Stat303: Statistical MethodsHomework 5 (07/18/2009)Instructor: Anh DaoProblem 1(a) µcaught= nπcaught(b) Let n = 29 and if probability of errors caught is 0.70 then probability of errors missed= 1 −0.70 =0.3. Thus µmissed= n ∗ (1 − πcaught) = 29 ∗ (1 − 0.7) = 29 ∗ 0.3(c) If π (or p in hw and ebook notation)=0.7 then, σ =p(nπ(1 − π)) =√29 ∗ 0.7 ∗ 0.3. If we changeπ (or p) σ will also change.Problem 2 In this problem we need to change a proportion (π) to Z in order to use z-table.(a) Suppose that the probability of correctly answering a question chosen at random from a universeof possible questions is 0.85 for John.That is π = 0.85. Then the probability that John scored 80% or lower is P (p ≤ 0.80) for a testwith 100 problems. Recall that p ∼ N(π, p(1 − p)/n). We need to change X to Z to use z-table.Thus,P [p ≤ 0.80] = P"p − πpπ(1 − π)/n<0.8 − πpπ(1 − π)/n#= P"Z <0.8 − 0.85p0.85(1 − 0.85)/100#= P [Z < −1.4] = 0.0808(b) Same as part a just use n = 250 instead of 100.(c) It is given that σ1=p0.85(1 − 0.85)/100 for John and if σ2=σ1/2 then what is n? Use thefollowing σ2=p0.85(1 − 0.85)/n, then solve for n in dependence of σ2. Hence n = {π(1−π)}/σ22.Problem 3 Use sampling distribution of ¯x. Suppose σ = 5.8 mg and the measurements are repeated 4times, n = 4.(a) σ¯x=σ√n.(b) If σ = 5.8 then what should be n so that σ¯x= 2.9. Using σ¯x=σ√nwe can write n =σσ¯x2=5.82.92. Round up to the next natural number.(c) Suppose the question was: What is the probability that Antonio misses the true mass by morethan 2.66 mg in either direction if he makes one measurement.We want to find: P [|X − µ| > 2.66] = Ph|X−µ|σ> 2.66/σi= P[|Z| > 2.66/5.8] = 2P [Z >2.66/5.8](d) If he had 8 independent samples , then we want to find P [|¯X−µ| > 2.66] = Ph|¯X−µ|σ/√n> 2.66/(5.8/√8)i=P [|Z| > 2.66/(5.8/√8)] = 2P [Z > 2.66/(5.8/√8)].12 Homework 5Problem 4 Note that¯X ∼ Nµ,pσ2/n2. So µ¯x= µ and σ¯x= σ/√n.What is the probability that the diameter of a single randomly chosen axle differs from the target valueby 0.0053 mm or more?We want to compute P (|X − µ| > 0.0053). Proceed as in Problem 3.What is the probability that the mean diameter of an SRS of 4 axles differs from the target value by0.0053 mm or more?We want to compute P (|¯X − µ| > 0.0053). Proceed as in Problem 3.Problem 5 Using Central Limit Theorem we have, ¯x ∼ Nµ,σ√n2, then find probability that themean number of flaws exceed 1.6 (for example) ,i.e., P [¯X > 1.6]. So change¯X to Z by using Z =¯X−µσ/√n.Problem 6(a)(b) The probability that untreated is at least 25 lbs greater than treated is the same as the probabilitythat the random variable¯Xuntreated−¯Xtreatedis greater than 25. Note that¯Xuntreated−¯Xtreatedisdistributed Nµuntreated− µtreated, σ2untreated/4 + σ2treated/4. Let S =¯Xuntreated−¯Xtreated. Now wewant to find P (S > 25). Compute z-score for S and use z-Table to find P (S > 25).Problem 7(a) First you have to check if the condition for large sample holds to have normal distribution i.e.,check if nπ ≥ 10 and n(1 −π) ≥ 10. Only when these conditions hold true compute probabilitiesby using the normal approximation for p.(b) Note that the distribution for p is Nπ,pπ(1 − π)/n2.Suppose, the question was “If you choose 100 vehicles at random, what is the probability thatmore than half (that is, 51 or more) carry just one person?”That means we should compute P (p > 0.5). Now compute z-score for p. Recall the mean for p isπ which is already given and standard deviation ispπ(1 − π)/n.Problem 8 Let µ for 1 egg=65.1 and σ2=52then µ for a carton having 12 eggs is µcarton=12*65.1and σ2for carton is σ2carton= 12 ∗ σ2. Knowing mean and variance for a carton we can find theP [775 < Carton < 825] =P [775−µcartonσcarton< Z <825−µcartonσcarton].Problem 9 same as problem 2.Problem 10 From Central limit theorem,¯X ∼ N(µ,σ√n) so that Empirical rule works for¯X also, lowerlimit= µ −3 ∗σ√nand upper limit= µ + 3 ∗σ√n.Problem 11(a) Just find the mean and s.d. for sampling distribution of ¯x and ¯y.(b) Use the linear combination property that µ¯y−¯x= µ¯y− µ¯xand σ2¯y−¯x= σ2¯y+ σ2¯xif X and Y areindependent.Problem 12(a) What is the approximate distribution of the proportion pFof women who worked last summer?N (πF, πF(1 − πF)/n)(b) Use the linear combination property that πM-F= πM−πF. and σ2M-F= πM(1−πM)/n+πF(1−πF)/nif pFand pMare
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