Suppose that A and B are events for which Р(A| B) — 0.8, Р(В A) — 0.15, and Р(A) — 0.2. P(B) =
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A: Given that P(W1) = 0.10, P(W2) = 0.90 P(X | W1) = 0.80, P(X | W2) = 0.30
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- Sometimes experiments involving success or failure responses are run in a paired or before/after manner. Suppose that before a major policy speech by a political candidate, n individuals are selected and asked whether (S) or not (F) they favor the candidate. Then after the speech the same n people are asked the same question. The responses can be entered in a table as follows: After S F Sx₁₂ Before FX3 X4 where x1 + x2 + x3 + x4 = n. Let P1, P2, P3, and P4 denote the four cell probabilities, so that p₁ = P(S before and S after), and so on. We wish to test the hypothesis that the true proportion of supporters (S) after the speech has not increased against the alternative that it has increased. a. State the two hypotheses of interest in terms of P1, P2, P3, and P4. b. Construct an estimator for the after/before difference in success probabilities. c. When n is large, it can be shown that the rv (X; - X;)/n has approximately a normal distribution with variance given by [P; + Pj - (P₁ -…8.why the correct answer is letter C?
- A random group of customers at a fast food chain were asked whether they preferred hamburgers, chicken sandwiches, or fish sandwiches. The restaurant's marketing department claims that 45% of customers prefer hamburgers, 45% of the customers prefer chicken sandwiches, and 10% of the customers prefer fish sandwiches. Is there evidence to reject this hypothesis at = 0.05? Plans Hamburgers Chicken Fish Number of students 34 18 8 Group of answer choices A. There is evidence to reject the claim that the customers' preferences are distributed as claimed because the test value 17.2> 5.481 B. There is not evidence to conclude the claim that the customers' preferences are distributed as claimed because the test value 5.481 < 7.815 C. There is not evidence to reject the claim that the customers' preferences are distributed as claimed because the test value 5.481 < 5.991 D. There is evidence to conclude the claim that the customers' preferences are…Brian and Jennifer are planning to get married at the end of Brian's blood type Jennifer's blood type the year. They read that certain diets are healthier for specific A 0.29 blood types and hope that they both have the same blood type. Unfortunately, neither Brian nor Jennifer can remember their blood type. - B -0.35- -0.24. - AB 0.12. Suppose a recent study shows that 29% of adults have blood type A, 35% of adults have blood type B, 24% of adults have blood type AB, and 12% of adults have blood type O. The tree A 0.29 0.29 В -0.35 -0.24. AB diagram shows the possible outcomes for Brian and Jennifer 0.12 0.35 along with their associated probabilities. What is the probability that Brian and Jennifer share the same 0.29 0.24. -B blood type? Give your answer as a decimal, precise to at least 0.35- AB -0.24. two decimal places. -AB 0.12. 0.12 P(Brian and Jennifer have the same blood type) = 0.29 B -0.35 -0.24 AB 0.12.The CDC (Center for Disease Control) did a study to see if there is an association between lung cancer and exposure to asbestos (this data set has been made up. However, my understanding is that the association has already been proven). The doctors at the CDC took a SRS of 20 people with lung cancer and a SRS of 30 people without lung cancer. Suppose the incidence of lung cancer is 1 out of every 1000 people (I made this number up).Let C = a randomly selected person has lung cancerLet E = a randomly selected person has been exposed to asbestosUse 3 decimal places for the following questions.P(C | E) =
- If P(A)-=0.7, P(B)=0.3, and P(A and B)=0.07, are A and B independent? A. No B. Yes C. Not enough informationA teacher believes that students who study more than four hours for her tests will do better than students who do not study for her tests. To test this belief, the teacher recruited 16 students and randomly assigned them to two groups: G1: a group of n1=8 students that studied more than four hours for her test, and G2: a group of n2=8 students that did not study for her test. The following are the data from G1, who studied more than four hours for the test: n1=8 M1=85 s1=5 (this is the standard deviation of the sample, dividing the sum of squares by n1) The following are the data from G2, who did not study for the test: n2=8 M2=75 s2=4 (this is the standard deviation of the sample, dividing the sum of squares by n2) Perform a t-test by answering the questions below. Use an alpha-level of α=.05. 0. Using formulas from Section 4, compute the estimates of the population variances, est. σ12 and est. σ22 (from s1 and s2 above). 1. What is the research…There is a chance that a bit transmitted through a digital transmission channel is received in error. Let X equal the number of bits in error in the next four bits transmitted. The possible values of X are {0,1,2,3}. Suppose that P(X=0)=0.6, P(X=1)= 0.3, P(X=2)= 0.05, P(X=3)= 0.05 What is the CDF of X at 3? i.e., what is F(3)? a). 0.6 b). 0.9 c). 0.95 d). 1 e). none of above For the random variable defined in question 12, if we know that Var(X)=0.6475, what is the standard deviation of X? a). 0.55 b). 0.6475 c). 0.419 d). 0.8 e). none of above
- An automobile service facility specializing in engine tune-ups knows that 60% of all tune-ups are done on four-cylinder automobiles, 30% on six-cylinder automobiles, and 10% on eight-cylinder automobiles. Let X = the number of cylinders on the next car to be tuned. What is the pmf of X?Suppose E= "sum is 5" and F= "exactly one of the dice is even". Are E and F independent? Explain why or why not.Suppose the probability that a male develops some form of cancer in his lifetime is 0.4573. Suppose the probability that a male has at least one false positive test result (meaning the test comes back for cancer when the man does not have it) is 0.51, and that this probability is independent of the probability that a male will develop cancer in his lifetime. • C = a man develops cancer in his lifetime• F = a man has at least one false positive Part (a) Construct a tree diagram of the situation. P(C) = P(F | C) = P(F | C' ) If a test comes up positive, based upon numerical values, can you assume that man has cancer? Justify numerically and explain why or why not. -You cannot assume the man has cancer because there is not enough information given. -You cannot assume the man has cancer because there is a 51% chance that the test is false. -You cannot assume the man has cancer because both the probability of developing cancer in his lifetime and the probability of…