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1. Every sample of size 6 has equal chance (randomly choose 6 students out of 30). Answer following questions: (a) How many possible samples? We have a total of 30 students and want a sample of 6, therefore we must calculate how many combinations of 6 we can make out of 30, 30πΆ6 = 593,775 possible samples. (b) p(s) = ? The probability of selecting a specific sample is 1/593,775 = 0.000001684 1 π(π ) = {153,775 0 ππ π  βππ  6 πππππππ‘π  ππ‘βπππ€ππ π (c) ns = ? The size of the sample is ππ  = 6 (d) Number of sample containing unit k? If we include unit k in the sample, we have 5 more spots to fill from the 29 students left. That is 29C5=118,755. (e) First order inclusion probabilities? The probability that unit k is included in the sample, π(π β π ), denoted as ππ , is ππ = 6 πππ π = 1,2, β¦ 30 30 ππ = 1 πππ π = 1,2, β¦ 30 5 (f) Second order inclusion probabilities? The probability that units k and l are included in the sample, π(π, π β π ), denoted as πππ , is πππ = 6(6 β 1) πππ π β  π = 1,2, β¦ 30 30(30 β 1) πππ = 30 πππ π β  π = 1,2, β¦ 30 870 πππ = 1 πππ π β  π = 1,2, β¦ 30 29 2. If the design π(. ) has a fixed sample size n, then πππ β₯ ππ + ππ β π. We know that for simple size n from a population N, we ...
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