QUESTION 5 In a test of ESP (extrasensory perception), the experimenter looks at cards that are hidden from the subject. Each card contains either a star, a circle, a wavy line, or a square. An experimenter looks at each of 100 cards in turn, and the subject tries to read the experimenter's mind and name the shape on each. What is the probability that the subject gets 30 or more correct if the subject does not have ESP and is just guessing? A. 0.1251 B. less than 0.0001 C. 0.25 D.0.31 QUESTION 6 An article in Parenting magazine (December/January 2004) reported that 60% of Americans surveyed say that they need a vacation after visiting family for the holidays. Suppose the true proportion of all Americans who need a vacation after visiting family for the holidays is indeed 60%. A simple random sample is taken of 150 Americans. What is the probability that 50% or less of the people in the sample say that they need a vacation after visiting family for the holidays? A. 0.0062 B. 0.50 C. 0.0072 D.0.60
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This can be calculated using the normal distribution formula for proportions. We first calculate the z-score: \[ z = \frac{0.3 - 0.25}{\sqrt{\frac{0.25 \times 0.75}{100}}} = 1.15 \] Show more…
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In a test for ESP (extrasensory perception), a subject is told that cards only the experimenter can see contain either a star, a circle, a wave, or a square. As the experimenter looks at each of 20 cards in turn, the subject names the shape on the card. A subject who is just guessing has probability 0.25 of guessing correctly on each card. a. The count of correct guesses in 20 cards has a binomial distribution. What are n and p? b. What is the mean number of correct guesses in 20 cards for subjects who are just guessing? c. What is the probability of exactly 5 correct guesses in 20 cards if a subject is just guessing?
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In an experiment on extrasensory perception (ESP) a subject in one room is asked to state the color (red or blue) of a card chosen from a deck of 50 well-shuffled cards by an individual in another room. It is unknown to the subject how many red or blue cards are in the deck. If the subject identifies 32 cards correctly, determine whether the results are significant at the 0.05 0.01 level of significance. Find and interpret the P value of the test. Use the following hypotheses when answering the above: H0: p = 0.5, and the subject is simply guessing, i.e., results are due to chance H1: p > 0.5, and the subject has powers of ESP.
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An investigator wishes to estimate the proportion of students at a certain university who have violated the honor code. Having obtained a random sample of $n$ students, she realizes that asking each, "Have you violated the honor code?" will probably result in some untruthful responses. Consider the following scheme, called a randomized response technique. The investigator makes up a deck of 100 cards, of which 50 are of Type I and 50 are of Type II. Type I: Have you violated the honor code (yes or no)? Type II: Is the last digit of your telephone number a $0,1,$ or 2 (yes or no)? Each student in the random sample is asked to mix the deck, draw a card, and answer the resulting question truthfully. Because of the irrelevant question on Type II cards, a yes response no longer stigmatizes the respondent, so we assume that responses are truthful. Let $p$ denote the proportion of honor-code violators (i.e., the probability of a randomly selected student being a violator), and let $\lambda=P($ yes response). Then $\lambda$ and $p$ are related by $\lambda=.5 p+(.5)(.3) .$ (a) Let $Y$ denote the number of yes responses, so $Y \sim \operatorname{Bin}(n, \lambda) .$ Thus $Y / n$ is an unbiased estimator of $\lambda .$ Derive an estimator for $p$ based on $Y .$ If $n=80$ and $y=20,$ what is your estimate? [Hint: Solve $\lambda=.5 p+.15$ for $p$ and then substitute $Y / n$ for $\lambda . ]$ (b) Use the fact that $E(Y / n)=\lambda$ to show that your estimator is unbiased for $p$ . (c) If there were 70 Type I and 30 Type II cards, what would be your estimator for $p ?$
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