Practice Final A WI 2011 SOLNS

# Which situation gives you less bias a randomly

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28. Which situation gives you less bias? a. Randomly sampling 100 people and getting 100 responses b. Randomly sampling 1,000 people and getting 100 responses c. Both give the same amount of bias. 29. Let X = the number of coin flips you need to get a tail. X has a binomial distribution. 30. I took a random sample of size n and got a mean 80 and standard deviation 10. I used it to find a confidence interval for the population mean, and my margin of error was 1. What size sample did I take? 4

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For #31-33: Suppose test scores on an accounting exam have a normal distribution with mean 80 and standard deviation 8. 31. You select one student at random. What is the chance their score is more than 83? X normal with mean 80 and sd = 8 Want P(X>83) = P(Z > (83-80)/8) = P(Z> ) = 1 - _____ from Z table 32. What exam score is at the 80 th percentile? 80 th percentile of Z is ______________ (look up closest # to .8000 in body of Z table and findZ) Change to X value using Z formula: ______ = (X – 80)/8 so X = 80 + ____(8)=__________ is the 80 th percentile for exam scores 33. Select 49 students at random. What’s the chance the average score is more than 83? Want P(X-bar > 83) = P(Z > (80-83)/(8/square root of 49)) = P(Z > ____) = 1 - _______ from Z table -- No CLT needed since X has a normal distribution. Do not check n > 30 For #34: Quiz scores (10 points each) for two quizzes are analyzed for 6 students. The regression analysis is below. What is the residual for the point where X = 4? Use proper units and show all work for credit. DATA: ANALYSIS: Quiz 1 Quiz 2 Predictor Coef SE Coef T P 5 5 Constant 1.533 2.431 -0.63 0.562 2 3 Quiz 1 0.9429 0.6242 -1.51 0.205 3 3 S = 2.61133 R-Sq = 36.3% R-Sq(adj) = 20.4% 4 10 6 6 1 2 Equation of the line is 1.533 + .9429X. Predicted Y when X = 4 is Y = 1.533+.9429(4)= (plug X = 4 into the equation) Observed Y when X = 4 is 10 (given) Residual = Observed Y – Predicted Y = 10 - ___________ 6

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For #35-36: Bob heard on the news that 30% of all employees would prefer to work from home one day a week. SHOW WORK! 35. Bob surveys 8 of his employees at random and counts the number that would prefer to work from home one day a week. What’s the probability that 2 of Bob’s employees would prefer to work from home one day per week? X = number preferring to work from home one day per week X has a binomial distribution with n = 8 and p = .30 We want P(X = 2). Use the binomial table to get ______________ 36. Bill also hears the same newscast as Bob and he conducts the same survey on a random sample of 80 of his employees. What’s the chance that at least 25 of Bill’s employees would prefer to work from home?
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