Question

8. A Union-Intersection Test

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Let X1,…,Xn be i.i.d. Bernoulli random variables with unknown parameter p∈(0,1). Suppose we want to test

H0:p∈[0.48,0.51]vsH1:p∉[0.48,0.51]

We want to construct an asymptotic test ψ for these hypotheses using X¯¯¯¯n. For this problem, we specifically consider the family of tests ψc1,c2 where we reject the null hypothesis if either X¯¯¯¯n<c1≤0.48 or X¯¯¯¯n>c2≥0.51 for some c1 and c2 that may depend on n, i.e.

ψc1,c2=1((X¯¯¯¯n<c1)∪(X¯¯¯¯n>c2))where c1<0.48<0.51<c2.

Throughout this problem, we will discuss possible choices for constants c1 and c2, and their impact to both the asymptotic and non-asymptotic level of the test.

Answer 1

0 unknown parameter pecc.. The Xn is @ the type Consider a random sample X,X2---Xn, where each random Variable is an lid Bernoulli random Varia-te with - NCOI) nin- PN(O) The hypothesis is to be tested is given as folloos Ho : PE (0.48,0.51] H: PE [048,0.51) asymptotic test w for testing the ruell hypothesis using given by Yesuch + I[(#seci)0 (Fo>ca)] where ,<O46cosiąc I event to reject the null hypothesis when the null hypothesis is correct. The maximun Type I error rate is given as follows :- Pe [0.48,0.51) Thus the correct answer is © According to Central limit thesiem, in-Elx). PC1-P) because E(X)=p and var(x)=pl-p)) Also it is given that VPC1-P) was where p€[048,0.51] paramo) (C#o 2c,)U (87120) [PCkon<ci)+ P(7) >ca)] max * = prefer Plitncci]+ plxnses] N Vasil) vn

Now, P(InCC 1) =P non-p) (-P) 12 1/2 mo Con Placo) Z= * Pfz< vané 7) (where 112 [aro (ci--)) (where ø is the cdf of 2) Thus, P(Xn-cı) = °CQva (cv-p)) Since [a rocci-p)] is a decreasing function of p, thus plin<er) is maximum at P=0.48 Similarly, (*) (6-P) p({n}(2) =P LPCI-P) PCI-P3 roncin-P) v ce-P) 1/2 112 Z. r Can-p) UN(0.1) 1/2 ? (where (C2-P) 142 (C2-P) - P[2> = 1-p[zz :1-9 [255 (C2-p)] (where & is the caf of 2) o won (P-C,)). Since Cavo (p-ca)) s an increasing function of p. thus,

3 P(xn>(2) is maximum at P=0.5 @ max It is given that P(En<C2) =0.005 PE (0:48,0-51] » Qlao (, -0.48)) = 0.025 CE -a(0.095) 28 +0.48 Also, max PE (0.48,0.51] P(x >C») =0.025 → ♡ (@ja (0.51-62)) =0.025 W(0.025) С, +0.51 210 * please Do Like, "Thank you you *