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Z score and P value?
A noted psychic was tested for extrasensory perception. The psychic was presented with 200 cards face down and asked to determine if the card were one of five symbols. 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.
Suppose you wished to see if there were evidence that the psychic is doing better than just guessing. To do this, you test the hypotheses H_0 : p = .20 versus H_{alternative} : p > .20.
What is the Z score?
What is the p-value for the test statistic? (You can use Q function in the answer)
2 Answers
- Anonymous7 years agoFavorite Answer
Let X = total number of successes in 200 trials. Since X can either be a success or failure, X is a binomial random variable. Randomly guessing the face of a card will be correct with a probability of 0.20.
The expected value of X is np = 200(0.20) = 40, and the standard deviation is sqrt(np(1 - p)) = sqrt(40(.80)) = 4 sqrt(2). By the Central Limit Theorem, (X - mu) / sigma is a standard normal distribution when n is large. Hence the test statistic,
Z = (X - 40) / 4 sqrt(2),
has a standard normal distribution.
The actual value of X is 50, so,
Z = (50 - 40) / 4 sqrt(2) = 5 / 2 sqrt(2) = 1.77.
This is the Z-score. The P-value is 0.077. This isn't good enough evidence to establish that the psychic did better than chance with 95% confidence.
- scrubbagLv 77 years ago
Bob, this somehow got into the Religion and Spirituality Category, it probably should have been Science and Mathematics.
