How to find a confidence interval of a binomial distribution using a simulated random sample? ...
Inductor and Capacitor in Parallel
Is it convenient to ask the journal's editor for two additional days to complete a review?
Is there an equivalent of cd - for cp or mv
Why is information "lost" when it got into a black hole?
Plausibility of squid whales
What CSS properties can the br tag have?
What difference does it make using sed with/without whitespaces?
how one can write a nice vector parser, something that does pgfvecparse{A=B-C; D=E x F;}
The Ultimate Number Sequence Puzzle
Is it ok to trim down a tube patch?
My ex-girlfriend uses my Apple ID to login to her iPad, do I have to give her my Apple ID password to reset it?
Audio Conversion With ADS1243
How to avoid supervisors with prejudiced views?
Is there a difference between "Fahrstuhl" and "Aufzug"?
Asymptote: 3d graph over a disc
From jafe to El-Guest
(How) Could a medieval fantasy world survive a magic-induced "nuclear winter"?
How to use ReplaceAll on an expression that contains a rule
Would a grinding machine be a simple and workable propulsion system for an interplanetary spacecraft?
How to Implement Deterministic Encryption Safely in .NET
Do scriptures give a method to recognize a truly self-realized person/jivanmukta?
Can Sneak Attack be used when hitting with an improvised weapon?
Why am I getting "Static method cannot be referenced from a non static context: String String.valueOf(Object)"?
Is it correct to say moon starry nights?
How to find a confidence interval of a binomial distribution using a simulated random sample?
The Next CEO of Stack OverflowHow to calculate a confidence interval for $p$, the parameter of the binomial distribution?Uniform Distribution vs Exponential Distribution + Confidence intervalHow to find the confidence interval without the variance?Finding confidence interval for a binomial process using the normal distribution?Confidence interval of a binomial distribution?Confidence interval for sampleObtain a $(1-alpha) 100$% confidence interval for $theta$ using the moment estimatorHypothesis testing using the 95% Confidence Interval of Sample MeanIs Confidence Interval taken on one Random Sample or A Sampling DistributionBinomial distribution confidence interval using a random sample?
$begingroup$
I have a random sample of 1000 values of deviates from binomial distribution with n = 52 and p^
So I have 1000 values from the distribution.
How can I find a 95% confidence interval for the true value of p? (Without using normal distribution approximations).
It seems I just have random values as a sample to find two values from within which have a 95% probability of containing p?
statistics statistical-inference binomial-distribution
asked Mar 17 at 17:46
PumpkinpeachPumpkinpeach
578
$endgroup$
add a comment |
$begingroup$
I have a random sample of 1000 values of deviates from binomial distribution with n = 52 and p^
So I have 1000 values from the distribution.
How can I find a 95% confidence interval for the true value of p? (Without using normal distribution approximations).
It seems I just have random values as a sample to find two values from within which have a 95% probability of containing p?
statistics statistical-inference binomial-distribution
asked Mar 17 at 17:46
PumpkinpeachPumpkinpeach
578
$endgroup$
add a comment |
$begingroup$
I have a random sample of 1000 values of deviates from binomial distribution with n = 52 and p^
So I have 1000 values from the distribution.
How can I find a 95% confidence interval for the true value of p? (Without using normal distribution approximations).
It seems I just have random values as a sample to find two values from within which have a 95% probability of containing p?
statistics statistical-inference binomial-distribution
asked Mar 17 at 17:46
PumpkinpeachPumpkinpeach
578
$endgroup$
I have a random sample of 1000 values of deviates from binomial distribution with n = 52 and p^
So I have 1000 values from the distribution.
How can I find a 95% confidence interval for the true value of p? (Without using normal distribution approximations).
It seems I just have random values as a sample to find two values from within which have a 95% probability of containing p?
statistics statistical-inference binomial-distribution
statistics statistical-inference binomial-distribution
asked Mar 17 at 17:46
PumpkinpeachPumpkinpeach
578
asked Mar 17 at 17:46
PumpkinpeachPumpkinpeach
578
edited Mar 19 at 15:56
Pumpkinpeach
asked Mar 17 at 17:46
PumpkinpeachPumpkinpeach
578
asked Mar 17 at 17:46
PumpkinpeachPumpkinpeach
578
asked Mar 17 at 17:46
PumpkinpeachPumpkinpeach
578
578
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
By p^, I assume you mean $hat p = 44/52,$ the estimate of the cure rate $p$ as found in the experiment done by the manufacturer of the new drug.
