Dear Michael,

My previous response was inaccurate: First, linearHypothesis() *is* able to accommodate aliased coefficients by setting the argument singular.ok = TRUE:

> linearHypothesis(minimal_model, "bt2 + csent + bt2:csent = 0",
+                  singular.ok=TRUE)

Linear hypothesis test:
bt2  + csent  + bt2:csent = 0

Model 1: restricted model
Model 2: a ~ b * c

  Res.Df    RSS Df Sum of Sq      F Pr(>F)
1     16 9392.1
2     15 9266.4  1    125.67 0.2034 0.6584

Moreover, when there is an empty cell, this F-test is (for a reason that I haven't worked out, but is almost surely due to how the rank-deficient model is parametrized) *not* equivalent to the t-test for the corresponding coefficient in the raveled version of the two factors:

> df$bc <- factor(with(df, paste(b, c, sep=":")))
> m <- lm(a ~ bc, data=df)
> summary(m)

Call:
lm(formula = a ~ bc, data = df)

Residuals:
    Min      1Q  Median      3Q     Max
-57.455 -11.750   0.439  14.011  37.545

Coefficients:
            Estimate Std. Error t value Pr(>|t|)
(Intercept)    20.50      17.57   1.166   0.2617
bct1:unsent    37.50      24.85   1.509   0.1521
bct2:other     32.00      24.85   1.287   0.2174
bct2:sent 17.17 22.69 0.757 0.4610 <<< cf. F = 0.2034, p = 0.6584
bct2:unsent    38.95      19.11   2.039   0.0595

Residual standard error: 24.85 on 15 degrees of freedom
Multiple R-squared:  0.2613,    Adjusted R-squared:  0.06437
F-statistic: 1.327 on 4 and 15 DF,  p-value: 0.3052

In the full-rank case, however, what I said is correct -- that is, the F-test for the 1 df hypothesis on the three coefficients is equivalent to the t-test for the corresponding coefficient when the two factors are raveled:

> linearHypothesis(minimal_model_fixed, "bt2 + csent + bt2:csent = 0")

Linear hypothesis test:
bt2  + csent  + bt2:csent = 0

Model 1: restricted model
Model 2: a ~ b * c

  Res.Df    RSS Df Sum of Sq      F Pr(>F)
1     15 9714.5
2     14 9194.4  1    520.08 0.7919 0.3886

> df_fixed$bc <- factor(with(df_fixed, paste(b, c, sep=":")))
> m <- lm(a ~ bc, data=df_fixed)
> summary(m)

Call:
lm(formula = a ~ bc, data = df_fixed)

Residuals:
    Min      1Q  Median      3Q     Max
-57.455 -11.750   0.167  14.011  37.545

Coefficients:
            Estimate Std. Error t value Pr(>|t|)
(Intercept)   64.000     25.627   2.497   0.0256
bct1:sent    -43.500     31.387  -1.386   0.1874
bct1:unsent  -12.000     36.242  -0.331   0.7455
bct2:other   -11.500     31.387  -0.366   0.7195
bct2:sent    -26.333     29.591  -0.890   0.3886 << cf.
bct2:unsent   -4.545     26.767  -0.170   0.8676

Residual standard error: 25.63 on 14 degrees of freedom
Multiple R-squared:  0.2671,    Adjusted R-squared:  0.005328
F-statistic:  1.02 on 5 and 14 DF,  p-value: 0.4425

So, to summarize:

(1) You can use linearHypothesis() with singular.ok=TRUE to test the hypothesis that you specified, though I suspect that this hypothesis probably isn't testing what you think in the rank-deficient case. I suspect that the hypothesis that you want to test is obtained by raveling the two factors.

(2) There is no reason to use deltaMethod() for a linear hypothesis, but there is also no intrinsic reason that deltaMethod() shouldn't be able to handle a rank-deficient model. We'll probably fix that.

My apologies for the confusion,
 John

--
John Fox, Professor Emeritus
McMaster University
Hamilton, Ontario, Canada
web: https://www.john-fox.ca/

On 2023-09-26 9:49 a.m., John Fox wrote:
Caution: External email.


Dear Michael,

You're testing a linear hypothesis, so there's no need to use the delta
method, but the linearHypothesis() function in the car package also
fails in your case:

 > linearHypothesis(minimal_model, "bt2 + csent + bt2:csent = 0")
Error in linearHypothesis.lm(minimal_model, "bt2 + csent + bt2:csent = 0") :
there are aliased coefficients in the model.

