You can still use a DiD-like regression to calculate the four group $
\times$ time shares. These are random variables, with standard errors and covariances, and you will need to create a nonlinear combination of them to convert to percentage points:
$$
\begin{array}{rcl}
\frac{\bar{Y}(g=A,t=1) - \bar{Y}(g=A,t=0)}{\bar{Y}(g=A,t=0)} - \frac{\bar{Y}(g=B,t=1) - \bar{Y}(g=B,t=0)}{\bar{Y}(g=B,t=0)} & = & \frac{0.15 - 0.1}{0.1} - \frac{0.55 - 0.5}{0.5} \\[10pt]
& = & 0.5 - 0.1 \\[10pt]
& = & 0.4
\end{array}
$$
You will then need to apply the delta method to calculate the variances of this difference of percent changes. Here's how to do this in Stata:
. input str1 group int(t yes noobs)
group t yes noobs
1. "A" 0 0 900
2. "A" 0 1 100
3. "A" 1 0 850
4. "A" 1 1 150
5. "B" 0 0 50
6. "B" 0 1 50
7. "B" 1 0 45
8. "B" 1 1 55
9. end
. sencode group, replace
. regress yes i.group##i.t [fweight=noobs], vce(robust)
Linear regression Number of obs = 2,200
F(3, 2196) = 45.97
Prob > F = 0.0000
R-squared = 0.1023
Root MSE = .34885
------------------------------------------------------------------------------
| Robust
yes | Coefficient std. err. t P>|t| [95% conf. interval]
-------------+----------------------------------------------------------------
group |
B | .4 .0509384 7.85 0.000 .3001076 .4998924
1.t | .05 .0147613 3.39 0.001 .0210524 .0789476
|
group#t |
B#1 | -3.10e-16 .0721246 -0.00 1.000 -.1414396 .1414396
|
_cons | .1 .0094955 10.53 0.000 .081379 .118621
------------------------------------------------------------------------------
. margins group#t, post //coeflegend
Adjusted predictions Number of obs = 2,200
Model VCE: Robust
Expression: Linear prediction, predict()
------------------------------------------------------------------------------
| Delta-method
| Margin std. err. t P>|t| [95% conf. interval]
-------------+----------------------------------------------------------------
group#t |
A#0 | .1 .0094955 10.53 0.000 .081379 .118621
A#1 | .15 .0113019 13.27 0.000 .1278365 .1721635
B#0 | .5 .0500455 9.99 0.000 .4018585 .5981415
B#1 | .55 .0497947 11.05 0.000 .4523504 .6476496
------------------------------------------------------------------------------
. nlcom ///
> (pct_change_A:(_b[1.group#1.t] - _b[1.group#0.t])/_b[1.group#0.t]) ///
> (pct_change_B:(_b[2.group#1.t] - _b[2.group#0.t])/_b[2.group#0.t]), post //coeflegend
pct_change_A: (_b[1.group#1.t] - _b[1.group#0.t])/_b[1.group#0.t]
pct_change_B: (_b[2.group#1.t] - _b[2.group#0.t])/_b[2.group#0.t]
------------------------------------------------------------------------------
| Coefficient Std. err. z P>|z| [95% conf. interval]
-------------+----------------------------------------------------------------
pct_change_A | .5 .1818244 2.75 0.006 .1436307 .8563693
pct_change_B | .1 .148459 0.67 0.501 -.1909743 .3909743
------------------------------------------------------------------------------
. nlcom ///
> (A_vs_B_pp_change:_b[pct_change_A] - _b[pct_change_B]), post //coeflegend
A_vs_B_pp_~e: _b[pct_change_A] - _b[pct_change_B]
----------------------------------------------------------------------------------
| Coefficient Std. err. z P>|z| [95% conf. interval]
-----------------+----------------------------------------------------------------
A_vs_B_pp_change | .4 .2347343 1.70 0.088 -.0600707 .8600707
----------------------------------------------------------------------------------
. test A_vs_B_pp_change = 0
( 1) A_vs_B_pp_change = 0
chi2( 1) = 2.90
Prob > chi2 = 0.0884
The last command tests the null that the observed percentage point difference equals zero. The p-value is 8.8%, so we would fail to reject the null under the usual significance threshold of 5%. In other words, the observed difference could just be noise.
The above assumes that you are drawing new samples at each time rather than following the same users over time. If that is incorrect, you must cluster the standard errors by user in the regression. This will most likely raise the p-value since you have less data when you follow the same people over time. This requires a bit more than the summary statistics for me to do on your behalf, since I would need to know the number of transitions in each group:
group transition noobs
A 0->0 ?
A 0->1 ?
A 1->1 ?
A 1->0 ?
B 0->0 ?
B 0->1 ?
B 1->1 ?
B 1->0 ?
