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This text explores the challenges and applications of blind visual analysis, including the use of randomization. It presents examples of different design structures and the probability of correctly identifying treatment assignments.
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Challenges and Applications in Visual Analysis • Use of Randomization in Visual Analysis • Blind Visual Analysis Procedures
Example 1: Suppose there were 6 blocks of 5 observations and that three blocks were randomly assigned to Condition B C = 6!/((6-3)!3!) = 20 possible assignments AAABBB BBBAAA BAABBA ABBAAB BAAABB ABBBAA ABABBA BABAAB ABAABB BABBAA AABBBA BBAAAB AABABB BBABAA BAABAB ABBABA BBAABA AABBAB ABABAB BABABA
We could graph the data 20 different ways, one corresponding to each possible assignment. The visual could then be presented with this set of graphs with the task of determining which of the random assignments had been selected. Doing so keeps the visual analyst blind to the actual treatment assignment and allows us to control the Type I error rate.
Alternatively, the blind visual analyst could be presented with a single graph with the task of figuring out which blocks of observations were assigned to which condition.
If the treatment has no effect the same data series would be obtained for any of the possible treatment assignments. What is the probability that the assignment BAABBA would be randomly selected? p = 1/20 = .05
Example 2: Suppose that the researcher wants a reversal design with at least 3 opportunities to show and effect (ABAB type structure) and at least 5 observations per phase. Suppose they also want to use randomization and a “blind” visual analyst. Could imagine the following design: A ? B ? A ? B ? ?
A ? B ? A ? B ? ? How many possible assignments? 2 X 2 X 2 X 2 X 2 = 32 If the treatment has no effect, the probability that a “blind” visual analyst could identify each of the ? p = 1/32 = .03125
A ? ? ? A B B
Example 3: Suppose that the researcher conducts a multiple-baseline design with 4 participants. Suppose they also want to use randomization and a “blind” visual analyst. How many possible assignments of participants to treatment order? Who is 1st, 2nd, 3rd, and 4th? P = 4! = 24 possible assignments If the treatment has no effect, the probability that a “blind” visual analyst could identify the correct order p = 1/24 = .0417