A 23 experiment was performed by an industrial engineer and the results are give

Question

A 23 experiment was performed by an industrial engineer and the results are given in the following Table. Half of the experiment was done on one day and the remaining half on the next day a) Compute the main effect of task. (b) Use a half-normal plot to detect significant factorial effects. Compare your findings (c) With the timing mechanism operating properly, the standard deviation of the readings with those based on IER version of Lenth's method at an appropriate level of average response time is known to be about 1 (?-1). Subsequent to carrying out this experiment, however, it was discovered that the timing device, although it had been operating properly on the first day, had not been operating properly on the second day. On that second day, the standard deviation was ? 4. Given this information, what is the variance of the main effect of task computed in part (a)? Hint: See Lecture #5 for the variance of the factorial effect. Setup Flasher Type Inertia of Lever Task Avg. Response Time (secs) Randomized Order OW OW High High OW OW High High 12 10 16 14 15 19 Note: Flash A (low) and Flash B (high); Task Y (low) and Task Z (high); Randomized order is the order in which the experiments were performed

Explanation / Answer

Solution:

Suppose that in 23 Factorial Experiment, there are three factors X (Flasher Type), Y (Inertia of Lever) and Z (Task) each at two levels i.e. (Law and High).

We can denote them as

(1)= x0y0z0,   x=x1 y0z0, y= x0y1z0 , xy= x1y1z0

Z= x0y0z1 , xz= x1y0z1 , yz= x0y1z1 , and   xyz= x1y1z1

Assume that response values are

Y* =[(1), x, y, xy, z, xz, yz, xyz]’

(a) Main effects are

X             =             ¼ ([xyz] –[yz]+[xy]-[y]+[xz]-[z]+[x]-[I])

{Here [xyz]=19, [yz]=15, [xy]=11, [y]= 10, [xz]=14, [z]=16, [x]=12 and [I]= 11}

                =             ¼(19-15+11-10+14-16+12-11)

                =             1

Y              =            ¼ ([xyz] +[yz]+[xy]+[y]-[xz]-[z]-[x]-[I])

                =             ¼(19+15+11+10-14-16-12-11)

                =             0.5

Z              =             ¼ ([xyz] +[yz]-[xy]-[y]+[xz]+[z]-[x]-[I])

                =             ¼ (19+15-11-10+14+16-12-11)

                =             5

(b)

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