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AMS 572 Lecture NotesSept. 1 6 , 200 7 .Chapter 7: Inference on one population mean 1 Motivation & simple random sampleEg) We wish to estimate the average height of adult US males Take a random sample.- “Simple” random sample: every subject in the population has the same chance to be selected.2 Introduction to statistical inference on one population meanFor a “random sample” of size n: nXXX ,,,21<i> Point estimatior X → sample mean (nXnXXXniin121...)Other estimators: median, mode, trimmed mean, …<ii> Confidence Interval (C.I.)Eg) 95% C.I. for μ 99.9999% C.I. (‘6-9’ in the manufacture industry)<iii> Hypothesis TestEg) 0H: μ "'651H: μ "'65Point Estimator, C.I., Test  Statistical Inference- Draw some conclusion on the population (parameters of interest) based on a random sample.3 Normal Distribution <i> Probability Density Function (p.d.f.)),(~2NX(X follows normal distribution of mean  and variance 2)222)(21)(xexf, Rxx  ,badxxfbXaP )()(= area under the pdf curve bounded by a and b<ii> Cumulative Density Function (c.d.f.)xdttfxXPxF )()()()()]'([)( xFdxdxFxf <iii> Z-scoreStandard Normal Distribution)1,0(~ NZ),(~2NX)1,0(~ NXZ)()(bXaPbXaP)(bZaPProof) 1> Use c.d.f.)()()()()( aFaXPaXPaZPaFxz )()()()( afaFdadaFdadafxxzz22)(2222121aaee →the pdf for N(0,1)2> p.d.f. → Derive the p.d.f. directly (Jacobian) (Homework #1)3> m.g.f.)()(tXXeEtM  dxxfetx)(, if x is continuoussxtxxfe')(, if x is discreteThm If two ‘RV’s have the same MGF, they are of the same distribution. Normal Distributions ),(~2NX2221)(ttXetMEg) If the mgf of X is 2107 tte, then )20,7(~ NX)(),(~,2tMNXXZz ?Linear transformation : bXaY , a&b are constants)()(][)()()( btatXbtatXbaXttYYeeEeEeEeEtM ]2)exp[()(2222222tatbaeeeEetatabtatXbt),(~22abaNY  baXZ ,122221)1(21)1()1()(ttttXtZeeetMetM → m.g.f. for )1,0(N<iv> eg) Sgt. Jones wishes to select one army recruit into his unit. It is known that the IQ distribution for all recruits in normal with mean 180 and standard deviation 10. What is the chance that Sgt. Jones would select a recruit with an IQ of at least 200?Sol) )10,180(~ NX%28.2)2()1018020010180()200(  ZPXPXPX is the IQ of a randomly selected recruit.<v> the distribution of the sample mean XEg) If Sgt. Jones wishes to select 9 recruits into his unit, what is the chance that their average IQ is at least 200?Thm Let nXXX ,,,21 be a random sample from a normal population with mean , variance 2. Then, the distribution of X is ),(~2nNXProof) Show that niNXdiii,...1),,(~2... ⇒ ),(~21nNnXXnii(i.i.d. : independently and identically distributed))()()()(ntXnXtXtXiieEeEeEtM),(~)2)(exp()2exp()(22212221nNtntntnteEniniXntiSol) )9100,180(~,9)100,180(~2NXnNX 0)6()310180200310180()200( 


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SBU AMS 572 - AMS 572 Lecture Notes

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