├── LICENSE ├── README.md └── src ├── 00-firstbaby.ipynb ├── 03-cumulative-dist-funcs.ipynb ├── 04-cont-distribs.ipynb ├── 05-probability.ipynb ├── 06-ops-dists.ipynb ├── 07-hypothesis-testing.ipynb ├── 08-estimation.ipynb └── 09-correlation.ipynb /LICENSE: -------------------------------------------------------------------------------- 1 | Apache License 2 | Version 2.0, January 2004 3 | http://www.apache.org/licenses/ 4 | 5 | TERMS AND CONDITIONS FOR USE, REPRODUCTION, AND DISTRIBUTION 6 | 7 | 1. Definitions. 8 | 9 | "License" shall mean the terms and conditions for use, reproduction, 10 | and distribution as defined by Sections 1 through 9 of this document. 11 | 12 | "Licensor" shall mean the copyright owner or entity authorized by 13 | the copyright owner that is granting the License. 14 | 15 | "Legal Entity" shall mean the union of the acting entity and all 16 | other entities that control, are controlled by, or are under common 17 | control with that entity. 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The author builds up a small Python library that he uses throughout the book, almost like a Domain Specific Language (DSL). 6 | 7 | Having used scientific Python (Numpy, Scipy, Matplotlib, Pandas) for a while now, I found the need to learn the DSL a bit of a pain, and decided to just use these libraries instead. Besides, I wanted to apply these ideas to my own projects without having to carry over or rebuild the core library, but rather to continue using the tools I am used to. 8 | 9 | CAVEAT: the answers are not necessarily correct. I have tried to ensure that they are, but I am learning this stuff, so there might be errors in there, so please use at your own risk! If you do find errors please raise an issue, preferably telling me what I am doing wrong and the solution. I will update and give you credit on this page. 10 | 11 | -------------------------------------------------------------------------------- /src/05-probability.ipynb: -------------------------------------------------------------------------------- 1 | { 2 | "cells": [ 3 | { 4 | "cell_type": "markdown", 5 | "metadata": {}, 6 | "source": [ 7 | "## Probability\n", 8 | "\n", 9 | "This notebook is for Chapter 5 (Probability) of [Think Stats - Probability and Statistics for Programmers](http://greenteapress.com/thinkstats/) by Allen B Downey.\n", 10 | "\n", 11 | "Objective: Introduction to probability.\n", 12 | "\n", 13 | "__Frequentism__ - probability defined in terms of frequencies.\n", 14 | "\n", 15 | "__Bayesianism__ - probability is the degree of belief that an event will occur.\n", 16 | "\n", 17 | "### Rules of Probability\n", 18 | "\n", 19 | " P(A and B) = P(A).P(B), where A and B are independent.\n", 20 | " P(A|B) = P(A and B) / P(B), where A and B are not independent.\n" 21 | ] 22 | }, 23 | { 24 | "cell_type": "code", 25 | "execution_count": 1, 26 | "metadata": { 27 | "collapsed": true 28 | }, 29 | "outputs": [], 30 | "source": [ 31 | "from __future__ import print_function, division\n", 32 | "from scipy.misc import comb\n", 33 | "from scipy.stats import norm\n", 34 | "import math\n", 35 | "import matplotlib.pyplot as plt\n", 36 | "import numpy as np\n", 37 | "%matplotlib inline" 38 | ] 39 | }, 40 | { 41 | "cell_type": "markdown", 42 | "metadata": {}, 43 | "source": [ 44 | "__Exercise 5.1:__ If I roll 2 dice and get 8, what is the chance that one of the dice is 6?\n", 45 | "\n", 46 | "(2,6), (3,5), (4,4), (5,3), (6,2) are the only ways of getting an 8. Of these, there are 2 cases where 1 dice has a 6. Thus the answer is 2/5 = 0.4\n", 47 | "\n", 48 | "__Exercise 5.2:__ If I roll 100 die, what is the chance of getting all sixes?\n", 49 | "\n", 50 | "Chance of getting 6 on 1 die is 1/6. Since the dice throws are mutually independent events, it is (1/6)\\*\\*100, ie, 1.530646707486498e-78.\n", 51 | "\n", 52 | "__Exercise 5.3 (a):__ If a family has 2 children, what is the chance that they will be both girls?\n", 53 | "\n", 54 | "Family with 2 children can have one of (B,B), (B,G), (G,B), (G,G), so its 0.25.\n", 55 | "\n", 56 | "__Exercise 5.3 (b):__ If a family has 2 children and we know at least one of them is a girl, what is the chance that they have 2 girls?\n", 57 | "\n", 58 | "They can have (B,G) or (G,G), so its 0.5.\n", 59 | "\n", 60 | "__Exercise 5.3 (c):__ If a family has 2 children and we know the older one is a girl, what is the chance they have 2 girls?\n", 61 | "\n", 62 | "Same as above.\n", 63 | "\n", 64 | "__Exercise 5.3 (d):__ If a family has 2 children and we know that at least one of them is a girl named Florida, what is the chance they have 2 girls?\n", 65 | "\n", 66 | "Same as above.\n", 67 | "\n", 68 | "#### Monty Hall\n", 69 | "\n", 70 | "__Exercise 5.4:__ Write a program that simulates the Monty Hall problem and use it to estimate the probability if you stick or switch." 71 | ] 72 | }, 73 | { 74 | "cell_type": "code", 75 | "execution_count": 2, 76 | "metadata": {}, 77 | "outputs": [ 78 | { 79 | "name": "stdout", 80 | "output_type": "stream", 81 | "text": [ 82 | "Probability of winning if you stick: 0.320\n", 83 | "Probability of winning if you switch: 0.680\n" 84 | ] 85 | } 86 | ], 87 | "source": [ 88 | "nbr_games = 1000\n", 89 | "nbr_wins_if_stick = 0\n", 90 | "nbr_wins_if_switch = 0\n", 91 | "for i in range(nbr_games):\n", 92 | " # door with the car behind it\n", 93 | " door_with_prize = np.random.randint(3)\n", 94 | " # make first guess\n", 95 | " first_guess = np.random.randint(3)\n", 96 | " # monty opens one of the \"other\" doors\n", 97 | " doors_available = [x for x in [0, 1, 2] if x != door_with_prize\n", 98 | " and x != first_guess]\n", 99 | " door_opened = doors_available[np.random.randint(len(doors_available))]\n", 100 | " second_guess = [x for x in [0, 1, 2] if x != door_opened\n", 101 | " and x != first_guess][0]\n", 102 | " if door_with_prize == first_guess:\n", 103 | " nbr_wins_if_stick += 1\n", 104 | " if door_with_prize == second_guess:\n", 105 | " nbr_wins_if_switch += 1\n", 106 | "print(\"Probability of winning if you stick: %.3f\" % (nbr_wins_if_stick / nbr_games))\n", 107 | "print(\"Probability of winning if you switch: %.3f\" % (nbr_wins_if_switch / nbr_games))" 108 | ] 109 | }, 110 | { 111 | "cell_type": "markdown", 112 | "metadata": {}, 113 | "source": [ 114 | "__Exercise 5.5:__ It is important to realize that by deciding which door to open, Monty is giving you information. Simulate the situation where this is not the case by making Monty open door B or C at random. If he opens the door with the car, the game is over and you cannot decide whether to stick or switch." 115 | ] 116 | }, 117 | { 118 | "cell_type": "code", 119 | "execution_count": 3, 120 | "metadata": {}, 121 | "outputs": [ 122 | { 123 | "name": "stdout", 124 | "output_type": "stream", 125 | "text": [ 126 | "Probability of winning if you stick: 0.323\n", 127 | "Probability of winning if you switch: 0.327\n" 128 | ] 129 | } 130 | ], 131 | "source": [ 132 | "nbr_games = 1000\n", 133 | "nbr_wins_if_stick = 0\n", 134 | "nbr_wins_if_switch = 0\n", 135 | "for i in range(nbr_games):\n", 136 | " # door with the car behind it\n", 137 | " door_with_prize = np.random.randint(3)\n", 138 | " # make first guess\n", 139 | " first_guess = np.random.randint(3)\n", 140 | " # monty opens one of the \"other\" doors\n", 141 | " doors_available = [x for x in [0, 1, 2] if x != first_guess]\n", 142 | " door_opened = doors_available[np.random.randint(len(doors_available))]\n", 143 | " if door_opened == door_with_prize:\n", 144 | " continue\n", 145 | " second_guess = [x for x in [0, 1, 2] if x != door_opened\n", 146 | " and x != first_guess][0]\n", 147 | " if door_with_prize == first_guess:\n", 148 | " nbr_wins_if_stick += 1\n", 149 | " if door_with_prize == second_guess:\n", 150 | " nbr_wins_if_switch += 1\n", 151 | "print(\"Probability of winning if you stick: %.3f\" % (nbr_wins_if_stick / nbr_games))\n", 152 | "print(\"Probability of winning if you switch: %.3f\" % (nbr_wins_if_switch / nbr_games))" 153 | ] 154 | }, 155 | { 156 | "cell_type": "markdown", 157 | "metadata": {}, 158 | "source": [ 159 | "#### Poincare\n", 160 | "\n", 161 | "__Exercise 5.6:__ Write a program that simulates a baker who chooses n loaves from a distribution with mean 950g and standard deviation 50g and gives the heaviest one to Poincare. What value of n yields a distribution with mean 1000g? What is the standard deviation?