In [1]:

```
import numpy as np
```

In [3]:

```
treat = 330.8
treat_se = 99.7
placebo = 188.1
placebo_se = 55.5
n = 15
```

Formula for SE of difference:

$$SE = \sqrt{\frac{SD_T^2}{n_T} + \frac{SD_C^2}{n_C}}$$

and $SE = SD/\sqrt{n}$, so for this example:

$$SE = \sqrt{SE_T^2 + SE_C^2}$$

In [6]:

```
SE_diff = np.sqrt(99.7**2 + 55.5**2)
SE_diff
```

Out[6]:

114.10670444807351

Hence, a 95% confidence interval for the difference would be:

In [7]:

```
d = treat-placebo
d - 2*SE_diff, d + 2*SE_diff
```

Out[7]:

(-85.513408896147013, 370.91340889614708)

Which, given n=15 and the variability, is pretty unccertain.