Explanation Calculation: Step-by-step procedure to generate 200 random samples using the MINITAB software: Click on Calc > Random Data > Binomial Distribution . In Number of rows of data to generate section, enter 200 . In the Store in column , enter column C1 named as Random Numbers . For Number of trails , enter 20 . For Probability , enter 0.6 . Click OK .. Output using the MINITAB is given below: Random numbers 9 15 12 16 12 9 15 11 12 10 10 9 14 16 11 11 9 15 13 12 11 6 15 11 15 9 11 11 14 11 9 13 9 12 13 12 13 15 12 13 11 9 10 12 9 12 14 10 15 12 10 8 11 12 10 15 13 10 11 10 14 8 12 14 13 12 10 12 13 7 15 13 11 8 17 11 10 10 15 12 9 15 11 10 13 12 11 10 10 17 12 15 12 12 18 15 9 11 14 9 10 15 13 14 12 9 14 17 15 6 9 13 13 14 13 17 15 15 13 13 15 11 10 9 9 8 14 12 12 13 9 12 10 8 14 13 12 7 15 13 8 9 15 14 11 9 11 14 11 11 14 9 11 15 10 11 14 14 14 11 9 10 11 17 12 13 15 10 12 8 8 13 14 15 9 16 11 15 10 7 11 13 16 11 8 14 9 12 10 12 12 13 11 15 12 12 12 16 12 12 From the Standard Normal Table, the z value for 90% level is 1.645. The general formula for the confidence interval for the binomial parameter is, p ^ ± z 1 − α 2 p ^ q ^ n The lower bound and upper bound is tabulated below: Random Numbers p ^ q ^ = 1 − p ^ z 1 − α 2 S E = z 1 − α 2 p ^ q ^ n Lower bound Upper bound 9 0.45 0.55 1.645 0.1830 0.2670 0.6330 10 0.5 0.5 1.645 0.1839 0.3161 0.6839 11 0.55 0.45 1.645 0.1830 0.3670 0.7330 9 0.45 0.55 1.645 0.1830 0.2670 0.6330 11 0.55 0.45 1.645 0.1830 0.3670 0.7330 10 0.5 0.5 1.645 0.1839 0.3161 0.6839 14 0.7 0.3 1.645 0.1686 0.5314 0.8686 15 0.75 0.25 1.645 0.1593 0.5907 0.9093 9 0.45 0.55 1.645 0.1830 0.2670 0.6330 12 0.6 0.4 1.645 0.1802 0.4198 0.7802 10 0.5 0.5 1.645 0.1839 0.3161 0.6839 9 0.45 0.55 1.645 0.1830 0.2670 0.6330 15 0.75 0.25 1.645 0.1593 0.5907 0.9093 9 0.45 0.55 1.645 0.1830 0.2670 0.6330 8 0.4 0.6 1.645 0.1802 0.2198 0.5802 14 0.7 0.3 1.645 0.1686 0.5314 0.8686 9 0.45 0.55 1.645 0.1830 0.2670 0.6330 8 0.4 0.6 1.645 0.1802 0.2198 0.5802 11 0.55 0.45 1.645 0.1830 0.3670 0.7330 12 0.6 0.4 1.645 0.1802 0.4198 0.7802 15 0.75 0.25 1.645 0.1593 0.5907 0.9093 9 0.45 0.55 1.645 0.1830 0.2670 0.6330 6 0.3 0.7 1.645 0.1686 0.1314 0.4686 13 0.65 0.35 1.645 0.1754 0.4746 0.8254 9 0.45 0.55 1.645 0.1830 0.2670 0.6330 8 0.4 0.6 1.645 0.1802 0.2198 0.5802 8 0.4 0.6 1.645 0.1802 0.2198 0.5802 13 0.65 0.35 1.645 0.1754 0.4746 0.8254 15 0.75 0.25 1.645 0.1593 0.5907 0.9093 15 0.75 0.25 1.645 0.1593 0.5907 0.9093 15 0.75 0.25 1.645 0.1593 0.5907 0.9093 13 0.65 0.35 1.645 0.1754 0.4746 0.8254 11 0.55 0.45 1.645 0.1830 0.3670 0.7330 12 0.6 0.4 1.645 0.1802 0.4198 0.7802 9 0.45 0.55 1.645 0.1830 0.2670 0.6330 9 0.45 0.55 1.645 0.1830 0.2670 0.6330 10 0.5 0.5 1.645 0.1839 0.3161 0.6839 13 0.65 0.35 1.645 0.1754 0.4746 0.8254 13 0.65 0.35 1.645 0.1754 0.4746 0.8254 13 0.65 0.35 1.645 0.1754 0.4746 0.8254 12 0.6 0.4 1.645 0.1802 0.4198 0.7802 14 0.7 0.3 1.645 0.1686 0.5314 0.8686 15 0.75 0.25 1.645 0.1593 0.5907 0.9093 9 0.45 0.55 1.645 0.1830 0.2670 0.6330 10 0.5 0.5 1.645 0.1839 0.3161 0.6839 11 0.55 0.45 1.645 0.1830 0.3670 0.7330 12 0.6 0.4 1.645 0.1802 0.4198 0.7802 11 0.55 0.45 1.645 0.1830 0.3670 0.7330 11 0.55 0.45 1.645 0.1830 0.3670 0.7330 12 0.6 0.4 1.645 0.1802 0.4198 0.7802 13 0.65 0.35 1.645 0.1754 0.4746 0.8254 13 0.65 0.35 1.645 0.1754 0.4746 0.8254 10 0.5 0.5 1.645 0.1839 0.3161 0.6839 10 0.5 0.5 1.645 0.1839 0.3161 0.6839 15 0.75 0.25 1.645 0.1593 0.5907 0.9093 11 0.55 0.45 1.645 0.1830 0.3670 0.7330 11 0.55 0.45 1.645 0.1830 0.3670 0.7330 14 0.7 0.3 1.645 0.1686 0.5314 0.8686 16 0.8 0.2 1.645 0.1471 0.6529 0.9471 11 0.55 0.45 1.645 0.1830 0.3670 0.7330 16 0.8 0.2 1.645 0.1471 0.6529 0.9471 16 0.8 0.2 1.645 0.1471 0.6529 0.9471 11 0.55 0.45 1.645 0.1830 0.3670 0.7330 12 0.6 0.4 1.645 0.1802 0.4198 0.7802 12 0.6 0.4 1.645 0.1802 0.4198 0.7802 12 0.6 0.4 1.645 0.1802 0.4198 0.7802 14 0.7 0.3 1.645 0.1686 0.5314 0.8686 8 0.4 0.6 1.645 0.1802 0.2198 0.5802 10 0.5 0.5 1.645 0.1839 0.3161 0.6839 12 0.6 0.4 1.645 0.1802 0.4198 0.7802 14 0.7 0.3 1.645 0.1686 0.5314 0.8686 14 0.7 0.3 1.645 0.1686 0.5314 0.8686 9 0.45 0.55 1.645 0.1830 0.2670 0.6330 8 0.4 0.6 1.645 0.1802 0.2198 0.5802 14 0.7 0.3 1.645 0.1686 0.5314 0.8686 15 0.75 0.25 1.645 0.1593 0.5907 0.9093 17 0.85 0.15 1.645 0.1313 0.7187 0.9813 15 0.75 0.25 1.645 0.1593 0.5907 0.9093 11 0.55 0.45 1.645 0.1830 0.3670 0.7330 15 0.75 0.25 1.645 0.1593 0.5907 0.9093 12 0.6 0.4 1.645 0.1802 0.4198 0.7802 11 0

