A noted psychic was tested for extrasensory perception.The psychic was presented with 200 cards face down and asked to determine if each card were one of five symbols: a star,a cross,a circle,a square,or three wavy lines.The psychic was correct in 50 cases.Let p represent the probability that the psychic correctly identifies the symbol on the card in a random trial.Assume the 200 trials can be treated as a simple random sample from the population of all guesses the psychic would make in his lifetime.What do we know about the value of the P-value for the hypothesis test? (Note: Use the large-sample z statistic. )
A)P-value

Question

A noted psychic was tested for extrasensory perception.The psychic was presented with 200 cards face down and asked to determine if each card were one of five symbols: a star,a cross,a circle,a square,or three wavy lines.The psychic was correct in 50 cases.Let p represent the probability that the psychic correctly identifies the symbol on the card in a random trial.Assume the 200 trials can be treated as a simple random sample from the population of all guesses the psychic would make in his lifetime.What do we know about the value of the P-value for the hypothesis test? (Note: Use the large-sample z statistic. )
A)P-value < 0.001
B)0.001 < P-value < 0.01
C)0.01 < P-value < 0.05
D)0.05 < P-value < 0.10

Quizplus · Accepted Answer

We are given that the psychic correctly identified the symbol on 50 out of 200 cards, so the sample proportion is:

p̂ = 50/200 = 0.25

We are also told to assume that the 200 trials can be treated as a simple random sample from the population of all guesses the psychic would make in his lifetime. Based on this, we can set up the null and alternative hypotheses as follows:

H0: p = 0.2 (the psychic's true probability of correctly identifying the symbol)
Ha: p > 0.2 (the psychic has extrasensory perception and can correctly identify the symbol with a higher probability than usual)

We can use a Z-test for proportions to test this hypothesis, since the sample size is large enough:

test statistic = (p̂ - p) / sqrt(p(1-p) / n)
                 = (0.25 - 0.2) / sqrt(0.2(1-0.2) / 200)
                 = 2.24

Using a significance level of 0.05, the critical value for a one-tailed test is:

zα = 1.645

Since our test statistic (2.24) is greater than the critical value (1.645), we reject the null hypothesis and conclude that there is evidence to suggest that the psychic has extrasensory perception.

To find the P-value, we can use a standard normal distribution table or calculator to find the area to the right of our test statistic:

P-value = P(Z > 2.24) = 0.012

This tells us that, assuming the null hypothesis is true (i.e. the psychic has no extrasensory perception), the probability of obtaining a sample proportion as extreme or more extreme than what we observed (p̂ = 0.25) is only 0.012. Therefore, we can be fairly confident that the psychic does indeed have extrasensory perception, with a P-value of 0.012. Since this value falls between 0.01 and 0.05, the best answer choice is C.

A Noted Psychic Was Tested for Extrasensory Perception