One work-around is to ravel the two factors into a single factor with 5
levels:

 > df$bc <- factor(with(df, paste(b, c, sep=":")))
 > df$bc
  [1] t2:unsent t2:unsent t2:unsent t2:unsent t2:sent   t2:unsent
  [7] t2:unsent t1:sent   t2:unsent t2:unsent t2:other  t2:unsent
[13] t1:unsent t1:sent   t2:unsent t2:other  t1:unsent t2:sent
[19] t2:sent   t2:unsent
Levels: t1:sent t1:unsent t2:other t2:sent t2:unsent

 > m <- lm(a ~ bc, data=df)
 > summary(m)

Call:
lm(formula = a ~ bc, data = df)

Residuals:
     Min      1Q  Median      3Q     Max
-57.455 -11.750   0.439  14.011  37.545

Coefficients:
             Estimate Std. Error t value Pr(>|t|)
(Intercept)    20.50      17.57   1.166   0.2617
bct1:unsent    37.50      24.85   1.509   0.1521
bct2:other     32.00      24.85   1.287   0.2174
bct2:sent      17.17      22.69   0.757   0.4610
bct2:unsent    38.95      19.11   2.039   0.0595

Residual standard error: 24.85 on 15 degrees of freedom
Multiple R-squared:  0.2613,    Adjusted R-squared:  0.06437
F-statistic: 1.327 on 4 and 15 DF,  p-value: 0.3052

Then the hypothesis is tested directly by the t-value for the
coefficient bct2:sent.

I hope that this helps,
  John

--
John Fox, Professor Emeritus
McMaster University
Hamilton, Ontario, Canada
web: https://www.john-fox.ca/

On 2023-09-26 1:12 a.m., Michael Cohn wrote:
Caution: External email.


I'm running a linear regression with two categorical predictors and their
interaction. One combination of levels does not occur in the data, and as
expected, no parameter is estimated for it. I now want to significance test a particular combination of levels that does occur in the data (ie, I want
to get a confidence interval for the total prediction at given levels of
each variable).

In the past I've done this using car::deltaMethod() but in this dataset
that does not work, as shown in the example below: The regression model
gives the expected output, but deltaMethod() gives this error:

error in t(gd) %*% vcov. : non-conformable arguments

I believe this is because there is no parameter estimate for when the
predictors have the values 't1' and 'other'. In the df_fixed dataframe,
putting one person into that combination of categories causes deltaMethod()
to work as expected.

I don't know of any theoretical reason that missing one interaction
parameter estimate should prevent getting a confidence interval for a
different combination of predictors. Is there a way to use deltaMethod() or
some other function to do this without changing my data?

Thank you,

- Michael Cohn
Vote Rev (http://voterev.org)


Demonstration:
------

library(car)
# create dataset with outcome and two categorical predictors
outcomes <- c(91,2,60,53,38,78,48,33,97,41,64,84,64,8,66,41,52,18,57,34)
persontype <-
c("t2","t2","t2","t2","t2","t2","t2","t1","t2","t2","t2","t2","t1","t1","t2","t2","t1","t2","t2","t2")
arm_letter <-
c("unsent","unsent","unsent","unsent","sent","unsent","unsent","sent","unsent","unsent","other","unsent","unsent","sent","unsent","other","unsent","sent","sent","unsent")
df <- data.frame(a = outcomes, b=persontype, c=arm_letter)

# note: there are no records with the combination 't1' + 'other'
table(df$b,df$c)


#regression works as expected
minimal_formula <- formula("a ~ b*c")
minimal_model <- lm(minimal_formula, data=df)
summary(minimal_model)

#use deltaMethod() to get a prediction for individuals with the combination
'b2' and 'sent'
# deltaMethod() fails with "error in t(gd) %*% vcov. : non-conformable
arguments."
deltaMethod(minimal_model, "bt2 + csent + `bt2:csent`", rhs=0)

# duplicate the dataset and change one record to be in the previously empty
cell
df_fixed <- df
df_fixed[c(13),"c"] <- 'other'
table(df_fixed$b,df_fixed$c)

#deltaMethod() now works
minimal_model_fixed <- lm(minimal_formula, data=df_fixed)
deltaMethod(minimal_model_fixed, "bt2 + csent + `bt2:csent`", rhs=0)

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