From 83ced2d4cee2e46fe8d47e3e192b34efaa37bf0f Mon Sep 17 00:00:00 2001
From: Julian T <julian@jtle.dk>
Date: Tue, 16 Mar 2021 12:18:57 +0100
Subject: Added some assignments for prob

---
 sem6/prob/stat4/Opgaver.ipynb | 230 ++++++++++++++++++++++++++++++++++++++++++
 1 file changed, 230 insertions(+)
 create mode 100644 sem6/prob/stat4/Opgaver.ipynb

(limited to 'sem6/prob/stat4/Opgaver.ipynb')

diff --git a/sem6/prob/stat4/Opgaver.ipynb b/sem6/prob/stat4/Opgaver.ipynb
new file mode 100644
index 0000000..441d171
--- /dev/null
+++ b/sem6/prob/stat4/Opgaver.ipynb
@@ -0,0 +1,230 @@
+{
+ "cells": [
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "# Opgaver til stat4"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 34,
+   "metadata": {},
+   "outputs": [],
+   "source": [
+    "import numpy as np\n",
+    "from scipy.stats import norm\n",
+    "from IPython.display import display, Markdown"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Problem 1\n"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 9,
+   "metadata": {},
+   "outputs": [],
+   "source": [
+    "mu = 8.20\n",
+    "sigma = 0.02\n",
+    "samples = np.array([8.18, 8.17, 8.16, 8.15, 8.17, 8.21, 8.22, 8.16, 8.19, 8.18])\n",
+    "n = len(samples)"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "### Part A\n",
+    "\n",
+    "What conclusion can be made if $\\alpha = 0.10$ level of significance.\n",
+    "\n",
+    "We let our $H_0: \\mu = 8.20$.\n",
+    "We can conclude the probability of a *type 1* error, thus a false positive.\n",
+    "\n",
+    "We can calculate the stats \n",
+    "$$\n",
+    "v = \\frac {\\sqrt{n}} \\sigma | \\bar{X} - \\mu|\n",
+    "$$\n",
+    "then the *p-value* is the probability that the $H_0$ is falsely rejected when observing $v$.\n",
+    "If the *p-value* is less that $\\alpha$ then $H_0$ is rejected.\n",
+    "\n",
+    "The *p-value* can be calculated by finding the area outside $v$.\n",
+    "$$\n",
+    "p = 2 \\cdot P(Z > v) = 2 \\cdot (1 - \\Phi(V))\n",
+    "$$"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 37,
+   "metadata": {},
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "3.320391543176363\n",
+      "0.0008989127881156023 0.1 False\n"
+     ]
+    },
+    {
+     "data": {
+      "text/markdown": [
+       "Thus $H_0$ is rejected"
+      ],
+      "text/plain": [
+       "<IPython.core.display.Markdown object>"
+      ]
+     },
+     "metadata": {},
+     "output_type": "display_data"
+    }
+   ],
+   "source": [
+    "x_hat = np.mean(samples)\n",
+    "\n",
+    "def assign(alpha):\n",
+    "    v = np.sqrt(n) / sigma * np.abs(x_hat - mu)\n",
+    "    print(v)\n",
+    "    p_val = 2 * (1 - norm.cdf(v))\n",
+    "    print(p_val, alpha, p_val >= alpha)\n",
+    "    if p_val >= alpha:\n",
+    "        display(Markdown(\"Thus H_0 is accepted\"))\n",
+    "    else:\n",
+    "        display(Markdown(\"Thus $H_0$ is rejected\"))\n",
+    "        \n",
+    "assign(0.1)"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 39,
+   "metadata": {},
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "3.320391543176363\n",
+      "0.0008989127881156023 0.05 False\n"
+     ]
+    },
+    {
+     "data": {
+      "text/markdown": [
+       "Thus $H_0$ is rejected"
+      ],
+      "text/plain": [
+       "<IPython.core.display.Markdown object>"
+      ]
+     },
+     "metadata": {},
+     "output_type": "display_data"
+    }
+   ],
+   "source": [
+    "# Part b\n",
+    "\n",
+    "assign(0.05)"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Problem 2"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "### Part A\n",
+    "\n",
+    "We want a test where we handle the case of *type II* error where $H_0$ should be rejected with a probability of $0.95$ or larger if $|\\mu - 8.20| \\geq 0.03$.\n"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 42,
+   "metadata": {},
+   "outputs": [],
+   "source": [
+    "sigma_reject = 0.03"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "### Part B\n",
+    "\n",
+    "The $n$ required can be found with the following formula\n",
+    "$$\n",
+    "n \\approx \\frac {(z_{\\alpha / 2} + z_\\beta)^2 \\sigma^2} {(\\mu_s - \\mu)^2} \\,,\n",
+    "$$\n",
+    "where $n$ is the number of required samples to have $H_0$ rejected with probability $\\beta$ when the real mean is $\\mu_s$."
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 49,
+   "metadata": {},
+   "outputs": [
+    {
+     "data": {
+      "text/markdown": [
+       "Thus $n$ must be $6.0$"
+      ],
+      "text/plain": [
+       "<IPython.core.display.Markdown object>"
+      ]
+     },
+     "metadata": {},
+     "output_type": "display_data"
+    }
+   ],
+   "source": [
+    "beta = 0.05\n",
+    "alpha = 0.05\n",
+    "\n",
+    "def the_z_thing(p):\n",
+    "    return norm.ppf(1 - p)\n",
+    "\n",
+    "z_alpha2 = the_z_thing(alpha / 2)\n",
+    "z_beta = the_z_thing(beta)\n",
+    "\n",
+    "n = ((z_alpha2 + z_beta)**2 * sigma**2) / ((mu + sigma_reject - mu)**2)\n",
+    "display(Markdown(f\"Thus $n$ must be ${np.ceil(n)}$\"))"
+   ]
+  }
+ ],
+ "metadata": {
+  "kernelspec": {
+   "display_name": "Python 3",
+   "language": "python",
+   "name": "python3"
+  },
+  "language_info": {
+   "codemirror_mode": {
+    "name": "ipython",
+    "version": 3
+   },
+   "file_extension": ".py",
+   "mimetype": "text/x-python",
+   "name": "python",
+   "nbconvert_exporter": "python",
+   "pygments_lexer": "ipython3",
+   "version": "3.9.2"
+  }
+ },
+ "nbformat": 4,
+ "nbformat_minor": 4
+}
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