\n", 162 | "\n", 163 | "Compare this distribution to a normal distribution with the same mean and same standard deviation. Is the difference in the shape of the distribution big enough to convince the bread police?" 164 | ] 165 | }, 166 | { 167 | "cell_type": "code", 168 | "execution_count": 4, 169 | "metadata": {}, 170 | "outputs": [ 171 | { 172 | "name": "stdout", 173 | "output_type": "stream", 174 | "text": [ 175 | "Poincare's loaves: mean = 1003.365, sd = 35.146\n" 176 | ] 177 | }, 178 | { 179 | "data": { 180 | "text/plain": [ 181 | "" 182 | ] 183 | }, 184 | "execution_count": 4, 185 | "metadata": {}, 186 | "output_type": "execute_result" 187 | }, 188 | { 189 | "data": { 190 | "image/png": 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/+xkMHBi/a959t1ssc/jw+F0zRMI+GTM9PZ2XX36Zyy+/POhQfGGTMY1JJPFuLgObkGkS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\njAwyMjLiEKJJSAnSVDZ9Otx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191 | "text/plain": [ 192 | "" 193 | ] 194 | }, 195 | "metadata": {}, 196 | "output_type": "display_data" 197 | } 198 | ], 199 | "source": [ 200 | "nbr_days = 365\n", 201 | "n = 4\n", 202 | "ploaf = np.zeros(nbr_days)\n", 203 | "for day in range(nbr_days):\n", 204 | " loaf_wgts = 950 + 50 * np.random.randn(n)\n", 205 | " ploaf[day] = np.max(loaf_wgts)\n", 206 | "print(\"Poincare's loaves: mean = %.3f, sd = %.3f\" % (np.mean(ploaf), np.std(ploaf)))\n", 207 | "\n", 208 | "xs = np.linspace(950-100, 950+100, 200)\n", 209 | "loaf_ys = norm.pdf(xs, 950, 10)\n", 210 | "ploaf_ys = norm.pdf(xs, 1000, 34)\n", 211 | "plt.plot(xs, loaf_ys, color=\"red\", label=\"others\")\n", 212 | "plt.plot(xs, ploaf_ys, color=\"blue\", label=\"poincare\")\n", 213 | "plt.xlabel(\"weight (g)\")\n", 214 | "plt.ylabel(\"probability\")\n", 215 | "plt.legend(loc=\"best\")" 216 | ] 217 | }, 218 | { 219 | "cell_type": "markdown", 220 | "metadata": {}, 221 | "source": [ 222 | "__Exercise 5.7:__ If you go to a dance where partners are paired up randomly, what percentage of opposite sex couples will you see where the woman is taller than the man? Distribution of heights is roughly normal with mean of 178 and SD 59.4 for men and 163 and 52.8 for women." 223 | ] 224 | }, 225 | { 226 | "cell_type": "code", 227 | "execution_count": 5, 228 | "metadata": {}, 229 | "outputs": [ 230 | { 231 | "name": "stdout", 232 | "output_type": "stream", 233 | "text": [ 234 | "Probability of woman being taller: 0.408\n" 235 | ] 236 | } 237 | ], 238 | "source": [ 239 | "nbr_obs = 1000\n", 240 | "nbr_woman_taller = 0\n", 241 | "men_heights = np.random.normal(loc=178, scale=59.4, size=nbr_obs)\n", 242 | "women_heights = np.random.normal(loc=163, scale=52.8, size=nbr_obs)\n", 243 | "for i in range(nbr_obs):\n", 244 | " nbr_woman_taller += 1 if women_heights[i] > men_heights[i] else 0\n", 245 | "print(\"Probability of woman being taller: %.3f\" % (nbr_woman_taller / nbr_obs))" 246 | ] 247 | }, 248 | { 249 | "cell_type": "markdown", 250 | "metadata": {}, 251 | "source": [ 252 | "### Mutually Exclusive Events\n", 253 | "\n", 254 | " P(A|B) = P(B|A) = 0\n", 255 | " P(A or B) = P(A) + P(B) if A and B are mutually exclusive\n", 256 | " P(A or B) = P(A) + P(B) - P(A and B) in general\n", 257 | "\n", 258 | "__Exercise 5.8:__ If I roll 2 dice, what is the chance of rolling at least one six?\n", 259 | "\n", 260 | "Rolling at least 1 six = First dice roll 6 + Second dice roll 6 + Both dice roll 6 = (1/6)+(1/6)+(1/36) = 0.36.\n", 261 | "\n", 262 | "__Exercise 5.9:__ What is the general formula for probability of A or B but not both?\n", 263 | "\n", 264 | " P(A or B) = P(A) + P(B) - P(A and B)\n", 265 | " \n", 266 | "\n", 267 | "### Binomial Distribution\n", 268 | "\n", 269 | "Chance of getting exactly k successes from n trials, given p is the probability of success.\n", 270 | "\n", 271 | "$$PMF(k) = \\binom{n}{k}p^{k}(1-p)^{n-k}$$\n", 272 | "\n", 273 | "$$\\binom{n}{k} = \\frac{n!}{k!(n-k)!}$$\n", 274 | "\n", 275 | "__Exercise 5.10:__ If you flip a coin 100 times, you expect about 50 heads, but what is the probability of getting exactly 50 heads?" 276 | ] 277 | }, 278 | { 279 | "cell_type": "code", 280 | "execution_count": 6, 281 | "metadata": {}, 282 | "outputs": [ 283 | { 284 | "name": "stdout", 285 | "output_type": "stream", 286 | "text": [ 287 | "probability of exactly 50 heads in 100 tosses: 0.080\n" 288 | ] 289 | } 290 | ], 291 | "source": [ 292 | "p = comb(100, 50) * math.pow(0.5, 50) * math.pow(0.5, 50)\n", 293 | "print(\"probability of exactly 50 heads in 100 tosses: %.3f\" % (p))" 294 | ] 295 | }, 296 | { 297 | "cell_type": "markdown", 298 | "metadata": {}, 299 | "source": [ 300 | "### Streaks and hot spots\n", 301 | "\n", 302 | "__Example 5.11:__ If there are 10 players in a basketball game and each one takes 15 shots during the course of the game, and each shot has a 50% probability of going in, what is the probability that you will see, in a given game, at least 1 player who hits 10 shots in a row? If you watch a season of 82 games, what are the chances of seeing at least one string of 10 hits or misses?" 303 | ] 304 | }, 305 | { 306 | "cell_type": "code", 307 | "execution_count": 7, 308 | "metadata": {}, 309 | "outputs": [ 310 | { 311 | "name": "stdout", 312 | "output_type": "stream", 313 | "text": [ 314 | "chance of seeing at least 1 streak per game: 0.007\n", 315 | "chance of seeing at least 1 streak per season: 0.064\n" 316 | ] 317 | } 318 | ], 319 | "source": [ 320 | "def step(x, t):\n", 321 | " return 1 if x > t else 0\n", 322 | "\n", 323 | "nbr_sims = 1000\n", 324 | "nbr_players = 10\n", 325 | "nbr_succ = 0\n", 326 | "for g in range(nbr_sims):\n", 327 | " for p in range(nbr_players):\n", 328 | " nbr_streaks = 0\n", 329 | " shots = [step(x, 0.5) for x in np.random.random(15).tolist()]\n", 330 | " for i in range(len(shots)-10):\n", 331 | " window = shots[i:i+10]\n", 332 | " if sum(window) == 10:\n", 333 | " nbr_streaks += 1\n", 334 | " # we saw at least 1 streak (or not) in this game\n", 335 | " nbr_succ += 1 if nbr_streaks > 0 else 0\n", 336 | "print(\"chance of seeing at least 1 streak per game: %.3f\" % (nbr_succ / nbr_sims))\n", 337 | "\n", 338 | "nbr_games = 82\n", 339 | "nbr_succ = 0\n", 340 | "for s in range(nbr_sims):\n", 341 | " for g in range(nbr_games):\n", 342 | " nbr_streaks = 0\n", 343 | " for p in range(nbr_players):\n", 344 | " shots = [step(x, 0.5) for x in np.random.random(15).tolist()]\n", 345 | " for i in range(len(shots)-10):\n", 346 | " window = shots[i:i+10]\n", 347 | " if sum(window) == 10 or sum(window) == 0:\n", 348 | " nbr_streaks += 1\n", 349 | " nbr_succ += 1 if nbr_streaks > 0 else 0\n", 350 | "print(\"chance of seeing at least 1 streak per season: %.3f\" % (nbr_succ / nbr_sims))" 351 | ] 352 | }, 353 | { 354 | "cell_type": "markdown", 355 | "metadata": {}, 356 | "source": [ 357 | "__Exercise 5.12:__ In 1941 Joe DiMaggio got at least 1 hit in 56 consecutive games. Use Monte Carlo simulation to estimate the probability that any player in major league baseball will have a hitting streak of 57 or more games in the next century." 