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Ratling [72]
2 years ago
8

Researchers are investigating whether people who exercise with a training partner have a greater increase, on average, in target

ed exercise intensity compared with people who exercise alone. Two methods of collecting data have been proposed.
Method I: Recruit volunteers who are willing to participate. Randomly assign each participant to exercise with a training partner or to exercise alone.
Method II: Select a random sample of people from all the people who exercise at a community fitness center. Ask each person in the sample whether they use a training partner, and use the response to create the two groups.
(a) For each method, the researchers will record the change in targeted exercise intensity for each person in the investigation. They will compare the mean change in intensity between those who exercise with a training partner and those who do not.

(i) Describe the population of generalization if method I is used. Explain your answer.

(ii) Describe the population of generalization if method II is used. Explain your answer.



(b) Suppose the investigation produces a result that is statistically significant using both methods. What can be concluded if method I is used that cannot be concluded if method II is used? Explain your answer.
Mathematics
1 answer:
bulgar [2K]2 years ago
7 0

Answer:

1 - If method I is used, population of generalization will include all those people who may have varying exercising habits or routines.  They may or may not have a regular excersing habit.  In his case sample is taken from a more diverse population

2 - Population of generalization will include people who will have similar excersing routines and habits if method II is used since sample is choosen from a specific population

Step-by-step explanation:

Past excercising habits may affect the change in intensity to a targeted excersise in different manner. So in method I a greater diversity is included and result of excersing with or without a trainer will account for greater number of variables than method II.

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Answer:

a) P[ Pf | Ptd ]  = 0,428

b) P[ Pf | Nd ]  = 0,692

Step-by-step explanation:

We will call:

Pf  Probability of functioning correctly    60 %   or 0.6

Pfn  Probability of malfunctioning   40 %  or  0,4

PNd Probability of nondefective bulb

D₁  Probability of defective bulb when the process is correctly working

D₂  Probability of defective bulb when the process is in malfunction condition

Ptd Total probability of defective bulbs

Then applying theorem´s Bayes  have

P [ A | B ]  =  P(A) * P [ B | A ] / P(B)

a) Probabilty of the process is functioning correctly given that we pick up a defective bulb

The total probability of defective bulbs is equal to, probabilty of defective bulbs when the process is Ok, plus probability of defective bulbs when the process  is in malfunctioning condition, therefore

Ptd  = (0,6)*(0,25) + (0,40)*(0,50) = 0,15 * 0,20 = 0,35

P[ Pf | Ptd ]  =  Pf * P [ Pd | Pf ] / Ptd    =  0,6 * 0,25 / 0,35

P[ Pf | Ptd ]  = 0,428

b)  P [ Pf | Nd ]

P[ Pf | Nd ]  =  Pf *  P [Nd | Pf ] / PNd

PNd  = 1 - Ptd    =  1 - 0,35   =  0,65

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P[ Pf | Nd ]  = 0,692

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Answer:

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Step-by-step explanation:

a1,1=1

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a2,2=5

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Answer:

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Step-by-step explanation:

The Empirical Rule states that, for a normally distributed random variable:

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Step-by-step explanation:

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