In R, $B = 1000$ observations from $mathsf{Binom}(n=52,, p = 44/52)$ are generated as follows:
set.seed(2019); x = rbinom(1000, 52, 44/52)
Then we see that (a centrally located) 95% of the values $x/52$ lie
within the interval $(0.73, 0.94).$
set.seed(318); x = rbinom(1000, 52, 44/52)
quantile(x/52, c(.025, .975))
2.5% 97.5%
0.7307692 0.9423077
This is close to the same interval we get with
the Wald 95% confidence interval $hat p pm 1.96sqrt{frac{hat p(1-hat p)}{52}},$ which
amounts to $(0.748, 0.944).$
p.est = 44/52; pm=c(-1,1)
p.est + pm*1.96*sqrt(p.est*(1-p.est)/52)
[1] 0.7480870 0.9442207
Notes: (1) Of course, without simulation, we could have obtained a more accurate version of the first
interval in R, using the
quantile function (inverse CDF) of the appropriate
binomial distribution.
qbinom(c(.025,.975), 52, 44/52)/52
[1] 0.7500000 0.9423077
(2) Especially for $n$ as small as 52, the Agresti-Coull ("plus four") CI has better coverage properties than the (asymptotic) Wald interval. Perhaps see this Q & A.
n = 56; p.est = 46/n; pm=c(-1,1)
p.est + pm*1.96*sqrt(p.est*(1-p.est)/52)
[1] 0.7173299 0.9255273
answered Mar 18 at 22:37
BruceETBruceET
36.2k71540
$endgroup$
add a comment |
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3151825%2fhow-to-find-a-confidence-interval-of-a-binomial-distribution-using-a-simulated-r%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
By p^, I assume you mean $hat p = 44/52,$ the estimate of the cure rate $p$ as found in the experiment done by the manufacturer of the new drug.
In R, $B = 1000$ observations from $mathsf{Binom}(n=52,, p = 44/52)$ are generated as follows:
set.seed(2019); x = rbinom(1000, 52, 44/52)
Then we see that (a centrally located) 95% of the values $x/52$ lie
within the interval $(0.73, 0.94).$
set.seed(318); x = rbinom(1000, 52, 44/52)
quantile(x/52, c(.025, .975))
2.5% 97.5%
0.7307692 0.9423077
This is close to the same interval we get with
the Wald 95% confidence interval $hat p pm 1.96sqrt{frac{hat p(1-hat p)}{52}},$ which
amounts to $(0.748, 0.944).$
p.est = 44/52; pm=c(-1,1)
p.est + pm*1.96*sqrt(p.est*(1-p.est)/52)
[1] 0.7480870 0.9442207
Notes: (1) Of course, without simulation, we could have obtained a more accurate version of the first
interval in R, using the
quantile function (inverse CDF) of the appropriate
binomial distribution.
qbinom(c(.025,.975), 52, 44/52)/52
[1] 0.7500000 0.9423077
(2) Especially for $n$ as small as 52, the Agresti-Coull ("plus four") CI has better coverage properties than the (asymptotic) Wald interval. Perhaps see this Q & A.
n = 56; p.est = 46/n; pm=c(-1,1)
p.est + pm*1.96*sqrt(p.est*(1-p.est)/52)
[1] 0.7173299 0.9255273
answered Mar 18 at 22:37
BruceETBruceET
36.2k71540
$endgroup$
add a comment |
$begingroup$
By p^, I assume you mean $hat p = 44/52,$ the estimate of the cure rate $p$ as found in the experiment done by the manufacturer of the new drug.
In R, $B = 1000$ observations from $mathsf{Binom}(n=52,, p = 44/52)$ are generated as follows:
set.seed(2019); x = rbinom(1000, 52, 44/52)
Then we see that (a centrally located) 95% of the values $x/52$ lie
within the interval $(0.73, 0.94).$
set.seed(318); x = rbinom(1000, 52, 44/52)
quantile(x/52, c(.025, .975))
2.5% 97.5%
0.7307692 0.9423077
This is close to the same interval we get with
the Wald 95% confidence interval $hat p pm 1.96sqrt{frac{hat p(1-hat p)}{52}},$ which
amounts to $(0.748, 0.944).$
p.est = 44/52; pm=c(-1,1)
p.est + pm*1.96*sqrt(p.est*(1-p.est)/52)
[1] 0.7480870 0.9442207
Notes: (1) Of course, without simulation, we could have obtained a more accurate version of the first
interval in R, using the
quantile function (inverse CDF) of the appropriate
binomial distribution.
