Section 7.1: Basic Properties of ConfidenceInterval1• Let x1, ···, xnbe observations.• Suppose they are iid normal distributed, i.e.,x1, ···, xnare observed values of iid randomvariables X1, ···, Xnas we can writeX1, ···, Xn∼iidN(µ, σ2).• We define the sample mean¯x =1nnXi=1xiand the sample variances2=1n − 1nXi=1(xi− ¯x)2.We call s as the sample standard deviation.• Then,¯X ∼ N(µ,σ2n).• Thus, we haveP (−zα2≤¯X − µσ/√n≤ zα2) = 1 − α.2• Therefore, with probability 1 − α there is−zα2≤¯X − µσ/√n≤ zα2which is equivalent to say that with probability1 − α there is¯X − zα2σ√n≤ µ ≤¯X + zα2σ√n.• We may take the above as the confidence in-terval, i.e, when σ is known.• Therefore, we have the following formula: sup-pose x1, ···, xnare iid observations of a normalpopulation and assume the standard deviationσ is known, then the 1 −α level confidence in-terval for µ is¯x ± zα2σ√n= [¯x − zα2σ√n, ¯x + zα2σ√n].• The interpretation of the confidence intervalis that if we repeat the procedure many manytimes, with probability 1 − α the above confi-dence interval contains the true value of µ.3• An often asked question is about the length ofconfidence interval. How large is the samplesize n so that the 1−α level confidence intervalis less than w.• Note that the length of the 1 − α level confi-dence interval is2zα2σ√n.Thus, we have2zα2σ√n≤ w ⇒ n ≥ (2zα2σw)2=4z2α2σ2w2.4First example of Section 7.1: examples 7.1 and7.2 on textbook.• Study the preferred height for an experimentalkeyboard with large forearm-wrist support.• Take sample size n = 31. Observed samplemean ¯x = 80.0cm. We known the standarddeviation σ = 2.0. Assume height is normallydistributed.• The 95% confidence interval is¯x±z0.025σ√n= 80.0±1.96×2.0√31= [79.3, 80.7].5Second example of Section 7.1: example 7.3 ontextbook.• Study the hole diameters. Assume it is nor-mally distributed.• Historical data had suggested σ = 0.1mm.Assume it does not change.• A sample of n = 40 units reported the samplemean ¯x = 5.426.• The 90% confidence interval is5.426 ± 1.645 ×0.1√40= [5.400, 5.452].6Third example of Section 7.1: example 7.4 ontextbook. Suppose the response time is normallydistributed with σ = 25ms. How large the samplen we need so that the 95% confidence interval isnot greater than 10.Answer:n ≥ (2 × 1.96 × 25/10)2= 96.04Therefore, we should choose n = 97 since theleast integer ≥ 96.04 is
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