How to calculate a (possible) chance from a zero-incidence sample?
$begingroup$
This is probably a very simple problem, but I want to make sure I have a correct understanding. I have a sample of $500$ events, in which a complication $C$ didn't occur. How do I calculate a reasonably correct chance for $C$?
Would that be just $<1/500$? My intuition is that it would be a bit higher, as there is a sampling effect.
Obviously, it is impossible to calculate the chance exactly, but does something like some kind of confidence interval exist for these types of observations?
Much obliged,
Joris
probability
edited Jan 2 at 15:11
klirk
2,288631
asked Jan 2 at 15:07
JorisJoris
212
$endgroup$
add a comment |
$begingroup$
This is probably a very simple problem, but I want to make sure I have a correct understanding. I have a sample of $500$ events, in which a complication $C$ didn't occur. How do I calculate a reasonably correct chance for $C$?
Would that be just $<1/500$? My intuition is that it would be a bit higher, as there is a sampling effect.
Obviously, it is impossible to calculate the chance exactly, but does something like some kind of confidence interval exist for these types of observations?
Much obliged,
Joris
probability
edited Jan 2 at 15:11
klirk
2,288631
asked Jan 2 at 15:07
JorisJoris
212
$endgroup$
add a comment |
$begingroup$
This is probably a very simple problem, but I want to make sure I have a correct understanding. I have a sample of $500$ events, in which a complication $C$ didn't occur. How do I calculate a reasonably correct chance for $C$?
Would that be just $<1/500$? My intuition is that it would be a bit higher, as there is a sampling effect.
Obviously, it is impossible to calculate the chance exactly, but does something like some kind of confidence interval exist for these types of observations?
Much obliged,
Joris
probability
edited Jan 2 at 15:11
klirk
2,288631
asked Jan 2 at 15:07
JorisJoris
212
$endgroup$
This is probably a very simple problem, but I want to make sure I have a correct understanding. I have a sample of $500$ events, in which a complication $C$ didn't occur. How do I calculate a reasonably correct chance for $C$?
Would that be just $<1/500$? My intuition is that it would be a bit higher, as there is a sampling effect.
Obviously, it is impossible to calculate the chance exactly, but does something like some kind of confidence interval exist for these types of observations?
Much obliged,
Joris
probability
probability
edited Jan 2 at 15:11
klirk
2,288631
asked Jan 2 at 15:07
JorisJoris
212
edited Jan 2 at 15:11
klirk
2,288631
asked Jan 2 at 15:07
JorisJoris
212
edited Jan 2 at 15:11
klirk
2,288631
edited Jan 2 at 15:11
klirk
2,288631
edited Jan 2 at 15:11
klirk
2,288631
2,288631
asked Jan 2 at 15:07
JorisJoris
212
asked Jan 2 at 15:07
JorisJoris
212
asked Jan 2 at 15:07
JorisJoris
212
212
add a comment |
add a comment |
1 Answer
1
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oldest
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Here's a possible answer. I'm no statistician, but I remember this from many years ago when I worked for a large public accounting firm. When doing an audit, they might want to be say $95%$ confident that a particular procedure was followed at least $95%$ at the time. They would pick a certain number $n$ of transactions to sample so that if the procedure was followed exactly $95%$ of the time, then then probability of finding no exceptions would be less than $5%$. That is, determine $n$ so that $.95^n<.05$
Your situation seems a bit different, as you seem to have already made $500$ experiments, but following the logic above would give that we can be $99.07%$ confident that there is no complication at least $99.07%$ of the time, since $$.9907^{500}approx.009355$$ That is, with about $99%$ confidence, the chance of $C$ is less than $.0093,$ which is considerably higher than $1/500=.002$
As I say, I'm no statistician, but I would be more comfortable with this if the number of trials, and the interpretation in the event of no complication were determined in advance.