358 | ] 359 | }, 360 | { 361 | "cell_type": "code", 362 | "execution_count": 8, 363 | "metadata": {}, 364 | "outputs": [ 365 | { 366 | "name": "stdout", 367 | "output_type": "stream", 368 | "text": [ 369 | "chance of another hitting streak: 0.0000000\n" 370 | ] 371 | } 372 | ], 373 | "source": [ 374 | "nbr_sims = 1000\n", 375 | "nbr_years = 100 # next century\n", 376 | "nbr_games_per_season = 162 # from wikipedia\n", 377 | "nbr_succ = 0\n", 378 | "for sim in range(nbr_sims):\n", 379 | " for i in range(nbr_years):\n", 380 | " nbr_streaks = 0\n", 381 | " hits = [step(x, 0.5) for x in np.random.random(nbr_games_per_season).tolist()]\n", 382 | " for j in range(len(hits)-57):\n", 383 | " window = hits[j:j+57]\n", 384 | " if sum(window) == 57:\n", 385 | " nbr_streaks += 1\n", 386 | " if nbr_streaks > 0:\n", 387 | " nbr_succ += 1\n", 388 | "print(\"chance of another hitting streak: %.7f\" % (nbr_succ / nbr_sims))" 389 | ] 390 | }, 391 | { 392 | "cell_type": "markdown", 393 | "metadata": {}, 394 | "source": [ 395 | "__Exercise 5.13:__ Suppose that a particular cancer has an incidence of 1 case per thousand people per year. If you follow a particular cohort of 100 people for 10 years, you would expect to see about 1 case. If you saw two cases, that would not be very surprising, but more than than two would be rare. Write a program that simulates a large number of cohorts over a 10 year period and estimates the distribution of total cases.\n", 396 | "\n", 397 | "An observation is considered statistically significant if its probability by chance alone, called a p-value, is less than 5%. In a cohort of 100 people over 10 years, how many cases would you have to see to meet this criterion?\n", 398 | "\n", 399 | "Now imagine that you divide a population of 10000 people into 100 cohorts and follow them for 10 years. What is the chance that at least one of the cohorts will have a “statistically significant” cluster? What if we require a p-value of 1%?\n", 400 | "\n", 401 | "Now imagine that you arrange 10000 people in a 100x100 grid and follow them for 10 years. What is the chance that there will be at least one 10x10 block anywhere in the grid with a statistically significant cluster?\n", 402 | "\n", 403 | "Finally, imagine that you follow a grid of 10000 people for 30 years. What is the chance that there will be a 10-year interval at some point with a 10x10 block anywhere in the grid with a statistically significant cluster?" 404 | ] 405 | }, 406 | { 407 | "cell_type": "code", 408 | "execution_count": 9, 409 | "metadata": {}, 410 | "outputs": [ 411 | { 412 | "data": { 413 | "image/png": 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414 | "text/plain": [ 415 | "" 416 | ] 417 | }, 418 | "metadata": {}, 419 | "output_type": "display_data" 420 | }, 421 | { 422 | "data": { 423 | "image/png": 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424 | "text/plain": [ 425 | "" 426 | ] 427 | }, 428 | "metadata": {}, 429 | "output_type": "display_data" 430 | } 431 | ], 432 | "source": [ 433 | "p = 0.001 # one case per 1000 per year\n", 434 | "nbr_cohorts = 1000\n", 435 | "nbr_years = 10\n", 436 | "cohort_size = 100\n", 437 | "nbr_cases_dist = []\n", 438 | "for c in range(nbr_cohorts):\n", 439 | " nbr_cases = 0\n", 440 | " for y in range(nbr_years):\n", 441 | " for i in range(cohort_size):\n", 442 | " d = np.random.random()\n", 443 | " if d < p:\n", 444 | " nbr_cases += 1\n", 445 | " nbr_cases_dist.append(nbr_cases)\n", 446 | "nbr_case_dist = np.array(nbr_cases_dist)\n", 447 | "range_lb = np.min(nbr_case_dist)\n", 448 | "range_ub = np.max(nbr_case_dist)\n", 449 | "nbr_bins = range_ub - range_lb\n", 450 | "\n", 451 | "# compute PMF\n", 452 | "pmf = np.histogram(np.array(nbr_cases_dist), bins=nbr_bins, range=(range_lb, range_ub),\n", 453 | " normed=True)\n", 454 | "plt.bar(pmf[1][:-1], pmf[0])\n", 455 | "plt.xlabel(\"# of cases\")\n", 456 | "plt.ylabel(\"probability\")\n", 457 | "plt.show()\n", 458 | "\n", 459 | "# compute CDF\n", 460 | "ps = np.cumsum(np.sort(pmf[0]))\n", 461 | "xs = pmf[1][:-1]\n", 462 | "plt.step(xs, ps)\n", 463 | "plt.xlabel(\"# of cases\")\n", 464 | "plt.ylabel(\"cumulative probability\")\n", 465 | "plt.grid(True)\n", 466 | "plt.show()" 467 | ] 468 | }, 469 | { 470 | "cell_type": "markdown", 471 | "metadata": {}, 472 | "source": [ 473 | "Statistically significant at p-value of 5% means the # of cases at cumulative probability of 0.95, ie 5 cases. For statistical significance at p-value of 1% means the number of cases at cumulative probability 0.99, also 5 cases.\n", 474 | "\n", 475 | "Other cases are just about rearranging the loops so not doing them.\n", 476 | "\n", 477 | "### Bayes Theorem\n", 478 | "\n", 479 | "$$P(A|B) = \\frac{P(B|A)P(A)}{P(B)}$$\n", 480 | "\n", 481 | "Here P(A) is the __prior__, P(A|B) is the __posterior__, P(B|A) is the __likelihood__, and P(B) is the __normalizing constant__.\n", 482 | "\n", 483 | "__Exercise 5.14:__ Write a program that takes the actual rate of drug use, and the sensitivity and specificity of the test, and computes P(drug use|positive test).\n", 484 | "\n", 485 | "Suppose the same test is applied to a population where the actual rate of drug use is 1%. What is the probability that someone who tests positive is actually a drug user?" 486 | ] 487 | }, 488 | { 489 | "cell_type": "code", 490 | "execution_count": 10, 491 | "metadata": {}, 492 | "outputs": [ 493 | { 494 | "name": "stdout", 495 | "output_type": "stream", 496 | "text": [ 497 | "0.759493670886\n", 498 | "0.377358490566\n" 499 | ] 500 | } 501 | ], 502 | "source": [ 503 | "def compute_drug_use_given_positive_test(p_drug_use, # p(drug use)\n", 504 | " sensitivity, # p(positive result | drug use)\n", 505 | " specificity): # p(negative result | -drug use)\n", 506 | " norm = (p_drug_use * sensitivity) + ((1 - p_drug_use)*(1 - specificity))\n", 507 | " return (p_drug_use * sensitivity) / norm\n", 508 | "\n", 509 | "print(compute_drug_use_given_positive_test(0.05, 0.6, 0.99))\n", 510 | "print(compute_drug_use_given_positive_test(0.01, 0.6, 0.99))" 511 | ] 512 | }, 513 | { 514 | "cell_type": "markdown", 515 | "metadata": {}, 516 | "source": [ 517 | "Discussions for solutions to the exercises below can be found on Allen Downey's [blog post](http://allendowney.blogspot.com/2011/10/all-your-bayes-are-belong-to-us.html).\n", 518 | "\n", 519 | "__Exercise 5.15:__ Suppose there are two full bowls of cookies. Bowl 1 has 10 chocolate chip and 30 plain cookies, while Bowl 2 has 20 of each. Fred picks a bowl at random, and then picks a cookie at random. The cookie turns out to be a plain one. How probable is it that Fred picked it out of Bowl 1?" 520 | ] 521 | }, 522 | { 523 | "cell_type": "code", 524 | "execution_count": 11, 525 | "metadata": {}, 526 | "outputs": [ 527 | { 528 | "name": "stdout", 529 | "output_type": "stream", 530 | "text": [ 531 | "0.6\n" 532 | ] 533 | } 534 | ], 535 | "source": [ 536 | "p_bowl1 = 0.5\n", 537 | "p_bowl2 = 0.5\n", 538 | "p_plain_given_bowl1 = 30 / (10 + 30)\n", 539 | "p_plain_given_bowl2 = 20 / (20 + 20)\n", 540 | "p_bowl1_given_plain = p_plain_given_bowl1 * p_bowl1\n", 541 | "p_bowl1_given_plain /= (p_plain_given_bowl1 * p_bowl1) + (p_plain_given_bowl2 * p_bowl2)\n", 542 | "print(p_bowl1_given_plain)" 543 | ] 544 | }, 545 | { 546 | "cell_type": "markdown", 547 | "metadata": {}, 548 | "source": [ 549 | "__Exercise 5.16:__ The blue M&M was introduced in 1995. Before then, the color mix in a bag of plain M&Ms was (30% Brown, 20% Yellow, 20% Red, 10% Green, 10% Orange, 10% Tan). Afterward it was (24% Blue , 20% Green, 16% Orange, 14% Yellow, 13% Red, 13% Brown).