Chapter 3 Randomization Probability and Sampling Distributions 3 1 Assignments and Randomization Recall that Dawn s study of her cat Bob was presented in Chapter 1 Table 3 1 presents her data Reading from this table we see that in Dawn s study the chicken flavored treats were presented to Bob on days trials 1 5 7 8 9 11 13 15 16 and 18 Why did she choose these days How did she choose these days It will be easier to begin with the How question We have been looking at the data collected by Dawn We have listed the observations separated them by treatment sorted them within treatment and within treatments drawn dot plots and computed means medians variances and standard deviations But now we need to get into our time machine and travel back in time to before Dawn collected her data We go back to when Dawn had her study largely planned treatments selected trials defined response specified and the decision to have a balanced study with 20 trials We are at the point where Dawn pondered Which 10 trials should have chicken flavored treats assigned to them How should I decide Table 3 1 Dawn s data on Bob s consumption of cat treats C T is for chicken tuna flavored Day Flavor Number Consumed 1 C 4 2 T 3 Day 11 12 Flavor C T Number Consumed 6 3 3 T 5 4 T 0 5 C 5 6 T 4 13 14 15 16 C T C C 7 1 3 6 55 7 C 5 8 C 6 9 10 C T 1 7 17 18 19 20 T C T T 3 8 1 2 The answer is that Dawn did this by using a process called randomization I will explain what randomization is by showing you three equivalent ways to randomize First some terminology We call the list of 10 trials above an assignment of treatments to trials It tells us which trials were assigned to the first treatment chicken It also implies which trials were assigned to the second treatment tuna namely all of the trials not listed above If we are going to study assignments and we are it is easier if we make our assignments as simple to display as possible Thus an assignment will be presented by listing the trials that it assigns to treatment 1 A natural question is How many different assignments were possible for Dawn s study The answer is 184 756 I will give a brief digression into how I obtained this number You might recall from math the expression m which is read em factorial If m is a positive integer 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 later in these notes 0 1 Finally for any other value of m negatives non integers the expression m is not defined We have the following result You don t need to worry about proving it it is a given in these notes Result 3 1 The number of possible assignments For a total of n n1 n2 units the number of possible assignments of two treatments to the units with n1 units assigned to treatment 1 and the remaining n2 units assigned to treatment 2 is n n1 n2 3 2 I will evaluate 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 is 6 5 4 6 20 3 3 3 2 1 Notice that it is always possible to reduce the amount of arithmetic we do by canceling some terms in the numerator and denominator In particular the 6 in the numerator can be written as 6 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 is 10 10 9 8 7 6 252 5 5 5 4 3 2 1 For Dawn s study n 20 and n1 n2 10 Thus the number of possible assignments is 20 184 756 10 10 Notice that for Cathy s and Kymn s study I determined the answer by hand because the numbers are small enough to handle easily Dawn s study is trickier Many of you perhaps most perhaps all will consider it easy to determine the answer 184 756 But I will not require you to do so As a guide I will never have you evaluate m for any m 10 Sara s study is a real challenge The number of possible assignments is 80 40 40 This answer to four significant digits is 1 075 1023 Don t worry about how I obtained this answer If this issue however keeps you awake at night then send me an email and I will tell you If enough people email me then I will put a brief explanation in the next version of these Course Notes I now will describe three ways two physical and one electronic that Dawn could have performed her randomization 1 A box with 20 cards Take 20 cards of the same size shape texture etc and number them 1 2 20 with one number to each card Place the cards in a box mix the cards thoroughly and select 10 cards at random without replacement The numbers on the cards selected denoted the trials that will be assigned treatment 1 2 A deck of 20 ordinary playing cards This method is especially suited for units that are trials We need to have 10 black cards spades or clubs and 10 red cards diamonds or hearts We don t care about the rank ace king 3 etc of the cards The cards are thoroughly shuffled and placed in a pile face down Before each trial select the top card from the pile if it is a black card then treatment 1 is assigned to the trial if it is a red card then treatment 2 is assigned to the trial The selected card is set aside and the above process is repeated for the remaining trials 3 Using a website This method will be explained in Section 3 5 later in this chapter Are you familiar with the term black box I like the definition in Wikipedia http en wikipedia org wiki Black box 57 which is In science and engineering a black box is a device system or object which can be viewed solely in terms of its input output and transfer characteristics without any knowledge of its internal workings that is its implementation is opaque black Almost anything might be referred to as a black box a transistor an algorithm or the human mind Our website for randomization is a black box It executes a computer program that supposedly is mathematically equivalent to my two methods of randomization that involve using …
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