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Mathematics 11 Online
OpenStudy (anonymous):

A researcher looking for evidence of extrasensory perception (ESP) tests 500 subjects. Four of these subjects do significantly better (P < 0.01) than random guessing. a) Is it proper to conclude that these four people have ESP? Explain your answer. b) What should the researcher now do to test whether any of these four subjects have ESP?

OpenStudy (anonymous):

@perl

OpenStudy (anonymous):

@inkyvoyd

OpenStudy (anonymous):

so for a, we reject the null that they don't have ESP because the P value is less than .01? so it is proper to say these four people could have ESP

OpenStudy (perl):

lets make this more concrete, lets suppose the ESP test is to guess the correct flips when someone flips a coin in a room . and there are say 5 flips of a coin

OpenStudy (anonymous):

okay

OpenStudy (perl):

Lets say the ESP test consists of reading someones mind as they flip a coin eight times and correctly stating the sequence of flips heads or tails. If we suppose that the true order of flips is : HHTTHTHT it makes sense that someone could get it right due to guessing. We can make a histogram of all the guesses. There is .01 probability of 5 people or less getting it right, so 4 people is not impossible.

OpenStudy (anonymous):

okay

OpenStudy (perl):

I changed the number of flips from 5 to eight. Now let X = # of people out of 500 who guess the coin sequence correctly. This is a binomial random variable, which has probability n * p , here p is the probability of getting all coins right (1/2)^8 = .0039 and n = 500.

OpenStudy (perl):

but when you multiply 500*.0039 = 1.95, you expect around 2 people to get it right. thats the point im trying to make

OpenStudy (perl):

because the sample size is so large

OpenStudy (anonymous):

oh okay yeah that makes sense

OpenStudy (perl):

if the sample size was small, say 30 people, then it would be unusual for any of them to get it right (to accurately predict a sequence of 8 coin tosses). 30 * .0039 = .117 , which is less than 1 person.

OpenStudy (anonymous):

okay

OpenStudy (perl):

I think the `(P<.01)` in the problem stands for probability (not p value). So the ESP test involves something which has a low probability of getting right, less than .01, like flipping a coin 7 or eight times and guessing the correct sequence

OpenStudy (anonymous):

so what would we say for a?

OpenStudy (perl):

because of the size of the sample it is not unusual to get 5 or less people to pass the ESP test.

OpenStudy (anonymous):

okay what would we do for b?

OpenStudy (perl):

so after identifying the ones who got correct out of 500 people, you should do a new ESP test on them. if the probability P of passing the test is less than .01, then it is unlikely that any of these 4 people will pass it. here np = 4 * .01 = .04 , which is less than 1 person

OpenStudy (anonymous):

okay thank you! do you have time to help on a couple more?

OpenStudy (perl):

ok, please make a new post

OpenStudy (anonymous):

okay will do

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