Section 7.2: Large-Sample Confidence Intervalsfor a Population Mean and Proportion1Confidence Interval for expected value when n islarge (n > 40).• Let x1, ···, xnbe observations.• Suppose they are iid distributed, but we maynot have the normal assumption.• Assume n is large (e.g. n > 40). Then byCLT¯X ∼ N(µ,σ2n).• Even we may also not have the value of σ, weknow the sample variance s2≈ σ2. Thus, weapproximately have¯X − µs/√n∼approxN(0, 1).• Then, we haveP (−zα2≤¯X − µs/√n≤ zα2) ≈ 1 − α.which implies the 1 −α level confidence inter-val for µ is approximated by¯X − zα2s√n≤ µ ≤¯X + zα2s√nwhen n is large (n > 40).2Confidence Interval for binomial proportion whenn is large (np > 10 and n(1 − p) > 10).• Suppose X ∼ Bin(n, p) and X is observed.Then, the estimate of p is ˆp = X/n withˆp ∼approxN(p,p(1 − p)n)• Thus, approximatelyP (−zα2≤ˆp − pqp(1 − p)/n≤ zα2) ≈ 1 − α.• Solve the inequality−zα2≤ˆp − pqp(1 − p)/n≤ zα2.We have the 1−α level confidence interval forp has the lower boundˆp +z2α22n− zα2sˆp(1−ˆp)n+z2α24n21 +z2α2n,and the upper boundˆp +z2α22n+ zα2sˆp(1−ˆp)n+z2α24n21 +z2α2n.3Sample size determination for the length of con-fidence interval.• For the binomial proportion, we haven ≈4z2α2ˆp(1 − ˆq)w2and this attains its maximum when ˆp = 1/2.Thus, whenn ≈z2α/2w2the length of 1 −α level confidence interval isless than w (a given value).4First example of Section 7.2: Example 7.6 ontextbook. The data collected 48 breakdown volt-age of a particular circuit under certain condi-tions. Compute the 95% confidence interval forthe true breakdown voltage (µ).• Since n = 48, we can use the large sampleconfidence interval.• From the data, we have¯x =148(62 + 50 + ··· + 58) = 54.7ands2=14748Xi=1(xi− ¯x)2=147[(62 − 54.7)2+ (50 − 54.7)2+··· + (58 − 54.7)2]=5.232.• The 95% large sample confidence interval forµ is54.7 ± 1.965.23√48= [53.2, 56.2].5Second example of Section 7.2: Examples 7.8and 7.9 on textbook. An article reported that inn = 48 trials, 16 resulted in ignition of a partialtype of substrate by a lighted cigarette. Then,the observed value of X is x = 16. AssumeX ∼ Bin(48, p).Then, we haveˆp =1648= 0.3333.Put this in the formula. Note that z0.025= 1.96.We have the 95% large sample confidence intervalfor p is0.3333 +1.96296± 1.96 ×r0.3333×0.666748+1.9629621 +1.96248=[0.2167, 0.4746].Note that the length of the interval is 0.2579. Ifwe want the confidence interval less than 0.1, thesample size n is aroundn ≈1.9620.12= 380.25Then, we choose n =
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