Interval Estimation for a Binomial Proportion. This simulation will compare two different interval estimators for an unknown binomial proportion - a notoriously difficult and interesting problem. Consider observing n iid realizations from the following data generating process (DGP) X1, X2, n iid Xnd Ber(p) i=1 Consider estimator p EX₁. To form an interval estimator, recall that √(p-p) A √Var[X] N(0, 1) by the CLT. Thus, asymptotically P{%a/2 ≤ √(-) ≤ 21-1/a} = 1-a. Here, %, is the √Var[X] v-th percentile of the standard normal. Solving the inner inequality for p yields a (1-a)100% confidence interval for p. The Wald interval estimator does this inversion and estimates Var[X] by plugging in p wherever p appears in the expression for Var[X]. The Agresti-Coull interval "adds 2 success in 4 trials" and computes an interval using p = (2+1 X;), where ñ= n +4 (a) For a = .05, solve the inner inequality of P{2.025 ≤ √(1-1) ≤ 2.975} = .95 to obtain an √Var[X] interval that (claims to) contain p with 95% probability. Show that it has the form p(1 − p) n p±2.9751 The Wald interval by plugging in p: p±2.975 √p(1-P)/n
Interval Estimation for a Binomial Proportion. This simulation will compare two different interval estimators for an unknown binomial proportion - a notoriously difficult and interesting problem. Consider observing n iid realizations from the following data generating process (DGP) X1, X2, n iid Xnd Ber(p) i=1 Consider estimator p EX₁. To form an interval estimator, recall that √(p-p) A √Var[X] N(0, 1) by the CLT. Thus, asymptotically P{%a/2 ≤ √(-) ≤ 21-1/a} = 1-a. Here, %, is the √Var[X] v-th percentile of the standard normal. Solving the inner inequality for p yields a (1-a)100% confidence interval for p. The Wald interval estimator does this inversion and estimates Var[X] by plugging in p wherever p appears in the expression for Var[X]. The Agresti-Coull interval "adds 2 success in 4 trials" and computes an interval using p = (2+1 X;), where ñ= n +4 (a) For a = .05, solve the inner inequality of P{2.025 ≤ √(1-1) ≤ 2.975} = .95 to obtain an √Var[X] interval that (claims to) contain p with 95% probability. Show that it has the form p(1 − p) n p±2.9751 The Wald interval by plugging in p: p±2.975 √p(1-P)/n
MATLAB: An Introduction with Applications
6th Edition
ISBN:9781119256830
Author:Amos Gilat
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Chapter1: Starting With Matlab
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![Interval Estimation for a Binomial Proportion. This simulation will compare two
different interval estimators for an unknown binomial proportion - a notoriously difficult and
interesting problem. Consider observing n iid realizations from the following data generating
process (DGP)
Consider estimator p
~
√n(p-p)
√Var[X]
1X₁. To form an interval estimator, recall that √(-p) A
√Var[X]
N(0, 1) by the CLT. Thus, asymptotically P{%a/2 ≤ ≤ 21-1/a} 1-a. Here, z, is the
v-th percentile of the standard normal. Solving the inner inequality for p yields a (1-a)100%
confidence interval for p. The Wald interval estimator does this inversion and estimates
Var[X] by plugging in ô wherever p appears in the expression for Var[X]. The Agresti-Coull
interval "adds 2 success in 4 trials" and computes an interval using p = (2+Σ1 X₂), where
ñ = n +4
iid
X1, X2,... Xn Ber (p)
=
(a) For a = .05, solve the inner inequality of P{2.025 ≤ √(1-P) ≤ 2.975} = .95 to obtain an
✓Var[X]
interval that (claims to) contain p with 95% probability. Show that it has the form
p(1 - p)
n
=
p±2.9751
The Wald interval by plugging in p: p±2.975 √p(1-P)/n](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fda314194-c4cb-4c99-a706-843df994562e%2Ff2226cd6-1eac-4a0b-8d28-4ff2645996e6%2Fb3xoxpj_processed.png&w=3840&q=75)
Transcribed Image Text:Interval Estimation for a Binomial Proportion. This simulation will compare two
different interval estimators for an unknown binomial proportion - a notoriously difficult and
interesting problem. Consider observing n iid realizations from the following data generating
process (DGP)
Consider estimator p
~
√n(p-p)
√Var[X]
1X₁. To form an interval estimator, recall that √(-p) A
√Var[X]
N(0, 1) by the CLT. Thus, asymptotically P{%a/2 ≤ ≤ 21-1/a} 1-a. Here, z, is the
v-th percentile of the standard normal. Solving the inner inequality for p yields a (1-a)100%
confidence interval for p. The Wald interval estimator does this inversion and estimates
Var[X] by plugging in ô wherever p appears in the expression for Var[X]. The Agresti-Coull
interval "adds 2 success in 4 trials" and computes an interval using p = (2+Σ1 X₂), where
ñ = n +4
iid
X1, X2,... Xn Ber (p)
=
(a) For a = .05, solve the inner inequality of P{2.025 ≤ √(1-P) ≤ 2.975} = .95 to obtain an
✓Var[X]
interval that (claims to) contain p with 95% probability. Show that it has the form
p(1 - p)
n
=
p±2.9751
The Wald interval by plugging in p: p±2.975 √p(1-P)/n
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