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Unformatted text preview: Introduction, 1 sample ttest and tinterval Central Limit Theorem Let X 1 ,X 2 ,...,X n be a random sample from a distribution with mean μ and variance σ 2 .Then if n is sufficiently large (Rule of thumb > 30), ¯ X has approximately a normal distribution with μ ¯ X = μ and σ 2 ¯ X = σ 2 /n t distribution When ¯ X is the mean of a random sample of size n from a normal distribution with mean μ , the random variable t = ¯ X μ s/ √ n has a probability distribution called t distribution with n1 degrees of free dom. Experiment example 87 individuals were given a flu vaccination. After 28 days, blood samples were taken to assess the concentration of antibody ( X ) in their serum. Some summary statistics for the sample are as follows. n = 87, ¯ X = 1 . 689, s = 1 . 549. Assume that the sample is from a normal population with mean μ and vari ance σ 2 . 1 Confidence interval for the mean of a normal population The form of the 100(1 α ) % confidence interval for μ is ¯ x ± t α/ 2 ,n 1 s √ n where t α/ 2 ,n 1 is the upper α/ 2 percentage point of the t distribution with n 1 degrees of freedom. • To find a 99% CI, α = . 01, so α/ 2 = . 005 • n = 87, so there are 86 degrees of freedom • t . 005 , 86 ≈ t . 005 , 75 = 2 . 643 (using 75 degrees of freedom, which is the df nearest 86 from t table...
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 Spring '12
 daniel
 Normal Distribution, Statistical hypothesis testing, flu, CSB, Test of mu

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