Untitled.ipynb

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   "source": [
    "import numpy as np"
   ]
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    "treat = 330.8\n",
    "treat_se = 99.7\n",
    "\n",
    "placebo = 188.1\n",
    "placebo_se = 55.5\n",
    "\n",
    "n = 15"
   ]
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    "Formula for SE of difference:\n",
    "\n",
    "$$SE = \\sqrt{\\frac{SD_T^2}{n_T} + \\frac{SD_C^2}{n_C}}$$\n",
    "\n",
    "and $SE = SD/\\sqrt{n}$, so for this example:\n",
    "\n",
    "$$SE = \\sqrt{SE_T^2 + SE_C^2}$$"
   ]
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      "text/plain": [
       "114.10670444807351"
      ]
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     "execution_count": 6,
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   "source": [
    "SE_diff = np.sqrt(99.7**2 + 55.5**2)\n",
    "SE_diff"
   ]
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   "source": [
    "Hence, a 95% confidence interval for the difference would be:"
   ]
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    {
     "data": {
      "text/plain": [
       "(-85.513408896147013, 370.91340889614708)"
      ]
     },
     "execution_count": 7,
     "metadata": {},
     "output_type": "execute_result"
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   ],
   "source": [
    "d = treat-placebo\n",
    "d - 2*SE_diff, d + 2*SE_diff"
   ]
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   "cell_type": "markdown",
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   "source": [
    "Which, given n=15 and the variability, is pretty unccertain."
   ]
  }
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