ST3239: Survey Methodology - The Department of Statistics and ...

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4) An approximate (1 − α) confidence interval for τ is √ ˆτ ∓ z α/2 ̂V ar(ˆτ) = ˆτ ∓ z α/2 N √ s ( ) s √1 − f = N ȳ ∓ z α/2 √ √1 − f . n n 5) Minimum sample size n needed to have an error bound B with probability 1 − α n ≈ Nσ 2 B2 , where D = (N − 1)D + σ2 N 2 zα/2 2 Example 4.6. (Page 95 of the textbook). An investigator is interested in estimating the total weight gain in 0 to 4 weeks for N = 1000 chicks fed on a new ration. Obviously, to weigh each bird would be time-consuming and tedious. Therefore, determine the number of chicks to be sampled in this study in order to estimate τ within a bound on the error of estimation equal to 1000 grams with probability 95%. Many similar studies on chick nutrition have been run in the past. Using data from these studies, the investigator found that σ 2 , the population variance, was approximately 36.00 (grams) 2 . Determine the required sample size. Solution 15
2.4 Estimation of population proportion If we are interested in the proportion p of the population with a specified characteristic. Let y i = { 1 if the ith element has the characteristic 0 if not It is easy to see that E(y i ) = E(y 2 i ) = p (Why?). Therefore, we have µ = E(y i ) = p, σ 2 = var(y i ) = p − p 2 = pq, where q = 1 − p So the total number of elements in the sample of size n possessing the specified characteristic is ∑ n i=1 y i . Therefore, 1. An estimator of the population proportion p is ∑ ni=1 y i ȳ = = ˆp, say. n And an estimator of the population variance σ 2 = pq is s 2 1 n∑ = (y i − ȳ) 2 = 1 ( n ) ∑ yi 2 − n(ȳ) 2 n − 1 i=1 n − 1 i=1 ( 1 n ) ∑ = y i − nˆp 2 = 1 ( ) nˆp − nˆp 2 n − 1 i=1 n − 1 n = ˆpˆq where ˆq = 1 − ˆp n − 1 From Theorems 2.2.2 and 2.2.3, we have E(ˆp) = p, E(s 2 ) = N N − 1 σ2 = 2. Again, from Theorem 2.2.2, the variance of ˆp is V ar(ˆp) = σ2 n ( ) N − n = pq N − 1 n N pq. (4.1) N − 1 ( ) N − n . N − 1 3. From equation (4.1) and Theorem 2.2.5, an estimator of the variance of ˆp is ̂V ar(ˆp) = s2 n (1 − f) = ˆpˆq (1 − f) . n − 1 4. An approximate (1 − α) confidence interval for p is √ √ˆpˆq ˆp ∓ z α/2 ̂V ar(ˆp) = ˆp ∓ z α/2 √ √1 − f. n − 1 16
Page 1 and 2: ST3239: Survey Methodology by Wang
Page 3 and 4: 1.3 Why sample? If a sample is equa
Page 5 and 6: Proof. By the definition of s.r.s,
Page 7 and 8: 2.2 Estimation of population mean a
Page 9 and 10: Summary 1. How to draw a simple ran
Page 11 and 12: Central limit theorem: If n → ∞
Page 13 and 14: e.g. Suppose that a total of 1500 s
Page 15: 2.3.2 Estimation of population tota
Page 19 and 20: Solution 2.5 Comparing estimates Su
Page 21 and 22: Proof. Note that ˆp 1 = X/n and ˆ

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