Update 2014-01-15
I realize that I didn't answer Danica's original question about whether the margin of error for the indirectly adjusted proportion disabled would be larger or smaller than the margin of error for the same rate in ACS. The answer is: if the company category proportions do not differ drastically from the state ACS proportions, the margin of error given below will be smaller than the ACS margin of error.
The reason: the indirect rate treats organization job category person counts (or relative proportions) as fixed numbers. The ACS estimate of proportion disabled requires, in effect, an estimate of those proportions, and the margins of error will increase to reflect this.
To illustrate, write the disabled rate as:
$$
\hat{P}_{adj} = \sum \dfrac{n_i}{n} \hat{p_i} \\
$$
where $\hat{p}_i$ is the estimated disabled rate in category $i$ in the ACS.
On the other hand, the ACS estimated rate is, in effect:
$$
\hat{P}_{acs} = \sum\widehat{\left(\frac{N_i}{N}\right)} \hat{p_i} $$
where $N_i$ and $N$ are respectively the population category and overall totals and $N_i/N$ is the
population proportion in category $i$.
Thus, the standard error for the ACS rate will be larger because of the need to estimate $N_i/N$ in addition to $p_i$.
If the organization category proportions and population estimated proportions differ greatly, then the it is possible that $SE( \hat{P}_{adj} )>SE( \hat{P}_{acs} )$. In a two-category example that I constructed, the categories were represented in proportions $N_1/N= 0.7345$ and $N_2/N= 0.2655$. The standard error for the estimated proportion disabled
was $SE( \hat{P}_{acs} ) = 0.0677$.
If I considered 0.7345 and 0.2655 to be the fixed values $n_1/n$ and $n_2/n$ (the indirect adjustment approach), $SE(\hat{P}_{adj} )=0.0375$, much smaller. If instead, $n_1/n= 0.15$ and $n_2/n =0.85$, $SE( \hat{P}_{adj} )=0.0678$, about the same as $SE( \hat{P}_{acs} )$ At the extreme $n_1/n= 0.001$ and $n_2/n =0.999$, $SE( \hat{P}_{adj} )=0.079$.
I'd be surprised if organization and population category proportions differ so drastically. If they don't, I think that it's safe to use the ACS margin of error as a conservative, possibly very conservative, estimate of the true margin of error.
Update 2014-01-14
Short answer
In my opinion, it would be irresponsible to present such a statistic without a CI or margin of error (half CI length). To compute these, you will need to download and analyze the ACS Public Use Microdata Sample (PUMS) (http://www.census.gov/acs/www/data_documentation/public_use_microdata_sample/).
Long answer
This isn't really a re-weighting of the ACS. It is a version of indirect standardization, a standard procedure in epidemiology (google or see any epi text). In this case state ACS job (category) disability rates are weighted by organization job category employee counts. This will compute an expected number of disabled people in the organization E, which can be compared to the observed number O. The usual metric for the comparison is a standardized ratio R= (O/E). (The usual term is "SMR", for "standardized mortality ratio", but here the "outcome" is disability.). R is also the ratio of the observed disability rate (O/n) and the indirectly standardized rate (E/n), where n is the number of the organization's employees.
In this case, it appears that only a CI for E or E/n will be needed, so I will start with that:
If
n_i = the organization employee count in job category i
p_i = disability rate for job category i in the ACS
Then
E = sum (n_i p_i)
The variance of E is:
var(E) = nn' V nn
where nn is the column vector of organization category counts and V is the estimated variance-covariance matrix of the ACS category disability rates.
Also, trivially, se(E) = sqrt(var(E)) and se(E/n) = se(E)/n.
and a 90% CI for E is
E ± 1.645 SE(E)
Divide by n to get the CI for E/n.
To estimate var(E) you would need to download and analyze the ACS Public Use Microdata Sample (PUMS) data (http://www.census.gov/acs/www/data_documentation/public_use_microdata_sample/).
I can only speak of the process for computing var(E) in Stata. As I don't know if that's available to you, I'll defer the details. However someone knowledgeable about the survey capabilities of R or (possibly) SAS can also provide code from the equations above.
Confidence Interval for the ratio R
Confidence intervals for R are ordinarily based on a Poisson assumption for O, but this assumption may be incorrect.
We can consider O and E to be independent, so
log R = log(O) - log(E) ->
var(log R) = var(log O) + var(log(E))
var(log(E)) can be computed as one more Stata step after the computation of var(E).
Under the Poisson independence assumption:
var(log O) ~ 1/E(O).
A program like Stata could fit, say, a negative binomial model or generalized linear model and give you a more accurate variance term.
An approximate 90% CI for log R is
log R ± 1.645 sqrt(var(log R))
and the endpoints can be exponentiated to get the CI for R.
