Introduction Extra Sensory Perception (ESP) is a controversial ability where som

Question

Introduction Extra Sensory Perception (ESP) is a controversial ability where some people claim to be able to see things or make predictions about future events. Whether or not the existence of ESP is part of ones beliefs, the concept does pose a question that can and has been studied and analyzed using the same statistical methods you will be learning in this class. The classic study involved volunteers whose psychic abilities were tested using five cards with each with a different shape. We will use a standard deck of playing cards differentiating the particular card by suit. Take a deck of cards and select four cards each with a different suit, find a willing partner and use these cards to test each other's psychic ability Part 1 1. The Statistical Hypothesis. Write down the appropriate statistical hypothesis for deciding whether or not a particular subject might have ESP. Think about what the theory would be and what the status quo opinion would be. Ho: Hi 2 The variable. Let r = the number ofcorrect guess out of ten attempts. Write down the domain of this variable (hint: it's a set, so use set notation).

Explanation / Answer

1. Many experiments can be formulated using 4 cards, each of a different house, to test the ESP of a subject. A couple of them are as follows -

Experiment 1 - Hold up 4 cards at a time and ask the subject to guess the 4 houses simultaneously.

We must compute the expected 'correct guesses' out of the 4 cards for a person without ESP in order to establish the null hypothesis. Let x be the variable capturing number of correct gueses out of 4. Therefore, P(x=0) = 0.75^4, P(1) = 4*0.25*(0.75)^4, P(2) = 6*(0.25^2)*(0.75^2), P(3) = (0.25^3)*0.75*4, P(4) = 0.25^4. Taking the weighted average of the above probabilities (sum(x*p(x)), The expected value of x comes out to be 1.

Therefore, H0: Number of correct guesses = 1. H1: Number of correct guesses > 1

Experiment 2 - We could follow the experiment dscribed in subsequent parts, or hold up one card at a time and ask the subject to guess its house. Then repeat this procedure for 9 more trials and record the number of correct guesses out of 10. Since the base probability for each inpedendent trial (guess) would be 1/4, the number of expected correct guesses out of 10 would be 10/4 = 2.5. Therefore, H0: Number of correct guesses = 2.5. H1: Number of correct guesses > 2.5.

2. X is the number of correct guesses out of 10 attempts. Hence, domain of X = {0,1,2,3,4,5,6,7,8,9,10}. X(n) = n.

7. The parts 3 and 4 would be conducted by yourself

5. Binomial distribution with p(event) = 0.25, p(non-event) = 0.75

6. Conduct t test between 2 groups - 1. subjects with mean correct guesses = 2.5, 2. subjects with mean correct guesses = 2.5. Note that subjects would be classified into one of the 2 mentioned groups, after conducting a z test on each subject's distribution of scores.

7. Repeat for each subject -

Simulate 10 random trials, 100 times. Calculate the number of correct guesses in each set of 10 and test its mean over 100 iterations against 2.5, using a t-test. If p-val < 0.05, the subject has significantly better guessing power than random.

Related Questions

Navigate

• Browse (All)
• Browse I
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.