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Resampling Methods From Wikipedia: “Parametric statistics is a branch of statistics that assumes (that) data come from a type of probability distribution and makes inferences about the parameters of the distribution. Most well-known elementary statistical methods (e.g. the ones from our class) are parametric.” But there are alternative methods that don’t require any assumptions about the shape of the population’s probability distribution. Resampling methods are an example. There are three kinds of resampling methods: Permutation methods – used most commonly with correlations where the probability of the observed data is estimated by comparing the observed parings to a large number of random parings of the data. Monte Carlo methods – estimate the population probability distribution through simulation. Bootstrap methods – the population distribution of an observed statistic is estimated by repeatedly resampling the data with replacement and calculating the statistic. Example of a permutation method: Suppose you measured the IQ’s of 25 pairs of twins and found a correlation of r=0.36. The scatter plot of your data is shown below. Is the observed correlation significantly greater than zero? (use a = .01) Correlation r = 0.36 IQ Twin 2 100 80 60 60 80 100 IQ Twin 1 120 The (parametric) test used in our class would have found an rcrit value of 0.330 We would reject H0 and conclude that a correlation 0.36 is (barely) significantly greater than zero. The distribution under the null hypothesis can be estimated by repeatedly shuffling (or ‘permuting’) the relationship between the X and Y values and calculating the correlation: X Y 97 89 81 85 70 105 81 107 84 93 58 69 99 70 89 75 78 60 61 95 79 68 69 93 79 89 87 91 87 59 88 97 94 45 77 73 84 74 79 105 92 84 64 77 84 72 74 85 105 43 r = .36 X Y’ 97 89 81 85 70 105 81 107 84 93 58 69 99 70 89 75 78 60 61 95 79 68 69 93 79 59 91 45 72 43 84 74 77 105 64 87 73 79 89 84 77 87 84 94 92 97 88 85 74 105 r = -.12 X Y’ 97 89 81 85 70 105 81 107 84 93 58 69 99 70 89 75 78 60 61 95 79 68 69 93 79 79 85 88 84 77 84 105 72 43 77 73 97 89 92 45 64 91 94 87 74 74 84 87 59 105 r = -.26 X Y’ 97 89 81 85 70 105 81 107 84 93 58 69 99 70 89 75 78 60 61 95 79 68 69 93 79 64 72 105 73 91 97 84 92 77 74 77 59 85 105 84 43 74 84 45 87 94 89 87 79 88 r = .20 … This generates a distribution of correlations that should be centered around zero. r= 0.05 r=-0.01 r=-0.00 r=-0.22 r= 0.25 r= 0.05 r=-0.32 r=-0.47 r=-0.34 r=-0.18 r= 0.11 r=-0.12 r=-0.19 r=-0.01 r=-0.01 r= 0.31 r=-0.20 r=-0.25 r= 0.15 r=-0.37 r=-0.11 r=-0.24 r=-0.38 r=-0.36 r=-0.26 r=-0.30 r=-0.09 r=-0.24 r= 0.07 r= 0.05 r= 0.13 r=-0.05 r=-0.16 r= 0.02 r=-0.17 We can then use this distribution to calculate the probability of making our observed sample correlation. After 100000 reps, Pr(r> 0.36)= 0.0378 -0.6 -0.4 -0.2 0 0.2 0.4 Permuted correlation (r) 0.6 Only 3.78% of the correlations generated by permutation exceeds the observed correlation of 0.36, so we’d reject the null hypothesis using a = .05 Example of a Monte Carlo simulation: Liar’s dice This is a game where n players roll 40 6-sided dice and keep the outcome hidden under their own separate cups. The goal is to guess how many dice equal the mode. After a player makes a guess, the next player must decide if the guess is too high, or otherwise guess a higher number. If it is decided that the guess is too high, the cups are lifted and the number of dice equal to the mode is computed. If the he/she wins and the player that made the guess must drink (lemonade). Suppose there are eight players, each with 5 dice. The player to your right just guessed that the modal value is 14. What is the probability that the mode of the 40 dice is that high or higher? Here’s an example of 40 throws. The mode is 5, and 10 of these throws equals the mode. mode # 3 13 Example of 20 simulations. Each row is a throw of 40 dice. The last column is the number of throws that equal the mode. rep # mode # 1 5 12 2 1 8 3 1 8 4 2 9 5 2 11 6 3 9 7 2 10 8 3 12 9 3 8 10 3 9 11 2 10 12 4 12 13 6 13 14 2 9 15 2 11 16 1 11 17 6 10 18 2 12 19 3 10 20 2 9 A computer simulation of one million rolls generated this histogram. Shown in red are the examples when the number of dice equal to the mode is 14 or higher. Only 2.31% of the simulations found a count of 14 or higher. This small number means that the player should ask all players to lift their cups and calculate the value. Percent of rolls 30 20 10 0 10 15 Mode of 40 dice 20 Third method of resampling: bootstrapping to conduct a hypothesis test on medians. Suppose you measured the amount of time it takes for a subject to perform a simple mental rotation. Previous research shows that it should take a median of 2 seconds to conduct this task. Your subject conducts 500 trials and generates the distribution of response times below, which has a median of 2.15 seconds. Is this number significantly greater than 2? (use a = .05) median = 2.15 (sec) 0 5 10 15 Response Time (sec) 20 25 The trick to bootstrapping is to generate an estimate of the sampling distribution of your observed statistic by repeatedly sampling the data with replacement and recalculating the statistic. median = 2.22 0 5 10 15 20 median = 2.23 25 0 5 10 15 20 median = 2.22 25 0 5 10 15 20 25 median = 2.20 median = 2.10 0 5 10 15 20 median = 2.21 25 0 5 10 15 20 25 0 5 10 15 20 25 median = 2.08 median = 2.19 0 5 10 15 20 median = 1.98 25 0 5 10 median = 2.21 0 5 10 15 20 15 20 25 0 5 10 median = 2.07 25 0 5 10 15 20 15 20 25 median = 2.22 25 0 5 10 15 20 25 For our example, we can count the proportion of times that the median falls below 2. After 1000000 reps, Pr(median < 2.00)= 0.0620 1.6 1.8 2 2.2 2.4 Bootstrapped median 2.6 Since more than 5% of our bootstrapped medians fall below 2, we (just barely) cannot conclude that our observed median is significantly greater than 2.