Sometimes a great deal of background information is available about participants in experiments, so that a great many variables might possibly be used as covariates while testing the effect of some crucial independent variable. How many covariates should actually be used? This note suggests an answer for the case in which participants are randomly assigned to experimental treatments. First we briefly consider the case without random assignment.

Strategies like this have a very serious flaw in nonrandomized experiments. The very fact that a covariate correlates highly with the independent variable will tend to make both variables nonsignicant due to collinearity. But under the proposed strategy, the covariate will then be dropped. Thus the strategy drops precisely the covariates that are most important to keep--those that correlate highly with the independent variable. When an independent variable and covariate correlate highly and both are nonsignificant, the correct conclusion is that the collinearity has made it impossible to disentangle their effects, but the flawed strategy will resolve the uncertainty in favor of the independent variable. Thus if the substantive nature of a variable suggests that it should be used as a covariate, then it should not be dropped because of nonsignificance.

Specifically, the suggestion is to predict the dependent variable *Y* from the covariates alone,
first omitting *X* from the regression. Use the backward deletion procedure mentioned above.
That is, delete the least significant covariate from the regression, recompute the regression,
again delete the least significant covariate, and keep repeating the process until all the
*t* values for individual regressors are 1.42 or above, in absolute value. This
corresponds closely to a familiar rule that the two-tailed signicance levels should all be .15 or
below. However, advocates of that rule sometimes suggest that a different rule should be
used if high collinearity is observed, and I suggest using a cutting point of 1.42 regardless of
collinearity.

By Monte Carlo experiments, I have found that it is invalid to have the independent
variable in a regression while you are trying to determine which covariates to delete--because
then the covariates that happen, by chance, to correlate highly with *X* will tend to be deleted,
while those are precisely the ones it is important to leave in. Therefore we must consider
regressions predicting the dependent variable Y from the covariates alone. Once we have
used those regressions to determine which covariates to include, then we can add *X* to the
model.

Although we are working temporarily with regressions excluding *X*, we can
nevertheless estimate what Tol(X) would be if *X* were in the model. That's because random
assignment assures that RX is zero in the population. When a true *R* is zero, then
E(R^{2}) = P/(N-1). Therefore E(Tol(X)) = E(1-RX^{2})
= 1 - E(RX^{2}) = (N-P-1)/(N-1). Therefore we
can say that the expected tolerance for a randomly-assigned treatment variable is (N-P-1)/(N-1), where
*P* is the number of covariates. Since N-P-1 = df, where df denotes the residual
degrees of freedom in the regression, we can write the expected tolerance as df/(N-1).

If we then drop out *k* covariates, the numerator of this fraction increases
by *k*, so the expected tolerance in the new regression is (df+*k*)/(N-1),
where df is df for the original (larger) regression. Therefore the ratio of the two expected
tolerance values (before and after the deletion) is df/(df+*k*).

We want to determine under what conditions the estimated standard error of b(X) would be exactly the same in these two regressions. Then on one side of that line, the deletion would increase the standard error, and on the other side it would decease it. Therefore we will imagine for the moment that the two standard errors are exactly equal, and see what conditions are associated with that equality.

As described earlier, equality of the two standard errors means that MSE/Tol(X) is
the same in the two regressions. But since we expect the two Tol(X) values to be in the
ratio df/(df+*k*), that means the two MSE values must be in the same ratio. But
in any regression, MSE = SSE/df, where SSE is the sum of squared residuals. Let SSE and
SSE' refer to the larger and smaller regressions respectively. Then MSE for the larger
regression is SSE/df while MSE for the smaller regression is SSE'/(df+*k*), where
again df pertains to the larger regression. If we set the ratio of these two values to the
aforementioned value df/(df+*k*), we have

(SSE/SSE')*(df+*k*)/df = df/(df+*k*), so

SSE/SSE' = [df/(df+*k*)]^{2}

so SSE'/SSE = [(df+*k*)/df]^{2}

If we use the ordinary *F* test to test the significance of the regressors deleted from the larger
regression, we have

F = (SSE' - SSE)/SSE * df/*k* = (SSE'/SSE - 1)*df/*k*

Substituting the previous relation into this, we have

F = ([(df+*k*)/df]^{2} - 1)*df/*k* = (*k*^{2} +
2df*k)/(df**k*) = 2 + *k*/df

Thus to decide whether to delete some set of *k* covariates, the worker can
compute *F* and make the deletion if F < 2 + *k*/df.

Several considerations make it reasonable to ignore the *k*/df term. First,
*k* is most often 1 since we are deleting one covariate at a time--and even when
not 1, *k* is usually fairly small so that *k*/df is trivial in comparison to
2. *k*/df would rarely exceed .1, so the proper *F* is nearly always between 2 and
2.1. Second, in my own experience, deleting variables one at a time usually produces fairly
large jumps in the smallest remaining *F*. If *F* = 1.5 for one variable, then deleting it might
well yield 2.5 as the smallest *F* in the recomputed regression. Thus working with an exact
critical F is not very important. Third, when the difference between the two standard errors
of b(X) is plotted as a function of *F*, that difference is near 0 for a broad range of *F*-values around the exact value of 2 + *k*/df, so again hitting it exactly is not very
important. Thus for simplicity I suggest using a critical *F* of exactly 2, to avoid having to
recompute *k*/df after each deletion. Or if *k* = 1 and *t* =
sqrt(*F*) is being used, I suggest a *t* of 1.42, which is just above sqrt(2) =
1.414.