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← Revision 12 as of 2014-05-07 12:39:09 ⇥
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= How do I compare group means in a non-standard post-hoc contrast = |
= How do I compare group means in a non-standard post-hoc contrast? = |
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| Compare a group of controls (CO) with a group with general memory deficits (GM) and a group with delayed recall (DR) problems. control mean - 0.5(memory deficit group mean - delayed recall group mean) which therefore has contrast coefficients c(CO)=1, c(GM)=c(DR)=-0.5. |
Let us suppose we wish to compare the mean test performance of a group of controls (CO) with the average performance of a group with general memory deficits (GM) and a group with delayed recall (DR) problems. |
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| The variance of this contrast is | The difference we are interested in is, therefore, |
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| {{{ $$\frac{c(CO)^text{2}}{N(CO)} + \frac{c(GM)^text{2}}{N(GM)} + \frac{c(DR)^text{2}}{N(DR)} MSE }}} where N(.) is the number in each group and MSE is the mean square of the error term obtained from the full anova table. |
control mean - 0.5(memory deficit group mean - delayed recall group mean) |
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| For example if we are comparing the control mean with the average of the memory deficit and delayed recall groups the variance is $$\frac{1}{N(CO)} + \frac{1}{4N(GM)} + \frac{1}{4N(DR)} MSE}$$ |
which has contrast coefficients c(CO)=1, c(GM)=c(DR)=-0.5. |
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| If the control mean does not differ from the average of the two patient group means this difference in means divided by the square root of its variance approximately follows a t distribution with degrees of freedom equal to the error df from the anova table. | The variance of a contrast in group means may be written in general as |
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| {{{ $$\frac{control mean - 0.5(memory deficit group mean - delayed recall group mean)}{frac{1}{N(CO)} + \frac{1}{4N(GM)} + \frac{1}{4N(DR)} MSE} }}} |
sum(i=1 to k) [(c(i)*c(i))]/[N(i)]MSE where N(i) is the number in the i-th of k groups and MSE is the mean square of the error term obtained from the full anova table. For example if we are comparing the control mean with the average of the memory deficit and delayed recall groups the variance of this difference is 1/[N(CO)] + 1/[4N(GM)] + 1/[4N(DR)] MSE. If the control mean does not differ from the average of the two patient group means the difference in means divided by the square root of its variance approximately follows a t distribution with degrees of freedom equal to the error df from the anova table. So we compute [control mean - 0.5(memory deficit group mean - delayed recall group mean)]/ Sqrt[1/[N(CO)] + 1/[4N(GM] + 1/[4N(DR)] MSE] and compare with a t distribution. If the groups are all quite large (e.g. over 30) then we can compare the above test statistic with a z-value such as, for example, +/- 1.96 (for a two-tailed test at the 5% level). |
How do I compare group means in a non-standard post-hoc contrast?
There may be a specific group comparison you are interested in which is not routinely outputted with an analysis of variance.
Suppose we have a vector of contrast coefficients, c, with i-th term c(i) being the i-th group contrast coefficient.
Let us suppose we wish to compare the mean test performance of a group of controls (CO) with the average performance of a group with general memory deficits (GM) and a group with delayed recall (DR) problems.
The difference we are interested in is, therefore,
control mean - 0.5(memory deficit group mean - delayed recall group mean)
which has contrast coefficients c(CO)=1, c(GM)=c(DR)=-0.5.
The variance of a contrast in group means may be written in general as
sum(i=1 to k) [(c(i)*c(i))]/[N(i)]MSE
where N(i) is the number in the i-th of k groups and MSE is the mean square of the error term obtained from the full anova table.
For example if we are comparing the control mean with the average of the memory deficit and delayed recall groups the variance of this difference is
1/[N(CO)] + 1/[4N(GM)] + 1/[4N(DR)] MSE.
If the control mean does not differ from the average of the two patient group means the difference in means divided by the square root of its variance approximately follows a t distribution with degrees of freedom equal to the error df from the anova table.
So we compute
[control mean - 0.5(memory deficit group mean - delayed recall group mean)]/ Sqrt[1/[N(CO)] + 1/[4N(GM] + 1/[4N(DR)] MSE]
and compare with a t distribution. If the groups are all quite large (e.g. over 30) then we can compare the above test statistic with a z-value such as, for example, +/- 1.96 (for a two-tailed test at the 5% level).