# 5 - 4/17/2011 www.lecturebook.com/homeworks/sho Assignm ent...

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Assignment Stats 250 W11 HW 5 Graded Point Total: 25.5 out of 30 a. 0 out of 0.5 points Feedback: This is the expected value of xbar, not xbar. This is one of those important differences between a parameter and a statistic. See Textbook for Exercise: 9.a. Your Answer: =7.05 hours Solution: The mean would be the population mean μ = 7.05 hours b. 0.5 out of 1 points Feedback: Same as above. See Textbook for Exercise: 9.b. Your Answer: s= σ /sqrt(n) s=1.75/sqrt(190)=.1270 Solution: s.d. ( ) = σ /sqrt( n ) = 1.75/sqrt(190) = 0.127 hours c. 1 out of 1 points See Textbook for Exercise: 9.c. Your Answer: between 6.923 and 7.177 hours Solution: 6.923 and 7.177, calculated by going out one standard deviation each way from the mean, namely, 7.05 ± (1 x 0.127). a. Consider the procedure of selecting a random sample of 64 salaries and computing the sample mean salary, . Suppose you repeated that procedure many, many times. Describe and draw the distribution of the resulting values. Also provide the values along the axis that represent 1, 2, and 3 standard deviations from the mean. Required Assignment Stats 250 W11 HW 5 Question 1 (9.57) See Textbook for Exercise: 9.57. Question 2 (Chapter 9) Let X = the salary (in \$1000s) for a randomly selected full-time worker in a certain city. Suppose the expected (or mean) salary is \$62.0 and the standard deviation is \$12.0. 4/17/2011 www.lecturebook.com/homeworks/sho… lecturebook.com/homeworks/show/41 1/9

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3 out of 3 points Your Answer: Because the sample size is 64,which is greater than the required 25 for the Central Limit Theorem, is approximately N(62,1.5) ba Limit Theorem. Solution: b. 2 out of 2 points If you were to obtain a random sample of 64 salaries, what is the probability that the sample mean salary will exceed \$65? Your Answer: z score = (65-62)/1.5 = 2 P(z>2) = 1 - P(z<2) = 1 - .9772 = 0.0228 The probability that the sample mean salary will exceed \$65 is 0.0228. Solution: P( > \$65) = P(Z > (\$65 - \$62)/\$1.50) = P(Z > 2) = 1-P(Z 2) = 1-0.9772 = 0.0228 c. 0.5 out of 1 points Feedback: We don't have a sample yet in this type of question. We need to know about the population. Could you apply the same technique with a random sample of 9 salaries? Explain using 1 brief sentence. Your Answer: No, because with a sample size of 9 is not large enough of a sample to apply the Central Limit Theorem, and because the prompt does not say it is a normal sample, we would have to check for normality first. Solution: N W t i t h dl d it i bt i f l i f t h i lt i lh t h
No. We are not given the model or distribution for salaries for this population , we only have the mean and standard deviation. Thus we cannot apply the same technique because we do not have a large enough sample size (e.g. n > 25 or 30) to rely on the Central Limit Theorem, as we did to answer parts (a) and (b). 0 out of 1 points

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## This note was uploaded on 01/26/2012 for the course STATS 250 taught by Professor Gunderson during the Winter '10 term at University of Michigan.

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5 - 4/17/2011 www.lecturebook.com/homeworks/sho Assignm ent...

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