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Practice Final A WI 2011 SOLNS

# Want px bar 83 pz 80 838square root of 49 pz 1 from z

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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? X = number preferring to work from home one day per week X has a binomial distribution with n = 80 and p = .30 (n is too large for the table – need normal approximation) Check conditions for normal approximation: 1) np = 80x.3 = 24 is at least 10; and 2) n(1-p) = 80x.7 = 56 is at least 10. Use Z = (X – np)/(square root of np(1-p)) You want P(X at least 25) =P(Z at least (25-24)/square root (80x.3x.7) = =P(Z at least _________) = 1 – P(Z < ______) = ____________ This is an approximate answer. For #37: Bob measured the length (in cm) and weight (in grams) of a random sample of mice of a certain species. The results of his analysis are shown below. Variable Count Mean StDev Minimum Median Maximum Length 150 168.67 43.15 63.00 163.00 280.00 Weight 150 150.77 19.76 101.00 150.50 208.00 Predictor Coef SE Coef T P Constant 39.05 25.09 1.56 0.122 Weight 0.8597 0.1650 5.21 0.000 Pearson correlation of Length and Weight = 0.394 37. ESTIMATE the average weight of all mice of this species in the population. Show all work and use proper units WITH 95% CONFIDENCE. Best possible answer. Need a 95% confidence interval for average weight. Xbar plus/minus 1.96(standard deviation)/square root of n 7
X = weight, n = 150, mean = 150.77, stdev = 19.76 Estimate is: 150.77 plus/minus 19.76/square root of 150 = 150.77 plus/minus ______________ GRAMS = ( , ) GRAMS 8
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Want PX bar 83 PZ 80 838square root of 49 PZ 1 from Z table...

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