1 / 12

Resampling Methods

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.

tracey
Download Presentation

Resampling Methods

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript


  1. 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.

  2. There are three kinds of resampling methods: Permutationmethods – 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.

  3. Correlation r = 0.36 100 80 IQ Twin 2 60 60 80 100 120 IQ Twin 1 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) 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.

  4. 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 X Y’ X Y’ X Y’ 97 89 89 87 81 91 85 87 70 59 105 88 81 97 107 94 84 45 93 77 58 73 69 84 99 74 70 79 89 105 75 92 78 84 60 64 61 77 95 84 79 72 68 74 69 85 93 105 79 43 97 59 89 91 81 45 85 72 70 43 105 84 81 74 107 77 84 105 93 64 58 87 69 73 99 79 70 89 89 84 75 77 78 87 60 84 61 94 95 92 79 97 68 88 69 85 93 74 79 105 97 79 89 85 81 88 85 84 70 77 105 84 81 105 107 72 84 43 93 77 58 73 69 97 99 89 70 92 89 45 75 64 78 91 60 94 61 87 95 74 79 74 68 84 69 87 93 59 79 105 97 64 89 72 81 105 85 73 70 91 105 97 81 84 107 92 84 77 93 74 58 77 69 59 99 85 70 105 89 84 75 43 78 74 60 84 61 45 95 87 79 94 68 89 69 87 93 79 79 88 … r = .36 r = -.26 r = .20 r = -.12

  5. 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= 0.11 r=-0.12 r=-0.19 r=-0.01 r=-0.47 r=-0.34 r=-0.18 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.

  6. After 100000 reps, Pr(r> 0.36)= 0.0378 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 -0.6 -0.4 -0.2 0 0.2 0.4 0.6 Permuted correlation (r)

  7. 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

  8. 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 Example of 20 simulations. Each row is a throw of 40 dice. The last column is the number of throws that equal the mode.

  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.

  10. 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 20 25 Response Time (sec)

  11. median = 2.22 median = 2.23 median = 2.22 0 5 10 15 20 25 0 5 10 15 20 25 0 5 10 15 20 25 median = 2.20 median = 2.10 median = 2.21 0 5 10 15 20 25 0 5 10 15 20 25 0 5 10 15 20 25 median = 2.08 median = 2.19 median = 1.98 0 5 10 15 20 25 0 5 10 15 20 25 0 5 10 15 20 25 median = 2.21 median = 2.07 median = 2.22 0 5 10 15 20 25 0 5 10 15 20 25 0 5 10 15 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. For our example, we can count the proportion of times that the median falls below 2.

  12. After 1000000 reps, Pr(median < 2.00)= 0.0620 1.6 1.8 2 2.2 2.4 2.6 Bootstrapped median 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.

More Related