\n", 550 | "\n", 551 | "A friend has two bags of M&Ms, and he tells me that one is from 1994 and one from 1996. He won’t tell me which is which, but he gives me one M&M from each bag. One is yellow and one is green. What is the probability that the yellow M&M came from the 1994 bag?" 552 | ] 553 | }, 554 | { 555 | "cell_type": "code", 556 | "execution_count": 12, 557 | "metadata": {}, 558 | "outputs": [ 559 | { 560 | "name": "stdout", 561 | "output_type": "stream", 562 | "text": [ 563 | "0.740740740741\n" 564 | ] 565 | } 566 | ], 567 | "source": [ 568 | "# h1 = bag 1 is from 1994 and bag 2 is from 1996\n", 569 | "# h2 = bag 1 is from 1996 and bag 2 is from 1994\n", 570 | "# e = yellow and green M&M\n", 571 | "p_h1 = 0.5\n", 572 | "p_h2 = 0.5\n", 573 | "p_e_given_h1 = 0.2 * 0.2 # P(y from 1994 bag) * P(g from 1996 bag)\n", 574 | "p_e_given_h2 = 0.1 * 0.14 # P(y from 1996 bag) * P(g from 1994 bag)\n", 575 | "p_h1_given_e = p_e_given_h1 * p_h1\n", 576 | "p_h1_given_e /= ((p_e_given_h1 * p_h1) + (p_e_given_h2 * p_h2))\n", 577 | "print(p_h1_given_e)" 578 | ] 579 | }, 580 | { 581 | "cell_type": "markdown", 582 | "metadata": {}, 583 | "source": [ 584 | "__Exercise 5.17:__ Elvis Presley had a twin brother who died at birth. According to the\n", 585 | "Wikipedia article on twins - Twins are estimated to be approximately 1.9% of the world population, with monozygotic twins making up 0.2% of the total and 8% of all twins. What is the probability that Elvis was an identical twin?" 586 | ] 587 | }, 588 | { 589 | "cell_type": "code", 590 | "execution_count": 13, 591 | "metadata": {}, 592 | "outputs": [ 593 | { 594 | "name": "stdout", 595 | "output_type": "stream", 596 | "text": [ 597 | "0.148148148148\n" 598 | ] 599 | } 600 | ], 601 | "source": [ 602 | "# h1 = elvis had identical twin\n", 603 | "# h2 = elvis had fraternal twin\n", 604 | "# e = brother was male\n", 605 | "p_h1 = 0.08\n", 606 | "p_h2 = 1 - p_h1\n", 607 | "p_e_given_h1 = 1.0 # identical twins have same sex\n", 608 | "p_e_given_h2 = 0.5 # fraternal twins can have either sex\n", 609 | "p_h1_given_e = p_e_given_h1 * p_h1\n", 610 | "p_h1_given_e /= ((p_e_given_h1 * p_h1) + (p_e_given_h2 * p_h2))\n", 611 | "print(p_h1_given_e)" 612 | ] 613 | }, 614 | { 615 | "cell_type": "code", 616 | "execution_count": null, 617 | "metadata": { 618 | "collapsed": true 619 | }, 620 | "outputs": [], 621 | "source": [] 622 | } 623 | ], 624 | "metadata": { 625 | "kernelspec": { 626 | "display_name": "Python (bayes)", 627 | "language": "python", 628 | "name": "bayes" 629 | }, 630 | "language_info": { 631 | "codemirror_mode": { 632 | "name": "ipython", 633 | "version": 3 634 | }, 635 | "file_extension": ".py", 636 | "mimetype": "text/x-python", 637 | "name": "python", 638 | "nbconvert_exporter": "python", 639 | "pygments_lexer": "ipython3", 640 | "version": "3.8.1" 641 | } 642 | }, 643 | "nbformat": 4, 644 | "nbformat_minor": 1 645 | } 646 | -------------------------------------------------------------------------------- /src/07-hypothesis-testing.ipynb: -------------------------------------------------------------------------------- 1 | { 2 | "cells": [ 3 | { 4 | "cell_type": "markdown", 5 | "metadata": {}, 6 | "source": [ 7 | "# Hypothesis Testing\n", 8 | "\n", 9 | "This notebook is for Chapter 7 of [Think Stats - Probability and Statistics for Programmers](http://greenteapress.com/thinkstats/) by Allen B Downey.\n", 10 | "\n", 11 | "Objective: learn techniques to decide if apparent effects are __significant__. An apparent effect is __statistically significant__ if it unlikely to have occurred by chance." 12 | ] 13 | }, 14 | { 15 | "cell_type": "code", 16 | "execution_count": 1, 17 | "metadata": { 18 | "collapsed": true 19 | }, 20 | "outputs": [], 21 | "source": [ 22 | "from __future__ import division, print_function\n", 23 | "import math\n", 24 | "import matplotlib.pyplot as plt\n", 25 | "import numpy as np\n", 26 | "import pandas as pd\n", 27 | "import scipy.stats\n", 28 | "%matplotlib inline" 29 | ] 30 | }, 31 | { 32 | "cell_type": "markdown", 33 | "metadata": {}, 34 | "source": [ 35 | "## Testing Difference in Means\n", 36 | "\n", 37 | "In the NSFG data, we saw that the mean pregnancy length for first babies is slightly longer, and the mean weight at birth is slightly smaller. Now we will see if those effects are significant.\n", 38 | "\n", 39 | " H0: distribution of the two groups (first and other babies) are same\n", 40 | "\n", 41 | "\n", 42 | " " 43 | ] 44 | }, 45 | { 46 | "cell_type": "code", 47 | "execution_count": 2, 48 | "metadata": { 49 | "collapsed": false 50 | }, 51 | "outputs": [ 52 | { 53 | "data": { 54 | "text/html": [ 55 | "
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caseidnbrnalivbabysexbirthwgt_lbbirthwgt_ozprglengthoutcomebirthordagepregfinalwgt
01118133911336448.271112
11127143912396448.271112
22319239111412999.542264
32127039121712999.542264
42126339131812999.542264
\n", 140 | "
" 141 | ], 142 | "text/plain": [ 143 | " caseid nbrnaliv babysex birthwgt_lb birthwgt_oz prglength outcome \\\n", 144 | "0 1 1 1 8 13 39 1 \n", 145 | "1 1 1 2 7 14 39 1 \n", 146 | "2 2 3 1 9 2 39 1 \n", 147 | "3 2 1 2 7 0 39 1 \n", 148 | "4 2 1 2 6 3 39 1 \n", 149 | "\n", 150 | " birthord agepreg finalwgt \n", 151 | "0 1 33 6448.271112 \n", 152 | "1 2 39 6448.271112 \n", 153 | "2 1 14 12999.542264 \n", 154 | "3 2 17 12999.542264 \n", 155 | "4 3 18 12999.542264 " 156 | ] 157 | }, 158 | "execution_count": 2, 159 | "metadata": {}, 160 | "output_type": "execute_result" 161 | } 162 | ], 163 | "source": [ 164 | "pregnancies = pd.read_fwf(\"../data/2002FemPreg.dat\", \n", 165 | " names=[\"caseid\", \"nbrnaliv\", \"babysex\", \"birthwgt_lb\",\n", 166 | " \"birthwgt_oz\", \"prglength\", \"outcome\", \"birthord\",\n", 167 | " \"agepreg\", \"finalwgt\"],\n", 168 | " colspecs=[(0, 12), (21, 22), (55, 56), (57, 58), (58, 60),\n", 169 | " (274, 276), (276, 277), (278, 279), (283, 285), (422, 439)])\n", 170 | "pregnancies.head()" 171 | ] 172 | }, 173 | { 174 | "cell_type": "code", 175 | "execution_count": 3, 176 | "metadata": { 177 | "collapsed": false 178 | }, 179 | "outputs": [ 180 | { 181 | "name": "stdout", 182 | "output_type": "stream", 183 | "text": [ 184 | "# first babies: 4413, other babies: 4735\n", 185 | "Difference in means (data): 0.078 weeks\n" 186 | ] 187 | } 188 | ], 189 | "source": [ 190 | "# Look at the input data\n", 191 | "live_births = pregnancies[pregnancies[\"outcome\"] == 1]\n", 192 | "first_babies = np.array(live_births[live_births[\"birthord\"] == 1][\"prglength\"].dropna())\n", 193 | "other_babies = np.array(live_births[live_births[\"birthord\"] != 1][\"prglength\"].dropna())\n", 194 | "print(\"# first babies: %d, other babies: %d\" % (first_babies.shape[0], other_babies.shape[0]))\n", 195 | "delta = np.mean(first_babies) - np.mean(other_babies)\n", 196 | "print(\"Difference in means (data): %.3f weeks\" % (delta))" 197 | ] 198 | }, 199 | { 200 | "cell_type": "code", 201 | "execution_count": 4, 202 | "metadata": { 203 | "collapsed": false 204 | }, 205 | "outputs": [ 206 | { 207 | "data": { 208 | "text/plain": [ 209 | "" 210 | ] 211 | }, 212 | "execution_count": 4, 213 | "metadata": {}, 214 | "output_type": "execute_result" 215 | }, 216 | { 217 | "data": { 218 | "image/png": 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219 | "text/plain": [ 220 | "" 221 | ] 222 | }, 223 | "metadata": {}, 224 | "output_type": "display_data" 225 | } 226 | ], 227 | "source": [ 228 | "def compute_diff(model1, model2, replace=True):\n", 229 | " sample1 = np.random.choice(model1, size=model1.shape[0], replace=replace)\n", 230 | " sample2 = np.random.choice(model2, size=model2.shape[0], replace=replace)\n", 231 | " return np.mean(sample1) - np.mean(sample2)\n", 232 | " \n", 233 | "# Generate 2 random samples with the mean and std of the pooled distributio\n", 234 | "pooled_babies = np.array(pregnancies[pregnancies[\"outcome\"] == 1][\"prglength\"].dropna())\n", 235 | "sample_mean = np.mean(pooled_babies)\n", 236 | "sample_std = np.std(pooled_babies)\n", 237 | "np.random.seed(1)\n", 238 | "sample_first = np.random.normal(sample_mean, sample_std, 4413)\n", 239 | "sample_other = np.random.normal(sample_mean, sample_std, 4735)\n", 240 | "\n", 241 | "# draw samples from sample_first and sample_other and compute diffs.