125]
= 7.8125
dfC = c - 1
= 2-1
= 1
sC2 = SSC / dfC
= 7.8125 / 1
= 7.8125
In Step 7, calculate the sum-of-squares for the interaction (SSRC) by summing the sum-of-squares you calculated for between-groups, rows, and columns (SSB, SSR, and SSC, respectively):
SSRC = SSB - SSR - SSC
= 81.3375 - 63.0125 - 7.8125
= 10.5125
dfRC = (r - 1)(c - 1)
= (2-1)(2-1)
= 1
sRC2 = SSRC / dfRC
= 10.5125 / 1
= 10.5125
Now, in Step 8, we'll calculate the total sum-of-squares (SST) by summing all previously calculated sum-of-squares:
SST = SSB + SSW + SSR + SSC + SSRC
= 81.3375 + 24.8 + 63.0125 + 7.8125 + 10.5125
= 187.475
dfT = N - 1
= 20-1
= 19
At Step 9, we calculate the F values:
FR = sR2 / sW2
= 63.0125 / 1.55
= 40.65323
FC = sC2 / sW2
= 7.8125 / 1.55
= 5.040323
FRC = sRC2 / sW2
= 10.5125 / 1.55
= 6.782258
And, finally, at Step 10, we can organize all of the above into a table, along with the appropriate Fcrit values (looked up in a table like this one) that we'll use for comparison and interpretation of our computations:
Fcrit (1, 16) α=0.5 = 4.49
ANOVA TABLE |
||||||
Source |
SS |
df |
s2 |
Fobt |
Fcrit |
p |
rows |
63.0125 |
1 |
63.0125 |
40.65323 |
4.49 |
p < 0.05 |
columns |
7.8125 |
1 |
7.8125 |
5.040323 |
4.49 |
p < 0.05 |
r * c |
10.5125 |
1 |
10.5125 |
6.782258 |
4.49 |
p < 0.05 |
within |
24.8 |
16 |
1.55 |
-- |
-- |
-- |
total |
187.475 |
19 |
-- |
-- |
-- |
-- |
Both factors (treatment and age) are significant, as indicate by the fact that Fobt > Fcrit. Thus, we can reject all H01 and H02 and conclude that dendritic spine density is affected by prenatal cocaine exposure and age. The interaction between the two factors (r * c) is also significant. Thus, we can also reject H03 and conclude there is a significant interaction between treatment and age.
- To further interpret these results, we can plot the group means as follows:
NOTE: Remember that the statistics provided by the ANOVA quantify the effect of each factor (in this case, treatment and age). These statistics do not compare individual condition means, such as whether 4-week control differs from 12-week control. If k = the number of groups, the number of possible comparisons is k * (k-1) / 2. In the above example, we have 4 groups, so there are (4*3)/2 = 6 possible comparisons between these group means. Statistical testing of these individual comparisons requires a post-hoc analysis that corrects for experiment-wise error rate. If all possible comparisons are of interest, the Tukey's HSD (Honestly Significant Difference) Test is commonly used.
Tukey's HSD (Honestly Significant Difference) Test
Tukey's test is a single-step, multiple-comparison statistical procedure often used in conjunction with an ANOVA to test which group means are significantly different from one another. It is used in cases where group sizes are equal (the Tukey-Kramer procedure is used if group sizes are unequal) and it compares all possible pairs of means. Note that the Tukey's test formula is very similar to that of the t-test, except that it corrects for experiment-wise error rate. (When there are multiple comparisons being made, the probability of making a type I error (rejecting a true null hypothesis) increases, so Tukey's test corrects for this.) The formula for a Tukey's test is:
![\[
q_{obt} = (Y_{A} - Y_{B}) / \sqrt{ s_{W}^2 / n}
\]](/StatsWiki/MoreThanTwoVariables?action=AttachFile&do=get&target=latex_51b588f7c3115f85a29b1bd8bc32f7b8a737fe15_p1.png "\[
q_{obt} = (Y_{A} - Y_{B}) / \sqrt{ s_{W}^2 / n}
\]")
where YA is the larger of the two means being compared, YB is the smaller,sW2 is the mean squared error within, and n is the number of data points within each group . Once computed, the qobt value is compared to a q-value from the q distribution. If the qobt value is larger than the qcrit value from the distribution, the two means are significantly different.
So, if we wanted to use a Tukey's test to determine whether 4-week control significantly differs from 12-week control, we'd calculate it as follows
qobt =
12-week, control - 4-week, control / 
= 9.7 - 7 /
= 9.7 - 7 /
= 2.7 / 0.5567764
= 4.849343
The qcrit value may be looked up in a chart (like this one) using the appropriate values for k (which represents the number of group means, so 4 here) and dfw (16). So here, qcrit (4, 16) α=0.5= 4.05. Since qobt > qcrit (4.85 > 4.05), we can conclude that the two means are, in fact, (honestly) significantly different.
Additional group comparisons reveal that spine density does not change significantly with age in the cocaine-exposed group, but is significantly greater in the control group as compared to cocaine-exposed animals at both 4 and 12 weeks of age. Thus, the effect of prenatal drug exposure is magnified with time due to a developmental increase in spine density that occurs only in the control group.
IMPORTANT NOTE: There are various post-hoc tests that can be used, and the choice of which method to apply is a controversial area in statistics. The different tests vary in how conservative they are: more conservative tests are less powerful and have a lower risk of a type I error (rejecting a true null hypothesis), however this comes at the cost of increasing the risk of a type II error (incorrectly accepting the null hypothesis). Some common methods, listed in order of decreasing power, are: Fisher’s LSD, Newman-Keuls, Tukey HSD, Bonferonni, Scheffé. The following provides some guidelines for choosing an appropriate post-hoc procedure.
If all pairwise comparisons are truly relevant, the Tukey method is recommended.
If it is most reasonable to compare all groups against a single control, then the Dunnett test is recommended.
If only a subset of the pairwise comparisons are relevant, then the Bonferroni method is often utilized for those selected comparisons. For example, in the present example one would likely not be interested in comparing the 4-week control to the 12-week cocaine group, or the 12-week control to the 4-week cocaine group.
Multiple Regression
Another analysis technique you could use is a multiple regression. In multiple regression, we find coefficients for each group such that we are able to best predict the group means. The multiple regression computes standard errors on the coefficients, meaning that we can determine if a coefficient is significantly different from zero. On this simple example, multiple regression will give identical answers to ANOVA, but in more complex cases, multiple regression is a more powerful technique that allows you to include additional nuisance predictors, that the analysis controls for before testing for significance of your independent variables.