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I have conducted a controlled interrupted time series using Linden's formulation, where

$Y_t = β_0 + β_1T + β_2X_t + β_3TX_t + β_4Z + β_5ZT + β_6ZX_t + β_7ZX_tT$

$β_0...β_3$ represent the control (untreated) group and $β_4...β_7$ represent the difference between the treatment and control groups.

From this model, is it possible to calculate the values that represent the treatment and control groups separately?

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Let's assume you have two groups: Group 1 and Group 2. Then the variable Z in your model is defined as follows:

Z = 1 for Group 2  and Z = 0 for Group 1. 

For example, Group 2 could be California while Group 1 could be other states, as per the example used in Figure 2 in the paper. The intervention alluded to below would be the introduction of Proposition 99.

Now, all you have to do is to set the value of Z to 0 and 1, respectively, in your model.

For Group 1, setting Z to 0 yields the following model equation:

$Y_t = β_0 + β_1T + β_2X_t + β_3TX_t$.

For Group 2, setting Z to 1 yields the following model equation (after rearranging the terms):

$Y_t = (β_0 + β_4) + (β_1 + β_5)T + (β_2 +β_6)X_t + (β_3 + β_7)TX_t$.

In the above, $β_0$ and $β_0 + β_4$ represent starting levels of the outcome variable $Y$ in Group 1 and Group 2, respectively.

$β_1$ is the slope, or trajectory of the outcome variable $Y$ in Group 1 until the introduction of the intervention. On the other hand, $β_1 + β_5$ is the slope, or trajectory of the outcome variable $Y$ in Group 2 until the introduction of the intervention.

$β_2$ represents the starting level of the outcome variable $Y$ in Group 1 at the time of introduction of the intervention and indicates whether there was a change in the level of the outcome variable $Y$ in that group immediately following the introduction of the intervention. In contrast, $β_2 + β_6$ denotes the starting level of the outcome variable $Y$ in Group 2 at the time of introduction of the intervention.

Finally, $β_3$ represents the change in slope or trajectory of the outcome variable $Y$ in Group 1 after the introduction of the intervention until the end of the study, while $β_3 + β_7$ represents the change in slope or trajectory of the outcome variable $Y$ in Group 2 after the introduction of the intervention until the end of the study.

Addendum:

Let $b_0$, $b_2$, $b_3$, $b_4$, $b_5$, $b_6$ and $b_7$ be the estimated values of the parameters corresponding to your model.

The estimated standard errors for $b_0$, $b_1$, $b_2$ and $b_3$ (which refer to Group 1) should be reported directly in the model summary produced by your software.

For Group_2, estimated standard errors can be derived from the estimated variance-covariance matrix $V$ of $b_0$, $b_2$, $b_3$, $b_4$, $b_5$, $b_6$ and $b_7$. The $V$ matrix should be an 8x8 matrix. On its main diagonal, $V$ will contain the estimated variances of $b_0$, $b_1$, $b_2$, $b_3$, $b_4$, $b_5$, $b_6$ and $b_7$. Off the main diagonal, $V$ will contain the estimated covariances between pairs of these items.

To compute the standard error of $b_0 + b_4$ (i.e., the intercept for Group 2), for example, recall that:

$Var(b_0 + b_4) = Var(b_0) + Var(b_4) + 2Cov(b_0, b_4)$ 

Estimated values of the variances and covariance on the right-hand side of the above formula can be extracted from the $V$ matrix and plugged into the formula. Taking the square root of the resulting estimated value of $Var(b_0 + b_4)$ will yield the estimated standard error of $b_0 + b_4$.

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  • $\begingroup$ So, it really is just a matter of simple arithmetic? e.g., the intercept for the control group (b0) + the difference between the treatment and control groups (b4) = the intercept for the treatment group $\endgroup$ – C_H May 25 '18 at 4:12
  • $\begingroup$ Yes, you can just follow the arithmetic. $\endgroup$ – Isabella Ghement May 25 '18 at 14:56
  • $\begingroup$ I forgot to ask - are you able to comment on how to calculate appropriate standard errors? I would like to calculate CIs $\endgroup$ – C_H May 28 '18 at 9:34
  • $\begingroup$ @C_H: See the addendum to my original answer. $\endgroup$ – Isabella Ghement May 28 '18 at 13:47

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