\n", 242 | "# create a distribution of diffs\n", 243 | "diffs = np.array([compute_diff(sample_first, sample_other) for x in range(1000)])\n", 244 | "\n", 245 | "diffs_pmf = np.histogram(diffs, bins=50, normed=True)\n", 246 | "diffs_cdf_ps = np.cumsum(diffs_pmf[0])\n", 247 | "diffs_cdf_ps = diffs_cdf_ps / diffs_cdf_ps[-1]\n", 248 | "diffs_cdf_xs = diffs_pmf[1][:-1]\n", 249 | "\n", 250 | "# CDF plot of distribution difference of means\n", 251 | "plt.plot(diffs_cdf_xs, diffs_cdf_ps)\n", 252 | "plt.axvline(x=-delta, color='k')\n", 253 | "plt.axvline(x=delta, color='k')\n", 254 | "plt.xlabel(\"difference in mean (weeks)\")\n", 255 | "plt.ylabel(\"CDF(x)\")\n", 256 | "plt.title(\"Resampled Differences\")" 257 | ] 258 | }, 259 | { 260 | "cell_type": "code", 261 | "execution_count": 5, 262 | "metadata": { 263 | "collapsed": false 264 | }, 265 | "outputs": [ 266 | { 267 | "name": "stdout", 268 | "output_type": "stream", 269 | "text": [ 270 | "p-value = 0.172\n" 271 | ] 272 | } 273 | ], 274 | "source": [ 275 | "# p-value = fraction of diffs sample where value >= delta\n", 276 | "nbr_diffs_ge_delta = np.where(diffs >= delta)[0].shape[0]\n", 277 | "print(\"p-value = %.3f\" % (nbr_diffs_ge_delta / diffs.shape[0]))" 278 | ] 279 | }, 280 | { 281 | "cell_type": "markdown", 282 | "metadata": {}, 283 | "source": [ 284 | "__Exercise 7.1:__ In the NSFG dataset, the difference in mean weight for first births is 2.0 ounces. Compute the p-value of this difference. \n", 285 | "\n", 286 | "Hint: for this kind of resampling it is important to sample with replacement, so you should use random.choice rather than random.sample (see Section 3.8)." 287 | ] 288 | }, 289 | { 290 | "cell_type": "code", 291 | "execution_count": 6, 292 | "metadata": { 293 | "collapsed": false 294 | }, 295 | "outputs": [ 296 | { 297 | "name": "stderr", 298 | "output_type": "stream", 299 | "text": [ 300 | "/Users/palsujit/anaconda/lib/python2.7/site-packages/ipykernel/__main__.py:1: SettingWithCopyWarning: \n", 301 | "A value is trying to be set on a copy of a slice from a DataFrame.\n", 302 | "Try using .loc[row_indexer,col_indexer] = value instead\n", 303 | "\n", 304 | "See the caveats in the documentation: http://pandas.pydata.org/pandas-docs/stable/indexing.html#indexing-view-versus-copy\n", 305 | " if __name__ == '__main__':\n" 306 | ] 307 | } 308 | ], 309 | "source": [ 310 | "live_births[\"birthwgt\"] = 16 * live_births[\"birthwgt_lb\"] + live_births[\"birthwgt_oz\"]\n", 311 | "pooled = np.array(live_births[\"birthwgt\"].dropna())\n", 312 | "first = np.array(live_births[live_births[\"birthord\"] == 1][\"birthwgt\"].dropna())\n", 313 | "other = np.array(live_births[live_births[\"birthord\"] != 1][\"birthwgt\"].dropna())" 314 | ] 315 | }, 316 | { 317 | "cell_type": "code", 318 | "execution_count": 7, 319 | "metadata": { 320 | "collapsed": false 321 | }, 322 | "outputs": [ 323 | { 324 | "data": { 325 | "text/plain": [ 326 | "" 327 | ] 328 | }, 329 | "execution_count": 7, 330 | "metadata": {}, 331 | "output_type": "execute_result" 332 | }, 333 | { 334 | "data": { 335 | "image/png": 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2t6IrQnnhLiCrGvffD/fc464fs/bgs4Cs05g5E44+Ovvl74nezJacu4DKz5G8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\n", 336 | "text/plain": [ 337 | "" 338 | ] 339 | }, 340 | "metadata": {}, 341 | "output_type": "display_data" 342 | } 343 | ], 344 | "source": [ 345 | "# compute mean and std of pooled data and generate samples of size(first) and size(other)\n", 346 | "pooled_mean = np.mean(pooled)\n", 347 | "pooled_std = np.std(pooled)\n", 348 | "np.random.seed(1)\n", 349 | "sample_first = np.random.normal(pooled_mean, pooled_std, first.shape[0])\n", 350 | "sample_other = np.random.normal(pooled_mean, pooled_std, other.shape[0])\n", 351 | "\n", 352 | "# create a distribution of diffs by sampling from the above samples many times\n", 353 | "diffs = np.array([compute_diff(sample_first, sample_other, True) for x in range(1000)])\n", 354 | "\n", 355 | "# plot the distribution\n", 356 | "pmf_diffs = np.histogram(diffs, bins=100, normed=True)\n", 357 | "cdf_diffs_ps = np.cumsum(pmf_diffs[0])\n", 358 | "cdf_diffs_ps = cdf_diffs_ps / cdf_diffs_ps[-1]\n", 359 | "cdf_diffs_xs = pmf_diffs[1][:-1]\n", 360 | "\n", 361 | "plt.plot(cdf_diffs_xs, cdf_diffs_ps)\n", 362 | "plt.xlim(-2.5, 2.5)\n", 363 | "plt.axvline(x=-2.0, color='k')\n", 364 | "plt.axvline(x=2.0, color='k')\n", 365 | "plt.xlabel(\"difference in mean (oz)\")\n", 366 | "plt.ylabel(\"CDF(x)\")\n", 367 | "plt.title(\"Resampled Differences\")" 368 | ] 369 | }, 370 | { 371 | "cell_type": "code", 372 | "execution_count": 8, 373 | "metadata": { 374 | "collapsed": false 375 | }, 376 | "outputs": [ 377 | { 378 | "name": "stdout", 379 | "output_type": "stream", 380 | "text": [ 381 | "p-value = 0.00100\n" 382 | ] 383 | } 384 | ], 385 | "source": [ 386 | "diffs_ge_2 = np.where(diffs >= 2)[0]\n", 387 | "print(\"p-value = %.5f\" % (diffs_ge_2.shape[0] / diffs.shape[0]))" 388 | ] 389 | }, 390 | { 391 | "cell_type": "markdown", 392 | "metadata": {}, 393 | "source": [ 394 | "## Choosing a threshold\n", 395 | "\n", 396 | "Common threshold is $\\alpha$ = 5%. If the p-value is less than $\\alpha$, then difference is statistically significant, else it is not.\n", 397 | "\n", 398 | "So the difference in pregnancy lengths is __not statistically significant__ because 17.2% is greater than 5%, and the difference in birth weights is __statistically significant__ because 1% is less than 5%.\n", 399 | "\n", 400 | "Decreasing $\\alpha$ decreases the chance of false positive. But it also raises the standard of evidence, which increases the chance of rejecting a valid hypothesis.\n", 401 | "\n", 402 | "__Exercise 7.2:__ To investigate the effect of sample size on p-value, see what happens if you discard half of the data from the NSFG. Hint: use random.sample. What if you discard three-quarters of the data, and so on? What is the smallest sample size where the difference in mean birth weight is still significant with $\\alpha$ = 5%? How much larger does the sample size have to be with $\\alpha$ = 1%?" 403 | ] 404 | }, 405 | { 406 | "cell_type": "code", 407 | "execution_count": 9, 408 | "metadata": { 409 | "collapsed": false 410 | }, 411 | "outputs": [ 412 | { 413 | "name": "stdout", 414 | "output_type": "stream", 415 | "text": [ 416 | "(sample size, p-value): [(9087, 0.0), (4543, 0.0), (2271, 0.0), (2998, 0.004), (908, 0.0), (454, 0.0), (90, 0.005), (45, 0.092)]\n" 417 | ] 418 | } 419 | ], 420 | "source": [ 421 | "data_keep_fracs = [1.0, 0.5, 0.25, 0.33, 0.1, 0.05, 0.01, 0.005]\n", 422 | "spvs = []\n", 423 | "for data_keep_frac in data_keep_fracs:\n", 424 | " pool_sample = np.random.choice(pooled, int(pooled.shape[0] * data_keep_frac))\n", 425 | " pool_mean = np.mean(pool_sample)\n", 426 | " pool_std = np.std(pool_sample)\n", 427 | " first_sample = np.random.normal(pool_mean, pool_std, first.shape[0])\n", 428 | " other_sample = np.random.normal(pool_mean, pool_std, other.shape[0])\n", 429 | " diffs = np.array([compute_diff(first_sample, other_sample) for x in range(1000)])\n", 430 | " p_value = np.where(diffs >= 2.0)[0].shape[0] / diffs.shape[0]\n", 431 | " spvs.append((pool_sample.shape[0], p_value))\n", 432 | "print(\"(sample size, p-value):\", spvs)" 433 | ] 434 | }, 435 | { 436 | "cell_type": "markdown", 437 | "metadata": {}, 438 | "source": [ 439 | "## Defining the Effect\n", 440 | "\n", 441 | "__Two sided test__ - check both sides when computing the p-value.