Section 7.3: Interval based on a normalpopulation distribution1Definition of χ2and t distributions.• Suppose X1, ···, Xnare iid N(0, 1). Then,nXi=1X2i∼ χ2n,where χ2nrepresents the chi-square distribu-tion with n degrees of freedom.• If X ∼ N(0, 1), Y ∼ χ2n, and X and Y areindependent, thenT =XqY/n∼ tn,where tnrepresents the t-distribution with ndegrees of freedom.• Tables A.5 and A.7 gives some useful valuesfor quantile functions as denoted by tα,νandχ2α,νof tνand χ2νdistribution for selected ν,where α is the probability of the upper part.2Suppose we observed x1, ···, xnfrom a normalpopulation N(µ, σ2). Then•nXi=1[(Xi− µ)2] ∼ σ2χ2n•(n − 1)S2=nXi=1[(Xi−¯X)2] ∼ σ2χ2n−1.•¯X ∼ N(µ,σ2n)and¯X and S2are independent.• Therefore, we haveT =¯X − µqS2/n∼ tn−1.• We denote χ2α,νas the upper probability of χ2distribution with ν degrees of freedom.• We denote tα,νas the upper probability of t-distribution with ν degrees of freedom.3Suppose X1, ···, Xnare iid observed from a nor-mal population as N(µ, σ2). Both µ and σ2areunknown.• The 1 −α level one-sample t confidence inter-val is¯x ± tα2,n−1s√n.• The 1 − α level prediction interval for the ex-pected value of the next observation is¯x ± tα2,n−1s√nand the 1 − α level prediction interval for thevalue of the next observation is¯x ± tα2,n−1ss1 +1n.4First example of Section 7.3: example 7.11 ontextbook. In this example, the data collected 30observations on sweetgum lumber. We have¯x = 1203.191ands = 543.54.Then, the 95% confidence for the mean elasticityis1203.191±t0.025,29543.54√30= [7000.235, 7406.129].5Second example of Section 7.3: examples 7.12and 7.13 on textbook. We observed 10 valuesof fat content, with ¯x = 21.90 and s = 4.134.Then, the 95% confidence interval for the ex-pected value of next sample is21.90 ± t0.025,94.134√10= [18.94, 24.86].The 95% confidence interval for the next samplevalue is21.90 ± t0.025,94.134s1 +110= [12.09,
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