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UW-Madison STAT 371 - Ch. 4

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Approximating a Sampling DistributionTwo Computer Simulation ExperimentsHow Good are These Approximations?A Warning about Simulation ExperimentsComputingSummaryPractice ProblemSolution to Practice ProblemHomework ProblemsChapter 4Approximating a Sampling DistributionAt the end of the last chapter, we saw how tedious it is to find the sampli ng distribution of U evenwhen t here are only 20 pos sible assignments. We also experienced the limit of my comfort zon e:252 possible assignments. For studies like Dawn’s (184,756 possible assignments) and especiallySara’s (1.075 × 1023possible assignments) there are way too many possible assignments to s eekan exact answer. Fortunately, there is an extremely simple way to obtain a good approxi mation—subject to the caveats given below—to a sampling distribution regardless of how large the numberof possible assig nments.4.1 Two Computer Simulation ExperimentsLet’s return to Dawn’s study. Our goal is to create a table for Dawn that is analogous to Table 3.7on page 65 for Kymn’s study; i.e. we want to determine the value of u = ¯x − ¯y for every one ofthe 184,756 possible assignments. This is too big of a job for me!Instead of looking at all possible assignments, we look at some of them. We do this with acomputer simulation experiment. I wrote a computer program that selected 10,000 assignmentsfor Dawn’s study. For each selection , the program selected one assignment at rando m from the col-lection of 184,756 possi ble assignm ents. (You can visualize this as using our randomizer website10,000 times.) For each of the 10,000 simulated assignments, I det erm ined its value of u = ¯x − ¯y.My results are s ummarized in Table 4.1.Suppose that we want to know P (U = 0). By definition, it i s the proportion of the (184,756)possible assignments that would yield u = 0. We do not know this proportion because we havenot looked at all possible assignments. But we have—with the help of my computer—looked at10,000 assignments; of these 10,000 assignments, 732 gave u = 0 (see Table 4.1). The relativefrequency of u = 0 in the assignments we have examined is an in tuitively obvious approximationto the relative frequency of u = 0 among all possible assignments. The relative frequency of u = 0among all possibl e assig nments is, by definition, P (U = 0). To summarize, our approximation ofthe unknown P (U = 0) is the relative frequency of assignments that gave u = 0; which is, fromour table, 0.0732.The above argument for P (U = 0) can be extended to P ( U = u) for any of the possi ble75Table 4.1: The results of a 10,000 run simul at ion experiment for Dawn’s study.RelativeRelative Relativeu Freq. Freq. u Freq. Freq. u Freq. Freq.−3.6 1 0.0001 −1.2 419 0.0419 1.2 39 4 0.039 4−3.4 1 0.0001−1.0 506 0.050 6 1 .4 315 0.0315−3.2 3 0.0003 −0.8 552 0.0552 1.6 25 1 0.025 1−3.0 4 0.0004 −0.6 662 0.0662 1.8 15 0 0.015 0−2.8 16 0.0016−0.4 717 0.071 7 2 .0 108 0.0108−2.6 25 0.0025 −0.2 729 0.0729 2.2 93 0.0093−2.4 45 0.0045 0.0 732 0.0732 2.4 54 0.0054−2.2 78 0.00780.2 765 0.0765 2.6 23 0.0023−2.0 113 0.011 3 0.4 71 6 0.0716 2.8 17 0.0017−1.8 191 0.019 10.6 674 0.0674 3.0 8 0.0008−1.6 240 0.024 0 0.8 55 3 0.0553 3.2 3 0.0003−1.4 335 0.033 5 1.0 50 7 0.0507Table 4.2: Selected probabiliti es of interest for Dawn’s CRD and their approximations.Probability of Interest Its App roximationP (U ≥ 2.2) r.f. (U ≥ 2.2) = 0.019 8P (U ≤ 2.2) r.f. (U ≤ 2.2) = 1 − 0.0105 = 0.9895P (|U| ≥ 2.2) r.f. (|U| ≥ 2.2) = 0.0198 + 0.0173 = 0.0 371values u. But it will actually be more interesting to us to approximate probabilities of more com-plicated events than P (U = u). In particular, recall that Dawn’s actual u was 2.2. As we will seein Chapter 5, we will be interested in one or more of the probabilities given in Table 4.2. Let megive you some details on how the answers in this table were obtained.To o btain r.f. (U ≥ 2.2) we must sum the relative frequencies for the values 2.2, 2.4, 2.6, 2.8,3.0 and 3.2. From Table 4.1, we o btain0.0093 + 0.0054 + 0.00 23 + 0.0017 + 0.0008 + 0.0003 = 0.019 8.For r.f. (U ≤ 2.2) note that this value is 1− r.f. (U > 2.2) = 1 − 0.0105. Finally, r.f. (|U| ≥ 2.2) isthe sum of two relative frequencies: (U ≥ 2.2) and (U ≤ −2.2). The first of these has been foundto equal 0.0 198. The second of these is0.0078 + 0.0045 + 0.00 25 + 0.0016 + 0.0004 + 0.0003 + 0 .0001 + 0.00 01 = 0.0173.76Table 4.3: Selected probabiliti es of interest for Sara’s CRD and their approximations.Probability of InterestIts ApproximationP (U ≥ 8.700) r.f. (U ≥ 8.700) = 0.0903P (U ≤ 8.700) r.f. (U ≤ 8.700) = 0.9107P (|U| ≥ 8.700) r.f. (|U| ≥ 8.700) = 0.0903 + 0.0921 = 0.1824Adding these we get,r.f. (U ≥ 2.2) + r.f. (U ≤ −2.2) = 0.0198 + 0.0173 = 0.0371 .Next, I performed a computer simulation experiment for Sara’s CRD. As s tated earlier, thereare more than 1023different assignments for a b al anced stu d y with n = 80 total trials. Trying toenumerate all of these would be ri diculous, so we will use a computer simulation w ith 10,000 runs.My simulat ion study yielded 723 distinct values of u! T his is way too many to present in atable as I did for Dawn’s study. (The simulation study for Dawn, recall, yielded 35 distinct valuesfor u and that was unw ieldy.)Recall that for Sara’s data¯x = 106.875 and ¯y = 98.175, giving u = 8.700.Table 4.3 presents information that will be needed in Chapter 5.4.2 How Good are These Approximations?Tables 4.2 and 4.3 present six unknown probabilities and their respective approximations based onsimulatio n experiments with 10,000 runs. Lacking knowledge of the exact probabilities I cannotsay exactly how good any of th ese approximations are. What I can say, however, is that each ofthem is very likely to be very close to the exact (unknown) probability it is approximating. How canI know this? Wel l, we will see how later in this course when we learn about confidence intervals,so yo u will n eed to be patient.And, of course, the terms very likely and very close are quite vague. Here is what we willdo for now. First, the expression very close will be replaced by a specific n u mber, call it h, thatis comp uted from our simulation resul ts. In words, it is very likely that the simulation stud yapproximation is within h of its exact probability. (If this is confusing, see the numerical examplesbelow.)Next, the expression very likely will be


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