\n", 442 | "\n", 443 | "__One sided test__ - check only one side.\n", 444 | "\n", 445 | "For symmetric distributions the two sided p-value should be roughly twice the one-sided value, but in this case, the distribution is skewed right." 446 | ] 447 | }, 448 | { 449 | "cell_type": "code", 450 | "execution_count": 10, 451 | "metadata": { 452 | "collapsed": false 453 | }, 454 | "outputs": [ 455 | { 456 | "name": "stdout", 457 | "output_type": "stream", 458 | "text": [ 459 | "p-value (1 sided): 0.092, (2 sided): 0.092\n" 460 | ] 461 | } 462 | ], 463 | "source": [ 464 | "p_value_2_sided = np.where((diffs >= 2.0) | (diffs <= -2.0))[0]\n", 465 | "p_value_2_sided = p_value_2_sided.shape[0] / diffs.shape[0]\n", 466 | "p_value_1_sided = np.where(diffs >= 2.0)[0]\n", 467 | "p_value_1_sided = p_value_1_sided.shape[0] / diffs.shape[0]\n", 468 | "print(\"p-value (1 sided): %.3f, (2 sided): %.3f\" % (p_value_1_sided, p_value_2_sided))" 469 | ] 470 | }, 471 | { 472 | "cell_type": "markdown", 473 | "metadata": {}, 474 | "source": [ 475 | "## Interpreting the result\n", 476 | "\n", 477 | "We define event E, and compute the p-value, which is P(E|H0) and compare to the threshold $\\alpha$.\n", 478 | "\n", 479 | "__Classical:__ if p-value less than $\\alpha$, we can conclude effect is __statistically significant__, but can't conclude that its real.\n", 480 | "\n", 481 | "__Practical:__ in practice, people conclude its real. Lower the p-value, the higher the confidence in the conclusion.\n", 482 | "\n", 483 | "__Bayesian:__ what we want to know is:\n", 484 | "\n", 485 | "$$P(H_{A} | E) = \\frac {P(E | H_{A}) P(H_{A})} {P(E)}$$\n", 486 | "\n", 487 | "where $H_{A}$ is the alternate hypothesis that the effect is real. Here $P(H_{A})$ is the prior probability of $H_{A}$ before we see the effect, $P(E|H_{A})$ is the probability of seeing the effect assuming it is real, and P(E) is the probability of seeing E under any circumstances. Since the effect is either real or not real, P(E) is given by:\n", 488 | "\n", 489 | "$$P(E) = P(E|H_{A})P(H_{A}) + P(E|H_{0})P(H_{0})$$\n", 490 | "\n", 491 | "__Exercise 7.3:__ Using the NSFG data, what is the posterior probability that the distribution of birth weights is different for first babies than others?" 492 | ] 493 | }, 494 | { 495 | "cell_type": "code", 496 | "execution_count": 11, 497 | "metadata": { 498 | "collapsed": false 499 | }, 500 | "outputs": [ 501 | { 502 | "name": "stdout", 503 | "output_type": "stream", 504 | "text": [ 505 | "posterior probability that birth weights are different: 0.875\n" 506 | ] 507 | } 508 | ], 509 | "source": [ 510 | "# assume H(A) = first and other distributions are different and the difference is 2.0\n", 511 | "# compute P(E|HA)\n", 512 | "np.random.seed(1)\n", 513 | "first_sample = np.random.normal(pool_mean, pool_std, first.shape[0])\n", 514 | "other_sample = np.random.normal(pool_mean, pool_std, other.shape[0])\n", 515 | "diffs = np.array([compute_diff(first_sample, other_sample) for x in range(1000)])\n", 516 | "diffs_ge_2 = np.where(diffs >= 2.0)\n", 517 | "p_e_ha = diffs_ge_2[0].shape[0] / diffs.shape[0]\n", 518 | "# prior P(HA) = P(H0) = 0.5 since we assume no knowledge\n", 519 | "p_ha = 0.5\n", 520 | "p_h0 = 0.5\n", 521 | "# we already know H(E|H0) = 0.001 from before \n", 522 | "p_e_h0 = 0.001\n", 523 | "p_ha_e = p_e_ha * p_ha / (p_e_ha * p_ha + p_e_h0 * p_h0)\n", 524 | "print(\"posterior probability that birth weights are different: %.3f\" % (p_ha_e))" 525 | ] 526 | }, 527 | { 528 | "cell_type": "markdown", 529 | "metadata": {}, 530 | "source": [ 531 | "## Cross-validation\n", 532 | "\n", 533 | "Split the data into training and test OR partition and run over each partition, then average. Second method shown." 534 | ] 535 | }, 536 | { 537 | "cell_type": "code", 538 | "execution_count": 12, 539 | "metadata": { 540 | "collapsed": false 541 | }, 542 | "outputs": [ 543 | { 544 | "name": "stdout", 545 | "output_type": "stream", 546 | "text": [ 547 | "posterior probability (averaged) = 0.985\n" 548 | ] 549 | } 550 | ], 551 | "source": [ 552 | "p_ha = 0.5\n", 553 | "p_h0 = 0.5\n", 554 | "p_e_h0 = 0.001\n", 555 | "post_probs = []\n", 556 | "np.random.seed(1)\n", 557 | "for i in range(10):\n", 558 | " idxs = np.random.choice(pooled, size=8000)\n", 559 | " pool_split = pooled[np.array(idxs, dtype=int)]\n", 560 | " pool_mean = np.mean(pool_split)\n", 561 | " pool_std = np.mean(pool_split)\n", 562 | " first_sample = np.random.normal(pool_mean, pool_std, first.shape[0])\n", 563 | " other_sample = np.random.normal(pool_mean, pool_std, other.shape[0])\n", 564 | " diffs = np.array([compute_diff(first_sample, other_sample) for x in range(1000)])\n", 565 | " diffs_ge_2 = np.where(diffs >= 2.0)\n", 566 | " p_e_ha = diffs_ge_2[0].shape[0] / diffs.shape[0]\n", 567 | " post_probs.append(p_e_ha * p_ha / (p_e_ha * p_ha + p_e_h0 * p_h0))\n", 568 | "print(\"posterior probability (averaged) = %.3f\" % (sum(post_probs) / len(post_probs)))" 569 | ] 570 | }, 571 | { 572 | "cell_type": "markdown", 573 | "metadata": {}, 574 | "source": [ 575 | "## Reporting Bayesian Probabilities\n", 576 | "\n", 577 | "__Objection to Bayesian probabilities:__ arbitary priors. However, strong evidence will swamp the prior.\n", 578 | "\n", 579 | "__Likelihood Ratio:__ also called __Bayes Factor__.\n", 580 | "\n", 581 | "$$likelihood \\space ratio = \\frac {P(E|H_{A})}{P(E|H_{0})}$$\n", 582 | "\n", 583 | "__Exercise 7.4:__ If your prior probability for a hypothesis, HA, is 0.3 and new evidence becomes available that yields a likelihood ratio of 3 relative to the null hypothesis, H0, what is your posterior probability for HA?\n", 584 | "\n", 585 | "You can use Bayes theorem and the formula for LR to find that P(HA|E) = (9/7)P(H0|E). But P(HA|E) + P(H0|E) = 1, so P(HA|E) = 9/16.\n", 586 | "\n", 587 | "__Exercise 7.5:__ Two people have left traces of their own blood at the scene of a crime. A suspect, Oliver, is tested and found to have type O blood. The blood groups of the two traces are found to be of type O (a common type in the local population, having frequency 60%) and of type AB (a rare type, with frequency 1%). Do these data (the blood types found at the scene) give evidence in favor of the proposition that Oliver was one of the two people whose blood was found at the scene?\n", 588 | "\n", 589 | "Covered in [Allen Downey's blog post](http://allendowney.blogspot.com/2011/10/all-your-bayes-are-belong-to-us.html).\n", 590 | "\n", 591 | " H0 = Oliver is not one of the people\n", 592 | " H1 = Oliver is one of the people\n", 593 | " E = 2 blood samples found\n", 594 | "\n", 595 | " P(E|H0) = 2 samples * (0.6 * 0.01) = 0.12\n", 596 | " P(E|H1) = 0.01 (since Oliver already has O, we just need the other guy with AB)\n", 597 | " LR = P(E|H1) / P(E|H0) = 1/12\n", 598 | "\n", 599 | "## Chi-square test\n", 600 | "\n", 601 | "Instead of testing whether the apparent difference in mean pregnancy length is significant or not, alternative is to test the hypothesis as it appears - ie, first babies are more likely to be early, less likely to be on time and more likely to be late.\n", 602 | "\n", 603 | "Formula for chi-squared ratio is:\n", 604 | "\n", 605 | "$$\\chi^{2} = \\sum_{i} \\frac {(O_{i} - E_{i})^{2}}{E_{i}}$$\n", 606 | "\n", 607 | "H0 = pregnancy length distribution is the same for both groups. The p-value is the probability of seeing a chi-squared value as high as the one observed." 