Chapter 6Bernoulli Trials Revisited6.1 Is p Constant?Recall that there are three assumptions to BT. The first, that each trial yields a dichotomy, is easyto verify or refute. The other two assumptions can neither be verified nor refuted with certainty.In this Section we present methods for investigating the second assumption, that the probability ofsuccess remains constant. Our methods will consist of informal descriptive techniques and formaltests of hypotheses. Both of these categories of methods are very useful to a scientist.I begin with an extended example of data I collected circa 1990 on Tetris.Tetris is a video game (I fear that this expression—video game—is hopelessly dated; perhapsyou can give me a better name to use for the next version of these notes!) that can be played on avariety of ‘systems.’ If you have never played Tetris, then the following description of the gamemay be difficult to follow. Tetris rewards spatial reasoning and, not surprisingly, lightning reflexes.Essentially, one of seven possible geometrical shapes of four blocks ‘falls’ from the top of thescreen. The player translates and/or rotates the falling shape in an attempt to complete, with nogaps, a horizontal row of blocks. Each completed row of blocks disappears from the screen andblocks above, if any, drop down into the vacated space to allow room for more falling shapes. Aplayer’s score equals the number of rows completed before the screen overflows, which means thatthe current game is over. After every ten completed lines the speed of the falling shapes increasesmaking the game more difficult.Each drop of a shape by the computer can be viewed as a trial. Each trial has seven possibleoutcomes. For now, the shapes are divided into two types: a straight row (my favorite, labeled asuccess) and any other shape (a failure).I observed 1,872 trials during eight games of Tetris on my son’s Nintendo System. The methodsI present below are based on taking the total number of trials, 1,872 in the current case, and dividingthem into two or more segments. The choice of segments is a matter of taste and judgment andunless I have a good reason to do otherwise, I advocate dividing the data into segments of equalor nearly equal length, and I usually advocate having two segments of trials. For these Tetris data,however, there is a structure that I want my analysis to reflect, namely, every time a game endsthe system is ‘reset’ before the next game. (There was literally a button that said ‘Reset.’) Thus,65Table 6.1: A Comparison of Eight Games of Tetris.Frequencies Row ProportionsGame S F Total S F TotalFirst 22 171 193 0.114 0.886 1.000Second 33 185 218 0.151 0.849 1.000Third 36 215 251 0.143 0.857 1.000Fourth 25 206 231 0.108 0.892 1.000Fifth 30 198 228 0.132 0.868 1.000Sixth 42 220 262 0.160 0.840 1.000Seventh 33 215 248 0.133 0.867 1.000Eighth 33 208 241 0.137 0.863 1.000Total 254 1618 1872 0.136 0.864 1.000Figure 6.1: Plot of Proportion of Success for Eight Games of Tetris.••••••••..............................................................................................................................................................................................................................................................................................................................................................1 2345 6 7 8 Game0.000.040.080.120.16I divide my 1,872 trials into eight segments, with each segment consisting of the trials of a game.The outcomes, by game, are given in Table 6.1 and graphed in Figure 6.1. There is some variationin the proportion of successes from game to game, but nothing remarkable. Thus, I conclude,descriptively, that there is little evidence against the second assumption of Bernoulli Trials.Actually, I have gotten a little ahead of myself in telling this tale. I need to introduce somenotation. If we have BT, then there is a single p, which may be known or unknown to the researcher.We want, however, to investigate whether (or not) it is reasonable to make this assumption. Wedivide our data in r ≥ 2 segments. For the Tetris data, I have chosen r = 8 segments. I assumethat within each segment we have BT and that the probability of success in segment i is given bypi. Thus, for example, I assume that each of my eight games of Tetris were BT with game i haveprobability of success given by pi. And now here is the key idea. If all the pi’s are the same number,then we do have a single probability of success for the total of 1,872 trials; i.e. we have BT. But ifeven one pair of pi’s do not agree in value, then we do not have BT for the entire sequence.Thus, we want to decide whether or not all the pi’s are the same number. This suggests per-66forming a test of hypotheses and we will do that now.The null hypothesis isH0: p1= p2= . . . = pr.The alternative hypothesis isH1: Not H0; i.e. there exists at least one pair, i and j, such that pi6= pj.Of course, the researcher selects a significance level α.The test statistic is very similar to what we had in the previous chapter for the Goodness of FitTest. The first similar feature is that we once again have O ’s and E’s. The O’s are, not surprisingly,the observed counts in Table 6.1; i.e. the 16 counts in the S and F columns. The E’s, of course,are the expected values of those counts, calculated on the assumption that the null hypothesis istrue.Well, the null hypothesis states that the 1,872 trials all have the same p, but the p is unknown tothe researcher. Thus, we need to estimate that p. We want to estimate it as accurately as possible,so we use all 1,872 trials and get:ˆp = 254/1872 = 0.136.We now use this estimated p to obtain our E’s. For example, consider the upper left cell in thetable: the cell that counts the number of successes in the first game. There are 193 trials in game 1,so the expected number of successes is 193p which we estimate to be 193ˆp = 193(0.136) = 26.2,compared to O = 22 for this cell. There is another way to view the computation of E and thisalternate way is very helpful to us. Staying in the upper left cell, we see thatE = 193(0.136) = 193(254)/1872 = 26.2.Here is the key idea: Notice that the E for the upper left cell is obtained by first calculating theproduct of the row total for that cell (193) and the column total for that cell (254). Then we takethis product and divide by the grand total in the table (1872); in words E is row total
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