Survey
* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
Sampling Theory Exercises Use a standard normal table or the JMP calculator PROBABILITY function NORMAL DISTRIBUTION or NORMDIST to "compute" the standard normal probabilities. In fact, the entire exercise can be done on JMP. 1. Dwarf Apple Trees (This builds on the earlier exercise). The yearly growth of dwarf-apple-tree seedlings can be measured by the increase in the length of the central leader. Suppose that the second-year growth of such trees is normally distributed with a mean of 20 cm and a standard deviation of 6 cm. a. Compute the fraction of such dwarf-apple-tree seedlings that would be expected to have a second-year growth of between 18 and 22 cm. 22 20 18 20 P 18 Y 22 P Z 6 6 P 0.33 Z 0.33 P Z 0.33 P Z 0.33 0.6306 0.3694 0.2612 b. Now consider the sampling distribution of the sample mean for samples of size 4. Compute the expected value of the sampling distribution. E Y E Y 20 c. Compute the standard deviation of the sampling distribution of the mean for samples of size 4. SD Y Y Y n 6 3.0 4 d. For samples of size 5, compute the probability that the sample mean is between 18 and 22 cm. 22 20 18 20 P 18 Y 22 P Z 6 5 6 5 P 0.745 Z 0.745 P Z 0.745 P Z 0.745 0.7720 0.2280 0.5439 769815090, ©1998,2007 1 All rights reserved 4/29/2017 e. For samples of size 25, compute the probability that the sample mean is between 18 and 22 cm. 18 20 22 20 P 18 Y 22 P Z 6 25 6 25 P 1.67 Z 1.67 P Z 1.67 P Z 1.67 09522 0.0478 0.9044 f. For samples of size 100, compute the probability that the sample mean is between 18 and 22 cm. 18 20 22 20 P 18 Y 22 P Z 6 100 6 100 P 3.33 Z 3.33 P Z 3.33 P Z 3.33 099957 0.000429 0.99914 g. Graph the probability that the mean is between 18 and 22 cm as a function of the sample size for sample sizes of 1 to 100. 2. A veterinarian found that the average time it takes residents to perform a certain procedure is 12 minutes. Assume that the time it takes residents to perform the procedure is normally distributed with a mean of 12 minutes and a standard deviation of 2 minutes. 769815090, ©1998,2007 2 All rights reserved 4/29/2017 a. Compute the probability that a randomly selected resident would take between 11 and 13 minutes to perform the procedure, i.e., within 1.0 minute of the mean. 13 12 11 12 P 11 Y 13 P Z 2 2 P 0.50 Z 0.50 P Z 0.50 P Z 0.50 0.6915 0.3085 0.3829 b. Graph the probability that the sample mean would be between 11 and 13 minutes, for samples of size 1 to 100. 1 P(11<Time<13) 0.9 0.8 0.7 0.6 0.5 0.4 -10 0 10 20 30 40 50 60 70 80 90 100 n c. If you wanted the probability of being within 1.0 minute of the mean to be 95%, what is the minimum sample size that would be required? You can read this off the graph, or solve the appropriate formula for sample size. 1 P(11<Time<13) 0.9 0.8 0.7 0.6 0.5 0.4 -10 0 10 20 30 40 50 60 70 80 90 100 n 769815090, ©1998,2007 3 All rights reserved 4/29/2017 P 11 Y 13 0.95 11 12 13 12 P Z 2 n 2 n 1 1 P Z 0.95 2 n 2 n P 1.96 Z 1.96 0.95 implies 1 z0.975 1.96 2 n where z0.975 denotes the 0.975 quantile of the standard normal distribution. Solving the equation for n yields 1.96 2 n 15.4 1 2 indicating a sample of size 16 would be required. (notice that we always round up.) In general, the formula for prescribing the sample size is z1 2 n MOE 2 where z1 2 is the (1 – /2) quantile of the of the standard normal distribution, is the population standard deviation, and MOE is the desired margin of error. 769815090, ©1998,2007 4 All rights reserved 4/29/2017