608 | ] 609 | }, 610 | { 611 | "cell_type": "code", 612 | "execution_count": 13, 613 | "metadata": { 614 | "collapsed": false 615 | }, 616 | "outputs": [ 617 | { 618 | "name": "stdout", 619 | "output_type": "stream", 620 | "text": [ 621 | "observed chi-squared value: 92.12957073\n" 622 | ] 623 | }, 624 | { 625 | "data": { 626 | "text/plain": [ 627 | "" 628 | ] 629 | }, 630 | "execution_count": 13, 631 | "metadata": {}, 632 | "output_type": "execute_result" 633 | }, 634 | { 635 | "data": { 636 | "image/png": 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637 | "text/plain": [ 638 | "" 639 | ] 640 | }, 641 | "metadata": {}, 642 | "output_type": "display_data" 643 | } 644 | ], 645 | "source": [ 646 | "# (1) Find observed values and partition into 6 cells\n", 647 | "# |early | on-time | late\n", 648 | "# -------------+-------+---------+------\n", 649 | "# first babies | | |\n", 650 | "# others | | |\n", 651 | "observed = np.zeros(6)\n", 652 | "observed[0] = len(live_births[(live_births[\"birthord\"] == 1) & \n", 653 | " (live_births[\"prglength\"] <= 37)][\"prglength\"].dropna())\n", 654 | "observed[1] = len(live_births[(live_births[\"birthord\"] == 1) & \n", 655 | " ((live_births[\"prglength\"] == 38) |\n", 656 | " (live_births[\"prglength\"] == 39) |\n", 657 | " (live_births[\"prglength\"] == 40))][\"prglength\"].dropna())\n", 658 | "observed[2] = len(live_births[(live_births[\"birthord\"] == 1) & \n", 659 | " (live_births[\"prglength\"] >= 41)][\"prglength\"].dropna())\n", 660 | "observed[3] = len(live_births[(live_births[\"birthord\"] != 1) & \n", 661 | " (live_births[\"prglength\"] <= 37)][\"prglength\"].dropna())\n", 662 | "observed[4] = len(live_births[(live_births[\"birthord\"] != 1) & \n", 663 | " ((live_births[\"prglength\"] == 38) |\n", 664 | " (live_births[\"prglength\"] == 39) |\n", 665 | " (live_births[\"prglength\"] == 40))][\"prglength\"].dropna())\n", 666 | "observed[5] = len(live_births[(live_births[\"birthord\"] != 1) & \n", 667 | " (live_births[\"prglength\"] >= 41)][\"prglength\"].dropna())\n", 668 | " \n", 669 | "# (2) compute expected values for each cell. H0 says distributions are \n", 670 | "# same, so we use pooled probabilities P(early), P(ontime), P(late)\n", 671 | "# Actual expected numbers are n*P(...) for first and m*P(...) for other\n", 672 | "nbr_first_babies = len(live_births[live_births[\"birthord\"] == 1].dropna())\n", 673 | "nbr_other_babies = len(live_births[live_births[\"birthord\"] != 1].dropna())\n", 674 | "nbr_pooled = nbr_first_babies + nbr_other_babies\n", 675 | "nbr_early_pooled = len(live_births[live_births[\"prglength\"] <= 37])\n", 676 | "nbr_ontime_pooled = len(live_births[(live_births[\"prglength\"] == 38) | \n", 677 | " (live_births[\"prglength\"] == 39) | \n", 678 | " (live_births[\"prglength\"] == 40)].dropna())\n", 679 | "nbr_late_pooled = len(live_births[live_births[\"prglength\"] >= 41].dropna())\n", 680 | "expected = np.zeros(6)\n", 681 | "expected[0] = nbr_first_babies * nbr_early_pooled / nbr_pooled\n", 682 | "expected[1] = nbr_first_babies * nbr_ontime_pooled / nbr_pooled\n", 683 | "expected[2] = nbr_first_babies * nbr_late_pooled / nbr_pooled\n", 684 | "expected[3] = nbr_other_babies * nbr_early_pooled / nbr_pooled\n", 685 | "expected[4] = nbr_other_babies * nbr_ontime_pooled / nbr_pooled\n", 686 | "expected[5] = nbr_other_babies * nbr_late_pooled / nbr_pooled\n", 687 | "\n", 688 | "# (3) compute deviation, ie, difference between observed and expected\n", 689 | "# (4) compute chi-squared statistic\n", 690 | "deviation = np.power(observed - expected, 2) / expected\n", 691 | "chi2 = np.sum(deviation)\n", 692 | "print(\"observed chi-squared value:\", chi2)\n", 693 | "\n", 694 | "# (5) use Monte Carlo simulation to compute p-value, ie the probability of\n", 695 | "# seeing a chi-squared value as high as that observed under H0.\n", 696 | "pool_data = np.array(live_births[\"prglength\"].dropna())\n", 697 | "pool_mean = np.mean(pool_data)\n", 698 | "pool_std = np.std(pool_data)\n", 699 | "first_data = np.array(live_births[live_births[\"birthord\"] == 1][\"prglength\"].dropna())\n", 700 | "first_mean = np.mean(first_data)\n", 701 | "first_std = np.std(first_data)\n", 702 | "other_data = np.array(live_births[live_births[\"birthord\"] != 1][\"prglength\"].dropna())\n", 703 | "other_mean = np.mean(other_data)\n", 704 | "other_std = np.std(other_data)\n", 705 | "np.random.seed(1)\n", 706 | "sample_chi2s = []\n", 707 | "for i in range(10000):\n", 708 | " pool_sample = np.random.normal(pool_mean, pool_std, nbr_pooled)\n", 709 | " first_sample = np.random.normal(first_mean, first_std, nbr_first_babies)\n", 710 | " other_sample = np.random.normal(other_mean, other_std, nbr_other_babies)\n", 711 | " # compute sample observed\n", 712 | " sample_obs = np.zeros(6)\n", 713 | " sample_obs[0] = np.where(first_sample <= 37)[0].shape[0]\n", 714 | " sample_obs[1] = np.where((first_sample == 38) | \n", 715 | " (first_sample == 39) | \n", 716 | " (first_sample == 40))[0].shape[0]\n", 717 | " sample_obs[2] = np.where(first_sample >= 41)[0].shape[0]\n", 718 | " sample_obs[3] = np.where(other_sample <= 37)[0].shape[0]\n", 719 | " sample_obs[4] = np.where((other_sample == 38) | \n", 720 | " (other_sample == 39) | \n", 721 | " (other_sample == 40))[0].shape[0]\n", 722 | " sample_obs[5] = np.where(other_sample >= 41)[0].shape[0]\n", 723 | " # compute sample expected\n", 724 | " sample_pool_early = np.where(pool_sample <= 37)[0].shape[0]\n", 725 | " sample_pool_ontime = np.where((pool_sample == 38) | \n", 726 | " (pool_sample == 39) | \n", 727 | " (pool_sample == 40))[0].shape[0]\n", 728 | " sample_pool_late = np.where(pool_sample >= 41)[0].shape[0]\n", 729 | " sample_exp = np.zeros(6)\n", 730 | " sample_exp[0] = nbr_first_babies * sample_pool_early / nbr_pooled\n", 731 | " sample_exp[1] = nbr_first_babies * sample_pool_ontime / nbr_pooled\n", 732 | " sample_exp[2] = nbr_first_babies * sample_pool_late / nbr_pooled\n", 733 | " sample_exp[3] = nbr_other_babies * sample_pool_early / nbr_pooled\n", 734 | " sample_exp[4] = nbr_other_babies * sample_pool_ontime / nbr_pooled\n", 735 | " sample_exp[5] = nbr_other_babies * sample_pool_late / nbr_pooled\n", 736 | " # compute chi-squared, denom has a slight delta added for divide-by-zero\n", 737 | " sample_dev = np.power(sample_obs - sample_exp, 2) / (sample_exp + 1e-9)\n", 738 | " sample_chi2s.append(np.sum(sample_dev))\n", 739 | "\n", 740 | "pmf_chi2 = np.histogram(np.array(sample_chi2s), bins=1000, normed=True)\n", 741 | "cdf_chi2_ps = np.cumsum(pmf_chi2[0])\n", 742 | "cdf_chi2_ps = cdf_chi2_ps / cdf_chi2_ps[-1]\n", 743 | "cdf_chi2_xs = pmf_chi2[1][:-1]\n", 744 | "\n", 745 | "plt.plot(cdf_chi2_xs, cdf_chi2_ps)\n", 746 | "plt.xlabel(\"chi-squared value\")\n", 747 | "plt.ylabel(\"cumulative probability\")" 748 | ] 749 | }, 750 | { 751 | "cell_type": "code", 752 | "execution_count": 14, 753 | "metadata": { 754 | "collapsed": false 755 | }, 756 | "outputs": [ 757 | { 758 | "name": "stdout", 759 | "output_type": "stream", 760 | "text": [ 761 | "p-value: 0.000 (stat sig @ 99%)\n" 762 | ] 763 | } 764 | ], 765 | "source": [ 766 | "p_value_idx = np.where(cdf_chi2_xs > chi2)[0]\n", 767 | "p_value = 0 if len(p_value_idx) == 0 else cdf_chi2_xs[p_value_idx]\n", 768 | "stat_sig = \"\"\n", 769 | "if p_value < 0.01:\n", 770 | " stat_sig = \"stat sig @ 99%\"\n", 771 | "elif p_value < 0.05:\n", 772 | " stat_sig = \"stat sig @ 95%\"\n", 773 | "else:\n", 774 | " stat_sig = \"not stat sig\"\n", 775 | "print(\"p-value: %.3f (%s)\" % (p_value, stat_sig))" 776 | ] 777 | }, 778 | { 779 | "cell_type": "markdown", 780 | "metadata": {}, 781 | "source": [ 782 | "__Exercise 7.6:__ Suppose you run a casino and you suspect that a customer has replaced a die provided by the casino with a “crooked die;” that is, one that has been tampered with to make one of the faces more likely to come up than the others. You apprehend the alleged cheater and confiscate the die, but now you have to prove that it is crooked.