answered Jan 2 at 16:58
saulspatzsaulspatz
17.1k31435
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add a comment |
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1 Answer
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active
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1 Answer
1
active
oldest
votes
active
oldest
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active
oldest
votes
$begingroup$
Here's a possible answer. I'm no statistician, but I remember this from many years ago when I worked for a large public accounting firm. When doing an audit, they might want to be say $95%$ confident that a particular procedure was followed at least $95%$ at the time. They would pick a certain number $n$ of transactions to sample so that if the procedure was followed exactly $95%$ of the time, then then probability of finding no exceptions would be less than $5%$. That is, determine $n$ so that $.95^n<.05$
Your situation seems a bit different, as you seem to have already made $500$ experiments, but following the logic above would give that we can be $99.07%$ confident that there is no complication at least $99.07%$ of the time, since $$.9907^{500}approx.009355$$ That is, with about $99%$ confidence, the chance of $C$ is less than $.0093,$ which is considerably higher than $1/500=.002$
As I say, I'm no statistician, but I would be more comfortable with this if the number of trials, and the interpretation in the event of no complication were determined in advance.
answered Jan 2 at 16:58
saulspatzsaulspatz
17.1k31435
$endgroup$
add a comment |
$begingroup$
Here's a possible answer. I'm no statistician, but I remember this from many years ago when I worked for a large public accounting firm. When doing an audit, they might want to be say $95%$ confident that a particular procedure was followed at least $95%$ at the time. They would pick a certain number $n$ of transactions to sample so that if the procedure was followed exactly $95%$ of the time, then then probability of finding no exceptions would be less than $5%$. That is, determine $n$ so that $.95^n<.05$
Your situation seems a bit different, as you seem to have already made $500$ experiments, but following the logic above would give that we can be $99.07%$ confident that there is no complication at least $99.07%$ of the time, since $$.9907^{500}approx.009355$$ That is, with about $99%$ confidence, the chance of $C$ is less than $.0093,$ which is considerably higher than $1/500=.002$
As I say, I'm no statistician, but I would be more comfortable with this if the number of trials, and the interpretation in the event of no complication were determined in advance.
answered Jan 2 at 16:58
saulspatzsaulspatz
17.1k31435
$endgroup$
add a comment |
$begingroup$
Here's a possible answer. I'm no statistician, but I remember this from many years ago when I worked for a large public accounting firm. When doing an audit, they might want to be say $95%$ confident that a particular procedure was followed at least $95%$ at the time. They would pick a certain number $n$ of transactions to sample so that if the procedure was followed exactly $95%$ of the time, then then probability of finding no exceptions would be less than $5%$. That is, determine $n$ so that $.95^n<.05$
Your situation seems a bit different, as you seem to have already made $500$ experiments, but following the logic above would give that we can be $99.07%$ confident that there is no complication at least $99.07%$ of the time, since $$.9907^{500}approx.009355$$ That is, with about $99%$ confidence, the chance of $C$ is less than $.0093,$ which is considerably higher than $1/500=.002$
As I say, I'm no statistician, but I would be more comfortable with this if the number of trials, and the interpretation in the event of no complication were determined in advance.
answered Jan 2 at 16:58
saulspatzsaulspatz
17.1k31435
$endgroup$
Here's a possible answer. I'm no statistician, but I remember this from many years ago when I worked for a large public accounting firm. When doing an audit, they might want to be say $95%$ confident that a particular procedure was followed at least $95%$ at the time. They would pick a certain number $n$ of transactions to sample so that if the procedure was followed exactly $95%$ of the time, then then probability of finding no exceptions would be less than $5%$. That is, determine $n$ so that $.95^n<.05$
Your situation seems a bit different, as you seem to have already made $500$ experiments, but following the logic above would give that we can be $99.07%$ confident that there is no complication at least $99.07%$ of the time, since $$.9907^{500}approx.009355$$ That is, with about $99%$ confidence, the chance of $C$ is less than $.0093,$ which is considerably higher than $1/500=.002$
As I say, I'm no statistician, but I would be more comfortable with this if the number of trials, and the interpretation in the event of no complication were determined in advance.
answered Jan 2 at 16:58
saulspatzsaulspatz
17.1k31435
answered Jan 2 at 16:58
saulspatzsaulspatz
17.1k31435
answered Jan 2 at 16:58
saulspatzsaulspatz
17.1k31435
answered Jan 2 at 16:58
saulspatzsaulspatz
17.1k31435
17.1k31435
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