diff options
Diffstat (limited to 'sem6/prob')
| -rw-r--r-- | sem6/prob/stat4/Opgaver.ipynb | 230 |
1 files changed, 230 insertions, 0 deletions
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 +} |