\n", 783 | "\n", 784 | "You roll the die 60 times and get the following results: (1, 8), (2, 9), (3, 19), (4, 6), (5, 8), (6, 10). \n", 785 | "\n", 786 | "What is the chi-squared statistic for these values? What is the probability of seeing a chi-squared value as large by chance?" 787 | ] 788 | }, 789 | { 790 | "cell_type": "code", 791 | "execution_count": 15, 792 | "metadata": { 793 | "collapsed": false 794 | }, 795 | "outputs": [ 796 | { 797 | "name": "stdout", 798 | "output_type": "stream", 799 | "text": [ 800 | "observed chi-squared value: 10.600\n", 801 | "p-value: 0.001 (stat sig @ 99%)\n" 802 | ] 803 | } 804 | ], 805 | "source": [ 806 | "# compute the chi-squared value from the observed data\n", 807 | "observed = np.array([8, 9, 19, 6, 8, 10])\n", 808 | "expected = np.array([10, 10, 10, 10, 10, 10]) # each face equally likely\n", 809 | "observed_chi2 = np.sum(np.power(observed - expected, 2) / expected)\n", 810 | "print(\"observed chi-squared value: %.3f\" % (observed_chi2))\n", 811 | "\n", 812 | "# compute a distribution of chi-squared values by sampling\n", 813 | "def sample_by_cdf(cdf_xs, cdf_ps, n):\n", 814 | " \"\"\" Create a sample based on a CDF \"\"\"\n", 815 | " samples = []\n", 816 | " for i in range(n):\n", 817 | " prob = np.random.random()\n", 818 | " for j in range(cdf_ps.shape[0]):\n", 819 | " if cdf_ps[j] > prob:\n", 820 | " break\n", 821 | " samples.append(cdf_xs[j])\n", 822 | " return samples\n", 823 | "\n", 824 | "pmf_obs = observed / 60\n", 825 | "cdf_obs_ps = np.cumsum(pmf_obs)\n", 826 | "cdf_obs_xs = np.array([1, 2, 3, 4, 5, 6])\n", 827 | "sample_chi2s = np.zeros(1000)\n", 828 | "for i in range(1000):\n", 829 | " sample_obs = sample_by_cdf(cdf_obs_xs, cdf_obs_ps, 1000)\n", 830 | " sample_exp = np.random.uniform(low=1, high=6, size=1000)\n", 831 | " sample_chi2s[i] = np.sum(np.power(sample_obs - sample_exp, 2) / sample_exp)\n", 832 | " \n", 833 | "# compute the CDF for the chi-squared values distribution\n", 834 | "pmf_chi2 = np.histogram(sample_chi2s, bins=1000, normed=True)\n", 835 | "cdf_chi2_ps = np.cumsum(pmf_chi2[0])\n", 836 | "cdf_chi2_ps = cdf_chi2_ps / cdf_chi2_ps[-1]\n", 837 | "cdf_chi2_xs = pmf_chi2[1][:-1]\n", 838 | "\n", 839 | "# compute p-value = prob(seeing a chi-squared value high as that observed)\n", 840 | "p_value_idx = np.where(cdf_chi2_xs >= observed_chi2)[0][0]\n", 841 | "p_value = cdf_chi2_ps[p_value_idx]\n", 842 | "statsig = \"\"\n", 843 | "if p_value < 0.01:\n", 844 | " statsig = \"stat sig @ 99%\"\n", 845 | "elif p_value < 0.05:\n", 846 | " statsig = \"stat sig @ 95%\"\n", 847 | "else:\n", 848 | " statsig = \"not stat sig\"\n", 849 | "print(\"p-value: %.3f (%s)\" % (p_value, statsig))" 850 | ] 851 | }, 852 | { 853 | "cell_type": "markdown", 854 | "metadata": {}, 855 | "source": [ 856 | "## Efficient Resampling\n", 857 | "\n", 858 | "In the __Testing Difference in Means__ problem above, we can also use the formulas for normal distribution to reduce some of the work.\n", 859 | "\n", 860 | "We have n=4413 first babies and m=4735 other babies. Our H0 is that they are from the same distribution. So if $\\mu$ and $\\sigma$ are the mean and standard deviation for the pooled distribution, then a sample mean from the first baby distribution is $\\mathcal{N}(\\mu, \\frac {\\sigma^2}{n})$ and one from the other baby distribution is $\\mathcal{N}(\\mu, \\frac {\\sigma^2}{m})$. Using the formula for the difference of means:\n", 861 | "\n", 862 | "$$X - Y \\sim \\mathcal{N}(\\mu_{X}-\\mu_{Y}, \\sigma_{X}^2+\\sigma_{Y}^2) \\sim \\mathcal{N}(0, (\\frac{1}{m}+\\frac{1}{n})\\sigma^{2})$$" 863 | ] 864 | }, 865 | { 866 | "cell_type": "code", 867 | "execution_count": 16, 868 | "metadata": { 869 | "collapsed": false 870 | }, 871 | "outputs": [ 872 | { 873 | "name": "stdout", 874 | "output_type": "stream", 875 | "text": [ 876 | "p-value (evidence for H0): 0.168\n" 877 | ] 878 | } 879 | ], 880 | "source": [ 881 | "diff_dist = scipy.stats.norm(loc=0, \n", 882 | " scale=math.sqrt(((1 / 4413) + (1 / 4735)) * \n", 883 | " math.pow(pool_std, 2)))\n", 884 | "left = diff_dist.cdf(-0.078)\n", 885 | "right = 1 - diff_dist.cdf(0.078)\n", 886 | "print(\"p-value (evidence for H0): %.3f\" % (left + right))" 887 | ] 888 | }, 889 | { 890 | "cell_type": "markdown", 891 | "metadata": {}, 892 | "source": [ 893 | "## Power\n", 894 | "\n", 895 | "Negative hypothesis test does not necessarily imply effect is not real. Depends on __power__ of test, defined as the probability that the test will be positive if the null hypothesis is false. In general, the power of a test depends on the sample size, the magnitude of the effect, and the threshold $\\alpha$.\n", 896 | "\n", 897 | "Interesting tool for [Statistical Power Analysis](https://amarder.github.io/power-analysis/).\n", 898 | "\n", 899 | "__Exercise 7.7:__ What is the power of the test above using $\\alpha$ = 0.05 and assuming that the actual difference between the means is 0.078 weeks? You can estimate power by generating random samples from distributions with the given difference in the mean, testing the observed difference in the mean, and counting the number of positive tests.\n", 900 | "\n", 901 | "What is the power of the test with $\\alpha$ = 0.10?" 902 | ] 903 | }, 904 | { 905 | "cell_type": "code", 906 | "execution_count": 17, 907 | "metadata": { 908 | "collapsed": false 909 | }, 910 | "outputs": [ 911 | { 912 | "name": "stdout", 913 | "output_type": "stream", 914 | "text": [ 915 | "power with alpha 0.05 = 0.008\n", 916 | "power with alpha 0.1 = 0.998\n" 917 | ] 918 | } 919 | ], 920 | "source": [ 921 | "# assume H0 = both distributions are identical. So from above\n", 922 | "nbr_pos_at_5 = 0\n", 923 | "nbr_pos_at_1 = 0\n", 924 | "for i in range(1000):\n", 925 | " diff_std = math.sqrt(((1 / 4413) + (1 / 4735)) * math.pow(pool_std, 2))\n", 926 | " # generate a diff sample\n", 927 | " diff_sample = np.random.normal(0, diff_std, 1000)\n", 928 | " # calculate PMF and CDF\n", 929 | " pmf_diff = np.histogram(diff_sample, bins=100, normed=True)\n", 930 | " cdf_diff_ps = np.cumsum(pmf_diff[0])\n", 931 | " cdf_diff_ps = cdf_diff_ps / cdf_diff_ps[-1]\n", 932 | " cdf_diff_xs = pmf_diff[1][:-1]\n", 933 | " # calculate p-value\n", 934 | " left = np.where(cdf_diff_xs <= -0.078)[0][0]\n", 935 | " right = np.where(cdf_diff_xs >= 0.078)[0][0]\n", 936 | " p_left = cdf_diff_ps[left]\n", 937 | " p_right = 1.0 - cdf_diff_ps[right]\n", 938 | " p_value = p_left + p_right\n", 939 | " # null hypothesis is false if p-value < alpha\n", 940 | " # test is positive if null hypothesis is false\n", 941 | " if p_value < 0.05:\n", 942 | " nbr_pos_at_5 += 1\n", 943 | " if p_value < 0.1:\n", 944 | " nbr_pos_at_1 += 1\n", 945 | "print(\"power with alpha 0.05 = %.3f\" % (nbr_pos_at_5 / 1000))\n", 946 | "print(\"power with alpha 0.1 = %.3f\" % (nbr_pos_at_1 / 1000))\n", 947 | "# so null hypothesis is false at both levels, and there is a difference" 948 | ] 949 | }, 950 | { 951 | "cell_type": "code", 952 | "execution_count": null, 953 | "metadata": { 954 | "collapsed": true 955 | }, 956 | "outputs": [], 957 | "source": [] 958 | } 959 | ], 960 | "metadata": { 961 | "kernelspec": { 962 | "display_name": "Python 2", 963 | "language": "python", 964 | "name": "python2" 965 | }, 966 | "language_info": { 967 | "codemirror_mode": { 968 | "name": "ipython", 969 | "version": 2 970 | }, 971 | "file_extension": ".py", 972 | "mimetype": "text/x-python", 973 | "name": "python", 974 | "nbconvert_exporter": "python", 975 | "pygments_lexer": "ipython2", 976 | "version": "2.7.11" 977 | } 978 | }, 979 | "nbformat": 4, 980 | "nbformat_minor": 0 981 | } 982 | --------------------------------------------------------------------------------