Here is a simple probability model for multiple-choice tests. Suppose that each

Question

Here is a simple probability model for multiple-choice tests. Suppose that each student has probability p of correctly answering a question chosen at random from a universe of possible questions. (A strong student has a higher p than a weak student.) The correctness of answer to different questions are independent. Jodi is a good student for when p = 0.78. (a) Use the Normal approximation to find the probability that Jodi score 73% or lower on a 100-question test. (b) If the test contains 250 questions, what is the probability that Jodi will score 73% or lower? (c) How many questions must the test contain in order to reduce the standard deviation of Jodis proportion of correct answers to half its value for a 100-item test? questions d) Laura is a weaker student for whom p = 0 73 in order to reduce you gave in (c) for standard deviation of Jodis score apply to Lauras standard deviation also? Yes, the smaller p for Laura has no effect on the relationship between the number of questions and the standard deviation. No, the smaller p for Laura alters the relationship between the number of questions and the standard deviation. Does delaying oral practice hinder learning a foreign language? Researchers randomly assigned 23 beginning students of Russian to begin speaking practice immediately and another 23 to delay speaking for 4 weeks. At the end of the semester both groups took a standard test of comprehension of spoken Russian. Suppose that in the population of all beginning students, the test scores for early speaking vary according to the N(31, 6) distribution and scores for delayed speaking rave the N(25, 4) distribution. (a) What is the sampling distribution of the mean score x in the early speaking group in many receptions of the experiment? Mean = S = What is the sampling distribution of the difference y in the delayed speaking group? Mean = S = (b) If the experiment were repeated many time, what would be the sampling distribution of the y - x between the mean score in the two group? Mean = S = c) What is the probability that the experiment will find (misleadingly) that the mean score for delayed speaking is at least as large as that for early speaking?

Explanation / Answer

a)

good student for whom p = 0.78. Find the probability that student scores 73% or lower on a 100 question test.
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mean = 0.78
std = sqrt{0.78*0.22/100] = 0.04142
z(0.80) = (0.73-0.78)/0.04142 = -1.2071
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P(0 <- phat <= 0.73) = P(-inf < z < -1.2071) = 0.113697

b)

mean = 0.78
std = sqrt{0.78*0.22/250] = 0.02620
z(0.80) = (0.73-0.78)/0.02620= -1.9084
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P(0 <- phat <= 0.80) = P(-inf < z < -1.9084) = 0.02817

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(c)

How many questions must the test contain in order to reduce the standard deviation of Julie's proportion of correct answers to one-fourth its value for an 100-item test?
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Since std = sqrt(pq/n) = sqrt{0.78*0.22/100] = 0.04142
(1/2)sqrt(pq/n) = sqrt(pq/4n)

0.02071 = sqrt 0.78*0.22/4n

0.0004289 =0.78*0.22 /4n

0.0017156n = 0.1716

n=100.023 or 100

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