Asked by ramj
                Let  π1,β¦,ππ  be i.i.d. Bernoulli random variables with unknown parameter  πβ(0,1) . Suppose we want to test
π»0:πβ[0.48,0.51]vsπ»1:πβ[0.48,0.51]
 
We want to construct an asymptotic test π for these hypotheses using πβ―β―β―β―β―π. For this problem, we specifically consider the family of tests ππ1,π2 where we reject the null hypothesis if either πβ―β―β―β―β―π<π1β€0.48 or πβ―β―β―β―β―π>π2β₯0.51 for some π1 and π2 that may depend on π , i.e.
ππ1,π2=1((πβ―β―β―β―β―π<π1)βͺ(πβ―β―β―β―β―π>π2))where π1<0.48<0.51<π2.
 
Throughout this problem, we will discuss possible choices for constants π1 and π2 , and their impact to both the asymptotic and non-asymptotic level of the test.
b) Use the central limit theorem and the approximation π(1βπ)βΎβΎβΎβΎβΎβΎβΎβΎββ12 for πβ[0.48,0.51] to approximate ππ(πβ―β―β―β―β―π<π1) and ππ(πβ―β―β―β―β―π>π2) for large π. Express your answers as a formula in terms of π1, π2, π and π.
(Write Phi for the cdf of a Normal distribution, c_1 for π1, and c_2 for π2.)
ππ(πβ―β―β―β―β―π<π1)β ?
For what value of πβ[0.48,0.51] is the expression above for ππ(πβ―β―β―β―β―π<π1) maximized?
ππ(πβ―β―β―β―β―π<π1)is max at π= ?
ππ(πβ―β―β―β―β―π>π2)β ?
For what value of πβ[0.48,0.51] is the expression above for ππ(πβ―β―β―β―β―π>π2) maximized?
ππ(πβ―β―β―β―β―π>π2)is max at π= ?
d) Suppose that we wish to have a level πΌ=0.05. What π1 and π2 will achieve πΌ=0.05? Choose π1 and π2 by setting the expressions you obtained above for maxπβ[0.48,0.51]ππ(πβ―β―β―β―β―π<π1) and maxπβ[0.48,0.51]ππ(πβ―β―β―β―β―π>π2) to both be 0.025.
(If applicable, enter q(alpha) for ππΌ, the 1βπΌ-quantile of a standard normal distribution, e.g. enter q(0.01) for π0.01. )
π1= ?
 
π2=?
e) We will now show that the values we just derived for π1 and π2 are in fact too conservative.
Recall the expression from part (b) for ππ(πβ―β―β―β―β―π<π1) for large π. For π>0.48 (note the strict inequality), find limπββππ(πβ―β―β―β―β―π<π1).
limπββππ>0.48(πβ―β―β―β―β―π<π1)= ?
Similarly, for π<0.51 (note the strict inequality), find limπββππ(πβ―β―β―β―β―π>π2). Use the expression you found in part (b) for ππ(πβ―β―β―β―β―π>π2).
limπββππ<0.51(πβ―β―β―β―β―π>π2)= ?
f) Next, we analyze the asymptotic test given different possible values of π, in order to choose suitable and sufficiently-tight π1 and π2. Looking more closely at part (d), we may note that the asymptotic behavior of the expressions for the errors are different depending on whether π=0.48, 0.48<π<0.51, or π=0.51.
Based on your answers and work from the previous part, evaluate the asymptotic Type 1 error
π(πβ―β―β―β―β―π<π1)+π(πβ―β―β―β―β―π>π2).
 
on each of the three cases for the value of π in terms of π1, π2, and π, and determine in each case which component(s) of the Type 1 error will converge to zero.
This would allow you to come up with a new set of conditions for π1 and π2 in terms of π, given the desired level of 5%. Enter these values (in terms of π) below.
(If applicable, enter q(alpha) for ππΌ, the 1βπΌ-quantile of a standard normal distribution, e.g. enter q(0.01) for π0.01. Do not worry about the parser not rendering q(alpha) properly; the grader will work nonetheless. You could also enclose q(alpha) by brackets for the rendering to show properly.)
π1= ?
   
π2= ?
            