8th Edition

ISBN: 9781305465510

Author: Rosner

Publisher: YUZU

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Chapter 6, Problem 93P

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Check whether the large-sample binomial confidence limit formula is adequate for the given distribution.

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Chapter 6, Problem 92P

Chapter 6, Problem 94P

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26. A certain system can experience three different types of defects. Let A;(i = 1,2,3) denote the event that the sys- tem has a defect of type i. Suppose thatP(A1) = .12 P(A) = .07 P(A) = .05P(A, U A2) = .13P(A, U A3) = .14P(A2 U A3) = .10P(A, A2 A3) = .011Rshelfa. What is the probability that the system does not havea type 1 defect?b. What is the probability that the system has both type 1 and type 2 defects?c. What is the probability that the system has both type 1 and type 2 defects but not a type 3 defect? d. What is the probability that the system has at most two of these defects?

The following are suggested designs for group sequential studies. Using PROCSEQDESIGN, provide the following for the design O’Brien Fleming and Pocock.• The critical boundary values for each analysis of the data• The expected sample sizes at each interim analysisAssume the standardized Z score method for calculating boundaries.Investigators are evaluating the success rate of a novel drug for treating a certain type ofbacterial wound infection. Since no existing treatment exists, they have planned a one-armstudy. They wish to test whether the success rate of the drug is better than 50%, whichthey have defined as the null success rate. Preliminary testing has estimated the successrate of the drug at 55%. The investigators are eager to get the drug into production andwould like to plan for 9 interim analyses (10 analyzes in total) of the data. Assume thesignificance level is 5% and power is 90%.Besides, draw a combined boundary plot (OBF, POC, and HP)

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Chapter 6 Solutions

EBK FUNDAMENTALS OF BIOSTATISTICS

Ch. 6.5 - What is the difference between a standard...Ch. 6.5 - What does a 95% CI mean?Ch. 6 - Suppose we are asked to construct a list of...Ch. 6 - Suppose we are asked to construct a list of...Ch. 6 - Suppose we are asked to construct a list of...Ch. 6 - Suppose we are asked to construct a list of...Ch. 6 - The data in Table 6.10 concern the mean triceps...Ch. 6 - The data in Table 6.10 concern the mean triceps...Ch. 6 - The data in Table 6.10 concern the mean triceps...Ch. 6 - The data in Table 6.10 concern the mean triceps...

Ch. 6 - The data in Table 6.10 concern the mean triceps...Ch. 6 - The data in Table 6.10 concern the mean triceps...Ch. 6 - Refer to the data in Table 2.13. Regard this...Ch. 6 - Refer to the data in Table 2.13. Regard this...Ch. 6 - Refer to the data in Table 2.13. Regard this...Ch. 6 - Refer to the data in Table 2.13. Regard this...Ch. 6 - Refer to the data in Table 2.13. Regard this...Ch. 6 - Refer to the data in Table 2.13. Regard this...Ch. 6 - Refer to the data in Table 2.13. 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