Thus the airline company needs to select randomly

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

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...
View Full Document

This note was uploaded on 02/11/2014 for the course MATH 1390 taught by Professor Christopherstocker during the Spring '13 term at Marquette.

Ask a homework question - tutors are online