07+Sampling+distribution

Thus the airline company needs to select randomly

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Unformatted text preview: σ 2 . n= SE If σ is unknown but n is large we may replace it by the sample standard deviation zα/2 s 2 n= . SE If we know the range r of observations in the population, we can take σ ≈ r/4. Example. Consider previous example. Assume that the company wants the 90% confidence interval for estimate of mean with a sampling error SE=0.2. How many flights will have to be randomly selected? Solution. Using last formula we obtain: n= 1.645 · 4.1 0.2 2 ≈ 1, 137. Thus, the airline company needs to select randomly 1,137 flights. Small-sample estimation of a population mean. Now we try to estimate a population mean using small random samples. The population standard deviation σ is almost always unknown. The sample standard deviation s may provide a poor approximation for σ when the sample size is small. Therefore using the standard normal statistic z= x−µ x−µ = √, σx σ/ n 8 A. Zhensykbaev Sampling distributions we define the t-statistic x−µ √. s/ n For normal random variable t-statistic has mound-shaped symmetric sampling distribution with mean 0. It depends on the sample size n. This dependence we define as: t-statistic has (n − 1) degrees of freedom (df ). The number (n − 1) in the divisor in the formula of the sample variance. Small-sample confidence interval for a population mean µ is s s x − tα/2 √ , x + tα/2 √ , n n t= where tα/2 is defined similarly zα/2 from assumption s s P x − tα/2 √ ≤ x ≤ x + tα/2 √ = 1 − α, n n using the Table VI (McClave and Benson). For example, for α = 0.1 and df=5 we found from the Table VI (column t0.05 , row 5) that t0.05 = 2.015. The sample size needed to estimate µ within a given bound B is n= tα/2 s B 2 . Example. Suppose we get a sample of 5 expert opinions about next year’s earnings per share for a stock, x = 2.63, s = 0.72. What is the confidence interval of a level α = 0.05 for unknown µ? Solution. We have df = 5 − 1 = 4, P = 1 − α = 0.95, t0.05 = 2.132 and s s 0.72 0.72 µ ∈ x − tα/2 √ , x + tα/2 √ = 2.63 − 2.132 √ , 2.63 + 2.132 √ . n n 5 5 Thus, with the probability 95% µ ∈ [1.944, 3.316]. 9...
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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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