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Rescaling chi-square as an F-ratio is easy, just divide the chi-square value by its degrees of freedom. So a chi-square value of 6.9 with 3 df rescales to an F-ratio of 2.3 with 2 and ? degrees of freedom. The trick is to estimate a reasonable value for the denominator degrees of freedom.
Consider the following two-group (a design in which each subject receives four treatments (b) in a counterbalanced order.
use http://www.ats.ucla.edu/stat/data/repeated_missing, clear
tab s b
| b
s | 1 2 3 4 | Total
-----------+--------------------------------------------+----------
1 | 1 1 1 1 | 4
2 | 1 1 1 1 | 4
3 | 1 1 0 1 | 3
4 | 1 1 1 1 | 4
5 | 1 0 1 1 | 3
6 | 1 1 1 1 | 4
7 | 0 1 1 1 | 3
8 | 1 1 0 1 | 3
-----------+--------------------------------------------+----------
Total | 7 7 6 8 | 28
Due to random instrument failure one observation on each of four subjects is missing.
If we were to run this as a traditional repeated measures anova we would have to
drop all of the data for subjects 3, 5, 7 and 8. By running the analysis using
xtmixed we can retain all of the observations.
xtmixed y a##b || s:, var reml
Performing EM optimization:
Performing gradient-based optimization:
Iteration 0: log restricted-likelihood = -31.286348
Iteration 1: log restricted-likelihood = -31.28616
Iteration 2: log restricted-likelihood = -31.28616
Computing standard errors:
Mixed-effects REML regression Number of obs = 28
Group variable: s Number of groups = 8
Obs per group: min = 3
avg = 3.5
max = 4
Wald chi2(7) = 224.12
Log restricted-likelihood = -31.28616 Prob > chi2 = 0.0000
------------------------------------------------------------------------------
y | Coef. Std. Err. z P>|z| [95% Conf. Interval]
-------------+----------------------------------------------------------------
2.a | -.4684204 .703144 -0.67 0.505 -1.846557 .9097166
|
b |
2 | .25 .5749169 0.43 0.664 -.8768165 1.376816
3 | 3.263198 .6279667 5.20 0.000 2.032406 4.49399
4 | 4.25 .5749169 7.39 0.000 3.123184 5.376816
|
a#b |
2 2 | 1.861466 .8920747 2.09 0.037 .1130314 3.6099
2 3 | .7925351 .9271515 0.85 0.393 -1.024648 2.609719
2 4 | 2.71842 .8515934 3.19 0.001 1.049328 4.387513
|
_cons | 3.75 .4638207 8.09 0.000 2.840928 4.659072
------------------------------------------------------------------------------
------------------------------------------------------------------------------
Random-effects Parameters | Estimate Std. Err. [95% Conf. Interval]
-----------------------------+------------------------------------------------
s: Identity |
sd(_cons) | .4466089 .2491581 .1496413 1.332917
-----------------------------+------------------------------------------------
sd(Residual) | .8130553 .1495124 .567013 1.165862
------------------------------------------------------------------------------
LR test vs. linear regression: chibar2(01) = 1.45 Prob >= chibar2 = 0.1139
We can checkout what is and is not significant according to xtmixed using the
contrast introduced in Stata 12.
contrast a##b
Contrasts of marginal linear predictions
Margins : asbalanced
------------------------------------------------
| df chi2 P>chi2
-------------+----------------------------------
y |
a | 1 3.88 0.0489
|
b | 3 207.84 0.0000
|
a#b | 3 11.56 0.0091
------------------------------------------------
These results imply that the interaction and both main effects are statistically
significant. However, there are only four subjects nested in each level of variable b.
If there were no missing observations across time a repeated measures anova be our best bet.
But since there are missing observations we will rescale the chi-square values to F-ratios
and try to estimate the denominator degrees of freedom that can used with the F-distribution. The way we will do this is to run anova to obtain the between and within degrees of freedom. Although we are running anova we won't look at the anova results but only at the degrees of freedom. We also won't bother with the repeated option for anova. We will boldface the degrees of freedom of interest.
anova y a / s|a b a#b
Number of obs = 28 R-squared = 0.9446
Root MSE = .825177 Adj R-squared = 0.8932
Source | Partial SS df MS F Prob > F
-----------+----------------------------------------------------
Model | 162.574315 13 12.5057165 18.37 0.0000
|
a | 5.12395138 1 5.12395138 3.86 0.0971
s|a | 7.96717172 6 1.32786195
-----------+----------------------------------------------------
b | 136.083668 3 45.3612228 66.62 0.0000
a#b | 7.95033498 3 2.65011166 3.89 0.0325
|
Residual | 9.53282828 14 .680916306
-----------+----------------------------------------------------
Total | 172.107143 27 6.37433862
You didn't look at the F-ratios, did you? Just look at the two bolded degrees of
freedom.So now we know that the denominator degrees of freedom are 6 and 14. We can now rescale the chi-square vales for xtmixed as F-ratios and obtain p-values.
First the a#b interaction.
The main effect for b has the same denominator degrees of freedom as the interaction.chi-square = 11.56 df = 3 F = 11.56/3 = 3.8533333 df = 3 & 14 p-value = Ftail(3,14,3.8533333) = .03348207
Finally, the a main effect which has six degrees of freedom in the denominator.chi-square = 207.84 df = 3 F = 207.84/3 = 69.28 df = 3 & 14 p-value = Ftail(3,14,69.28) = 1.218e-08
While the conclusions for b and a#b do not change the F-ratio for the a main effect is not significant even thought the xtmixed chi-square suggested that it was.chi-square = 3.88 df = 3 F = 3.88/1 = 1 df = 1 & 6 p-value = Ftail(1,6,3.88) = .09638074
xtmixed & denominator degrees of freedom: myth or magic -- 2011 Chicago Stata Conference
16 Oct 2011
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