        π»0:πβ[0.48,0.51]vsπ»1:πβ[0.48,0.51]
We want to construct an asymptotic test π for these hypotheses using πβ―β―β―β―β―π. For this problem, we specifically consider the family of tests ππ1,π2 where we reject the null hypothesis if either πβ―β―β―β―β―π<π1β€0.48 or πβ―β―β―β―β―π>π2β₯0.51 for some π1 and π2 that may depend on π , i.e.
ππ1,π2=1((πβ―β―β―β―β―π<π1)βͺ(πβ―β―β―β―β―π>π2))where π1<0.48<0.51<π2.
Throughout this problem, we will discuss possible choices for constants π1 and π2 , and their impact to both the asymptotic and non-asymptotic level of the test.
b) Use the central limit theorem and the approximation π(1βπ)βΎβΎβΎβΎβΎβΎβΎβΎββ12 for πβ[0.48,0.51] to approximate ππ(πβ―β―β―β―β―π<π1) and ππ(πβ―β―β―β―β―π>π2) for large π. Express your answers as a formula in terms of π1, π2, π and π.
(Write Phi for the cdf of a Normal distribution, c_1 for π1, and c_2 for π2.)
ππ(πβ―β―β―β―β―π<π1)β ?
For what value of πβ[0.48,0.51] is the expression above for ππ(πβ―β―β―β―β―π<π1) maximized?
ππ(πβ―β―β―β―β―π<π1)is max at π= ?
ππ(πβ―β―β―β―β―π>π2)β ?
For what value of πβ[0.48,0.51] is the expression above for ππ(πβ―β―β―β―β―π>π2) maximized?
ππ(πβ―β―β―β―β―π>π2)is max at π= ?
d) Suppose that we wish to have a level πΌ=0.05. What π1 and π2 will achieve πΌ=0.05? Choose π1 and π2 by setting the expressions you obtained above for maxπβ[0.48,0.51]ππ(πβ―β―β―β―β―π<π1) and maxπβ[0.48,0.51]ππ(πβ―β―β―β―β―π>π2) to both be 0.025.
(If applicable, enter q(alpha) for ππΌ, the 1βπΌ-quantile of a standard normal distribution, e.g. enter q(0.01) for π0.01. )
π1= ?
π2=?
e) We will now show that the values we just derived for π1 and π2 are in fact too conservative.
Recall the expression from part (b) for ππ(πβ―β―β―β―β―π<π1) for large π. For π>0.48 (note the strict inequality), find limπββππ(πβ―β―β―β―β―π<π1).
limπββππ>0.48(πβ―β―β―β―β―π<π1)= ?
Similarly, for π<0.51 (note the strict inequality), find limπββππ(πβ―β―β―β―β―π>π2). Use the expression you found in part (b) for ππ(πβ―β―β―β―β―π>π2).
limπββππ<0.51(πβ―β―β―β―β―π>π2)= ?
f) Next, we analyze the asymptotic test given different possible values of π, in order to choose suitable and sufficiently-tight π1 and π2. Looking more closely at part (d), we may note that the asymptotic behavior of the expressions for the errors are different depending on whether π=0.48, 0.48<π<0.51, or π=0.51.
Based on your answers and work from the previous part, evaluate the asymptotic Type 1 error
π(πβ―β―β―β―β―π<π1)+π(πβ―β―β―β―β―π>π2).
on each of the three cases for the value of π in terms of π1, π2, and π, and determine in each case which component(s) of the Type 1 error will converge to zero.
This would allow you to come up with a new set of conditions for π1 and π2 in terms of π, given the desired level of 5%. Enter these values (in terms of π) below.
(If applicable, enter q(alpha) for ππΌ, the 1βπΌ-quantile of a standard normal distribution, e.g. enter q(0.01) for π0.01. Do not worry about the parser not rendering q(alpha) properly; the grader will work nonetheless. You could also enclose q(alpha) by brackets for the rendering to show properly.)
π1= ?
π2= ?
Answers
                    Answered by
            Anonymous
            
    8.(a). Ξ±=maxpβ[0.48,0.51](Pp(Xn<c1)+Pp(Xn>c2))
    
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