Randomization, Probability and Sampling DistributionsAssignments and RandomizationThe Skeptic's ArgumentThe Sampling Distribution of the Test Statistic for Cathy's StudyThe Sampling Distribution of U for Kymn's CRDComputingSummaryPractice ProblemsSolutions to Practice ProblemsHomework ProblemsChapter 3Randomization, Probability and SamplingDistributions3.1 Assignments and Randomiz ationRecall that Dawn’s study of h er cat Bob was presented in Chapter 1. Table 3.1 presents her data.Reading from this table, we s ee that in Dawn’s study, the chicken-flavored treats were presented toBob o n days (trials):1, 5, 7, 8, 9, 11, 13, 15, 16 and 18.Why did she choose these days? How di d she choose these days? It will be easier to begin withthe ‘How ’ question.We have been looking at the data collected by Dawn. We have li sted the ob servations; separatedthem by treatment; sorted them within treatment; and, within treatments, drawn dot plots andcomputed means, medians, variances and standard deviations. But now we need to get into ourtime machine and travel back in time to before Dawn collected her data. We go back to whenDawn had her study largely planned: treatments selected; trials defined; response specified; andthe decision to have a balanced study w ith 20 trials. We are at the point where Dawn pondered,“Which 10 trials should have chicken-flavored treats assigned to them ? How sho uld I decide?”Table 3.1: Dawn’s data on Bob’s consumption of cat treats. ‘C’ [‘T’] is for chicken [tuna] flavored.Day: 1 2 3 4 5 6 7 8 9 10Flavor: C T T T C T C C C TNumber Consumed: 4 3 5 0 5 4 5 6 1 7Day: 11 12 13 14 15 16 17 18 19 20Flavor: C T C T C C T C T TNumber Consumed: 6 3 7 1 3 6 3 8 1 255The answer is that Dawn did this by using a process called randomization. I wil l explain whatrandomization is by showing you three equivalent ways to randomize.First, some terminology. We call th e list of 10 trials above an assignment of treatments totrials. It tells us which trials were assigned to the first treatment (chicken). It also implies whichtrials were assigned to the second treatment (tuna); namely, all of t he t ri al s not li sted above. If weare going to study assignments—and we are—it is easier if we make our assignments as simple todisplay as possible. Thus, an assignment will be presented by listing the trials that it assignsto treatment 1.A natural question is: How many different assignments were possible for Dawn’s study? Theanswer is 184,7 56. I will give a brief d igression into how I ob tained thi s number.You might recall from math the expression m!, which is read em-factorial. If m is a posit iveinteger, then this expression is defined as:m! = m(m − 1)(m − 2) · · · 1 (3.1)Thus, for example,1! = 1; 2! = 2(1) = 2; 3! = 3(2)(1) = 6; and so on.By special definition (which will allow us to write more easily certain formulas that will arise laterin these notes), 0! = 1. Finally, for any other value o f m (negatives, non-integers), the expressionm! is not defined.We have the following result. You don’t need to worry about proving i t; it is a given in thesenotes.Result 3. 1 (The number o f possible assignments.) For a total of n = n1+ n2units, the numberof pos sible assignments of two treatments to the units, with n1units assigned to treatment 1 andthe remaining n2units assigned to treatment 2, isn!n1!n2!(3.2)I will eval uate Equation 3.2 for three of the studies presented in Chapters 1 and 2.• For Cathy’s study, n = 6 and n1= n2= 3. Thus, the number of possible assignments is6!3!3!=6(5)(4)3(2)(1)= 20.Notice t hat it is always possible to reduce the amount of arithm et ic we do b y canceling someterms in the numerator and denominator. In particular, th e 6 ! in the numerator can be writtenas6(5)(4)3!and its 3! cancels a 3 ! in the denominator.56• For Kymn’s study, n = 10 and n1= n2= 5. Thus, the number of possible assignments is10!5!5!=10(9)(8)(7 ) (6)5(4)(3)(2)( 1)= 252.• For Dawn’s study, n = 20 and n1= n2= 10. Thus, the number of po ssible ass ignments is20!10!10!= 184,756.Notice th at for Cathy’s and Kymn’s study, I determined t he answer by hand because the numb ersare small enough to handle easily. Dawn’s study is trickier. Many of you, perhaps mos t, perhapsall, will cons ider it easy to determi n e the answer: 184,756. But I w ill not require y o u to do so. Asa guide, I will never have you evaluate m! for any m > 10.Sara’s study is a real challenge. The number of possible assignments i s80!40!40!.This ans w er, to four significant dig its, is1.075 × 1023.Don’t worry about h ow I obtained this answ er. If this issue, however, keeps you awake at night,then send me an email and I wil l tell you. If enough people emai l me, then I wil l put a briefexplanation in the next version of t hese Course Notes.I now will describe three ways—two physical and one electronic—th at Dawn could have p er-formed her randomization.1. A box with 20 cards. Take 20 cards of the s ame s ize, shape, texture, etc. and numberthem 1, 2, . . . , 20, with one numb er to each card. Place the cards in a bo x; mix the cardsthoroughly and sel ect 10 cards at random wi thout replacement. The nu mbers on th e cardsselected denoted the trials that will be assigned treatm ent 1.2. A deck of 20 ordinary playing cards. (This method is especially suited for units thatare tri al s.) We n eed to have 10 black cards (spades or clubs) and 10 red cards (diamondsor hearts). We d on’t care about the rank (ace, king, 3, etc.) of the cards. The cards arethoroughly shuffled and placed in a pile, face down. Before each tri al select the top cardfrom the pile; if it is a black card, then treatm ent 1 is assigned to the tri al ; if it is a red card,then treatment 2 is assigned to the trial. The selected card is set aside and the above processis repeated for the remaining tri al s.3. Using a website. This method w ill be explained in Section 3.5 later in this chapter.Are yo u familiar with the term black box? I like the definition in Wikipediahttp://en.wikipedia.org/wiki/Black_box57which is :In science and engineering, a black box is a device, sys tem or object which can beviewed solely in terms of its input, output and transfer characteristics without anyknowledge of its internal workings, t h at is, i ts implementation is ”opaque” (black).Almost anything might be referred to as a black box: a transistor, an al gorithm, or thehuman mind.Our website for randomization
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