Practiceexam2solutions

practiceExam2solutions
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Statistics 651: Survey Sampling Methods Winter 2011 Practice Problems for Exam 2 Solutions 1. Nursery Workers (a) This is stratifed random sampling Followed by a single stage cluster sample. The probability that any one worker is selected can be calculated: π ij = P(worker selected | nursery selected)P(nursery selected) = 1 × n h N h Thus, For each oF the strata, the weight oF every worker in that stratum is N h n h , as shown in this table: State N h n h weight = N h /n n AZ 243 20 12.15000 TN 1383 23 60.13043 RI 87 2 43.50000 DE 131 10 13.10000 OH 1434 32 44.81250 MI 1525 12 127.08333 IN 437 10 43.70000 ±L 7703 30 256.76667 (b) With the Further subsampling, the probability oF selection now becomes: π ij = P(worker selected | nursery selected)P(nursery selected) = m ih M ih × n h N h Thus, the weight is w ij = N h M ih n h m ih . ±or the frst nursery: w ij = 87 × 21 2 × 10 = 91 . 35 ±or the second nursery: w ij = 87 × 34 2 × 10 = 147 . 9 Note that the weight increased over 43.5 For both nurseries, and it increased more For the larger nursery. This is because the 10 people selected From the
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Unformatted text preview: larger nursery need to represent more people than the 10 people From the smaller nursery. 2. Lockhart City (a) I used the SAS program using sampling probability proportional to size. ±irst, I read the data From fgure A2 into SAS, and then I used SURVEYSELECT with PPS, and SIZE = number oF households in the county. I have chosen districts 51, 71, 62, 65, and 63. 1 (b) P(select household ij on frst draw) = P(select household j on frst draw | Cluster i selected on frst draw) × P(Cluster i selected on frst draw) = 1 M i M i K = 1 K = 1 19664 . (c) Yes. (d) Variance Estimates and CIs (calculated in R): i. deF = 3. Ignoring the ±pc ±or the SRS sample, ˆ V ( SRS ) = 0 . 19 ˆ V ( Complex ) = deff * ˆ V ( SRS ) = 3 * . 19 = 0 . 575 95% CI: 11 . 047 ± 1 . 96 . 575 = (9 . 55 , 12 . 54) 2...
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