Question

In: Statistics and Probability

An area of research in biomechanics and gerontology concerns falls and fall-related injuries, especially for elderly...

An area of research in biomechanics and gerontology concerns falls and fall-related injuries, especially for elderly people. Recent studies have focused on how individuals respond to large postural disturbances (e.g., tripping, induced slips). One question is whether subjects can be instructed to improve their recovery from such disturbances. Suppose researchers want to compare two such recovery strategies, lowering (quickly stepping down with front leg and then raising back leg over the object) and elevating (lifting front leg over the object). Subjects will have first been trained on one of these two recovery strategies, and they will be asked to apply it after they feel themselves tripping. The researchers will then induce the subject to trip while walking (but harnessed for safety) using a concealed mechanical obstacle. Suppose the following 24 subjects have agreed to participate in such a study: Females: Alisha, Alice, Betty, Martha, Audrey, Mary, Barbie, Anna Males: Matt, Peter, Shawn, Brad, Michael, Kyle, Russ, Patrick, Bob, Kevin, Mitch, Marvin, Paul, Pedro, Roger, Sam

1A. (2 points) One way to design this study would be to assign the 8 females to use the elevating strategy and the 16 males to use the lowering strategy. Why won’t this be a reasonable strategy?

B. (2 points) Explain the best way to assign the 24 people to two groups.

C. (2 points) Let’s explore the process of random assignment to determine whether it does “work.” First, let’s focus on the sex (male vs. female) variable. Suppose we put each person’s name on a slip, put those slips in a hat and mix them up thoroughly, and then randomly draw out 12 slips for names of people to assign to the elevating strategy. What proportion of this group do you expect will be male? What proportion of the lowering strategy do you expect will be male? Do you think we will always get an 8/8 split (8 males in each treatmen

Solutions

Expert Solution

1

we migh run into selection bias as men and woen may have some inherent differencres which will affect the results

b)

we should use random sampling and assian equal numbr of men and women in both groups

3)

total males =16

E[male] = so in 12 we can have 12,11,10,9,8,7,6,5,4 males

so number of ways of selecting 12 people from 24 = 24c12

number of ways of selecting 12 males 0 females = 16C12 8C0/ 24C12 =1820/2704156

number of ways of selecting 11 males 1 females = 16C11 8C1/ 24C12 = 34944/2704156

number of ways of selecting 10 males 2 females = 16C10 8C2/ 24C12 =224224/2704156

number of ways of selecting 9 males 3 females = 16C9 8C3/ 24C12 = 640640/2704156

number of ways of selecting 8 males 4 females = 16C8 8C4/ 24C12= 900900/2704156

number of ways of selecting 7 males 5 females = 16C7 8C5/ 24C12 =640640/2704156

number of ways of selecting 6 males 6 females = 16C6 8C6/ 24C12= 224224/2704156

number of ways of selecting 5 males 7 females = 16C5 8C7/ 24C12 = 34944/2704156

number of ways of selecting 4 males 8 females = 16C4 8C8/ 24C12 = 1820/2704156

so expected number of males =

0.008076457 +0.1421456 +0.8291829+ 2.132185 +2.665231 +1.658366 +0.4975098 +0.06461166 +0.002692152

8

similarly in the lower level we will again get 8

. We may not always get a 8/8 split but we will get it on a verage i.e. in a large number of draws we will get 8/8 on average.


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