{
 "cells": [
  {
   "cell_type": "code",
   "execution_count": 10,
   "metadata": {},
   "outputs": [],
   "source": [
    "import numpy as np\n",
    "from scipy.stats import norm\n",
    "from IPython.display import display, Math, Markdown\n"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "# Opgaver\n",
    "\n",
    "Har lavet problem 1 på papir, men vil lave resten i python da det nok er lidt lettere."
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Problem 2\n",
    "\n",
    "Går ud fra at PCB er i ppm, og kalder den $\\theta$.\n",
    "\n",
    "$$\n",
    "X_i = \\theta + W_i\n",
    "$$\n",
    "hvor $W_i \\sim \\mathcal{N}(\\mu, \\sigma^2)$, $\\sigma = 0.08$.\n",
    "Her går jeg ud fra at $\\mu = 0$.\n",
    "\n",
    "Derfor er: \n",
    "$$\n",
    "E[X] = \\theta\n",
    "$$\n",
    "og\n",
    "$$\n",
    "Var[X] = (0.08)^2\n",
    "$$\n",
    "\n",
    "Kan sige at confidence level er:\n",
    "$$\n",
    "[\\bar{X} - z_{\\frac \\alpha 2} \\cdot \\frac \\sigma {\\sqrt{n}}, \\bar{X} + z_{\\frac \\alpha 2} \\cdot \\frac \\sigma {\\sqrt{n}}]\n",
    "$$\n",
    "with probability $1-\\alpha$.\n",
    "Her er \n",
    "$$\n",
    "z_p = \\Phi^{-1}(1 - p)$$"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 11,
   "metadata": {},
   "outputs": [],
   "source": [
    "samples = [11.2, 12.4, 10.8, 11.6, 12.5, 10.1, \n",
    "           11.0, 12.2, 12.4, 10.6]\n",
    "n = len(samples)\n",
    "\n",
    "X_bar = np.mean(samples)\n",
    "sigma = 0.08"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 15,
   "metadata": {},
   "outputs": [
    {
     "data": {
      "text/latex": [
       "$\\displaystyle P(11.430416397415636 \\leq \\theta \\leq 11.529583602584365) = 0.95$"
      ],
      "text/plain": [
       "<IPython.core.display.Math object>"
      ]
     },
     "metadata": {},
     "output_type": "display_data"
    }
   ],
   "source": [
    "# Del A\n",
    "alpha = 0.05\n",
    "Z = norm.ppf(1 - alpha/2)\n",
    "\n",
    "dist = Z * sigma / np.sqrt(n)\n",
    "lower = X_bar - dist\n",
    "upper = X_bar + dist\n",
    "\n",
    "display(Math(f\"P({lower} \\\\leq \\\\theta \\\\leq {upper}) = {1 - alpha}\"))"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 18,
   "metadata": {},
   "outputs": [
    {
     "data": {
      "text/latex": [
       "$\\displaystyle [-\\infty, 11.529583602584365]$"
      ],
      "text/plain": [
       "<IPython.core.display.Math object>"
      ]
     },
     "metadata": {},
     "output_type": "display_data"
    }
   ],
   "source": [
    "\n",
    "# Okay så lower confidence level er åbenbart at man går fra -infty til upper limit\n",
    "\n",
    "display(Math(f\"[-\\\\infty, {upper}]\"))\n"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 21,
   "metadata": {},
   "outputs": [
    {
     "data": {
      "text/latex": [
       "$\\displaystyle [11.430416397415636, \\infty]$"
      ],
      "text/plain": [
       "<IPython.core.display.Math object>"
      ]
     },
     "metadata": {},
     "output_type": "display_data"
    }
   ],
   "source": [
    "# Del C\n",
    "# Og igen for uppwer\n",
    "\n",
    "display(Math(f\"[{lower}, \\\\infty]\"))"
   ]
  }
 ],
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