07+Sampling+distribution

07 Sampling distribution

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Unformatted text preview: e coefficient (1 − α) (or confidence level 100(1 − α)%) for normal random variable we define as σ σ (x − zα/2 σx , x + zα/2 σx ) = x − zα/2 √ , x + zα/2 √ , n n where quantity zα/2 is founded from the requirement P (x − zα/2 σx ≤ x ≤ x + zα/2 σx ) = 1 − α. 6 A. Zhensykbaev Sampling distributions Using z -score we have P (x − zα/2 σx ≤ x ≤ x + zα/2 σx ) = P (−zα/2 ≤ z ≤ zα/2 ) = = 2P (0 ≤ z ≤ zα/2 ). Thus, P (x − zα/2 σx ≤ x ≤ x + zα/2 σx ) = 2P (0 ≤ z ≤ zα/2 ). (1) On the Pic. 2 the value of z = zα/2 such that the area α/2 lies to its right. T α/2 −zα/2 0 α/2 zα/2 E z Pic. 2 The z value we can find using Table IV (McClave and Benson). For example, if α = 0.1 in the Table IV we should find that z0.05 will be the z value corresponding to an area is equal to 0.5-0.05=0.45: z0.05 = 1.645. Example. In a large airline company 225 flights are randomly selected, and the number of unoccupied seats is noted for each sampled flights. The sample mean and standard deviation are x = 11.6, s = 4.1. Estimate the mean number of unoccupied seats per flight during the past year, using a 90% confidence interval. Solution. Find α/2: due to (1 − α)% = 90% we have α 1 − α = 0.9, = 0.05, z0.05 = 1.645. 2 Since n = 225, s = 4.1 and it is the best approximation of the standard deviation σ , we find 2σ 4.1 µ ≈ x ± √ = 11.6 ± 1.645 √ = 11.6 ± 0.45 n 225 7 A. Zhensykbaev Sampling distributions (it means that µ ∈ [11.15, 12.05] with the probability 90%). Sample size determination. Assume that we are given a 100(1 − α)% confidence interval for mean µ. We want to estimate µ to within a given boundary for sampling error SE. How much will be the sample size n to accomplish this? From the equality σ SE = zα/2 σx = zα/2 √ . n we find: zα/2...
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This note was uploaded on 02/11/2014 for the course MATH 1390 taught by Professor Christopherstocker during the Spring '13 term at Marquette.

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