qbinom(c(.025,.975), 52, 44/52)/52
[1] 0.7500000 0.9423077
(2) Especially for $n$ as small as 52, the Agresti-Coull ("plus four") CI has better coverage properties than the (asymptotic) Wald interval. Perhaps see this Q & A.
n = 56; p.est = 46/n; pm=c(-1,1)
p.est + pm*1.96*sqrt(p.est*(1-p.est)/52)
[1] 0.7173299 0.9255273
answered Mar 18 at 22:37
BruceETBruceET
36.2k71540
$endgroup$
add a comment |
$begingroup$
By p^, I assume you mean $hat p = 44/52,$ the estimate of the cure rate $p$ as found in the experiment done by the manufacturer of the new drug.
In R, $B = 1000$ observations from $mathsf{Binom}(n=52,, p = 44/52)$ are generated as follows:
set.seed(2019); x = rbinom(1000, 52, 44/52)
Then we see that (a centrally located) 95% of the values $x/52$ lie
within the interval $(0.73, 0.94).$
set.seed(318); x = rbinom(1000, 52, 44/52)
quantile(x/52, c(.025, .975))
2.5% 97.5%
0.7307692 0.9423077
This is close to the same interval we get with
the Wald 95% confidence interval $hat p pm 1.96sqrt{frac{hat p(1-hat p)}{52}},$ which
amounts to $(0.748, 0.944).$
p.est = 44/52; pm=c(-1,1)
p.est + pm*1.96*sqrt(p.est*(1-p.est)/52)
[1] 0.7480870 0.9442207
Notes: (1) Of course, without simulation, we could have obtained a more accurate version of the first
interval in R, using the
quantile function (inverse CDF) of the appropriate
binomial distribution.
qbinom(c(.025,.975), 52, 44/52)/52
[1] 0.7500000 0.9423077
(2) Especially for $n$ as small as 52, the Agresti-Coull ("plus four") CI has better coverage properties than the (asymptotic) Wald interval. Perhaps see this Q & A.
n = 56; p.est = 46/n; pm=c(-1,1)
p.est + pm*1.96*sqrt(p.est*(1-p.est)/52)
[1] 0.7173299 0.9255273
answered Mar 18 at 22:37
BruceETBruceET
36.2k71540
$endgroup$
By p^, I assume you mean $hat p = 44/52,$ the estimate of the cure rate $p$ as found in the experiment done by the manufacturer of the new drug.
In R, $B = 1000$ observations from $mathsf{Binom}(n=52,, p = 44/52)$ are generated as follows:
set.seed(2019); x = rbinom(1000, 52, 44/52)
Then we see that (a centrally located) 95% of the values $x/52$ lie
within the interval $(0.73, 0.94).$
set.seed(318); x = rbinom(1000, 52, 44/52)
quantile(x/52, c(.025, .975))
2.5% 97.5%
0.7307692 0.9423077
This is close to the same interval we get with
the Wald 95% confidence interval $hat p pm 1.96sqrt{frac{hat p(1-hat p)}{52}},$ which
amounts to $(0.748, 0.944).$
p.est = 44/52; pm=c(-1,1)
p.est + pm*1.96*sqrt(p.est*(1-p.est)/52)
[1] 0.7480870 0.9442207
Notes: (1) Of course, without simulation, we could have obtained a more accurate version of the first
interval in R, using the
quantile function (inverse CDF) of the appropriate
binomial distribution.
qbinom(c(.025,.975), 52, 44/52)/52
[1] 0.7500000 0.9423077
(2) Especially for $n$ as small as 52, the Agresti-Coull ("plus four") CI has better coverage properties than the (asymptotic) Wald interval. Perhaps see this Q & A.
n = 56; p.est = 46/n; pm=c(-1,1)
p.est + pm*1.96*sqrt(p.est*(1-p.est)/52)
[1] 0.7173299 0.9255273
answered Mar 18 at 22:37
BruceETBruceET
36.2k71540
edited Mar 18 at 23:11
answered Mar 18 at 22:37
BruceETBruceET
36.2k71540
answered Mar 18 at 22:37
BruceETBruceET
36.2k71540
answered Mar 18 at 22:37
BruceETBruceET
36.2k71540
36.2k71540
add a comment |
add a comment |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3151825%2fhow-to-find-a-confidence-interval-of-a-binomial-distribution-using-a-simulated-r%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
