Consistency of two measurements including means and standard deviations
$begingroup$
This is a simplified version of a real life experiment where we have done two experiments attempt to measure the same quantity and we obtained the results $0.8 pm 0.1$ and $1.2 pm 0.2.$ (That's all we know!)
How can we calculate the probability that these two measurements are consistent with each other (i.e. they are consistent with a single true value)?
probability statistics statistical-inference
asked Dec 7 '14 at 20:47
Ehsan M. KermaniEhsan M. Kermani
6,40412348
$endgroup$
add a comment |
$begingroup$
This is a simplified version of a real life experiment where we have done two experiments attempt to measure the same quantity and we obtained the results $0.8 pm 0.1$ and $1.2 pm 0.2.$ (That's all we know!)
How can we calculate the probability that these two measurements are consistent with each other (i.e. they are consistent with a single true value)?
probability statistics statistical-inference
asked Dec 7 '14 at 20:47
Ehsan M. KermaniEhsan M. Kermani
6,40412348
$endgroup$
$begingroup$
We don't know what $pm 0.1$ means. It seem likely to be a $95$% confidence region under an assumption of normal distribution, but it could be a $90$% confidence region; and if these measurements are from a "six-sigma" method then the confidence is much higher than $95$%. Are there really no other clues as to what was meant? (And, of course, what are your thoughts so far on how to do this?)
$endgroup$
– David K
Dec 7 '14 at 21:15
$begingroup$
Dear @DavidK, as far as I know, $pm 0.1$ is for one-sigma i.e. $68%$ confidence region. There are actually no other clues. I was thinking about t-test and F-test (F distribution) given one degree of freedom, but I'm not really sure!
$endgroup$
– Ehsan M. Kermani
Dec 8 '14 at 2:26
$begingroup$
Good, it sounds like you have sample standard deviations. If you don't know sample sizes, that's a bit of a hindrance.
$endgroup$
– David K
Dec 8 '14 at 5:24
add a comment |
$begingroup$
This is a simplified version of a real life experiment where we have done two experiments attempt to measure the same quantity and we obtained the results $0.8 pm 0.1$ and $1.2 pm 0.2.$ (That's all we know!)
How can we calculate the probability that these two measurements are consistent with each other (i.e. they are consistent with a single true value)?
probability statistics statistical-inference
asked Dec 7 '14 at 20:47
Ehsan M. KermaniEhsan M. Kermani
6,40412348
$endgroup$
This is a simplified version of a real life experiment where we have done two experiments attempt to measure the same quantity and we obtained the results $0.8 pm 0.1$ and $1.2 pm 0.2.$ (That's all we know!)
How can we calculate the probability that these two measurements are consistent with each other (i.e. they are consistent with a single true value)?
probability statistics statistical-inference
probability statistics statistical-inference
asked Dec 7 '14 at 20:47
Ehsan M. KermaniEhsan M. Kermani
6,40412348
asked Dec 7 '14 at 20:47
Ehsan M. KermaniEhsan M. Kermani
6,40412348
asked Dec 7 '14 at 20:47
Ehsan M. KermaniEhsan M. Kermani
6,40412348
asked Dec 7 '14 at 20:47
Ehsan M. KermaniEhsan M. Kermani
6,40412348
asked Dec 7 '14 at 20:47
Ehsan M. KermaniEhsan M. Kermani
6,40412348
6,40412348
$begingroup$
We don't know what $pm 0.1$ means. It seem likely to be a $95$% confidence region under an assumption of normal distribution, but it could be a $90$% confidence region; and if these measurements are from a "six-sigma" method then the confidence is much higher than $95$%. Are there really no other clues as to what was meant? (And, of course, what are your thoughts so far on how to do this?)
$endgroup$
– David K
Dec 7 '14 at 21:15
$begingroup$
Dear @DavidK, as far as I know, $pm 0.1$ is for one-sigma i.e. $68%$ confidence region. There are actually no other clues. I was thinking about t-test and F-test (F distribution) given one degree of freedom, but I'm not really sure!
$endgroup$
– Ehsan M. Kermani
Dec 8 '14 at 2:26
$begingroup$
Good, it sounds like you have sample standard deviations. If you don't know sample sizes, that's a bit of a hindrance.
$endgroup$
– David K
Dec 8 '14 at 5:24
add a comment |
$begingroup$
We don't know what $pm 0.1$ means. It seem likely to be a $95$% confidence region under an assumption of normal distribution, but it could be a $90$% confidence region; and if these measurements are from a "six-sigma" method then the confidence is much higher than $95$%. Are there really no other clues as to what was meant? (And, of course, what are your thoughts so far on how to do this?)
$endgroup$
– David K
Dec 7 '14 at 21:15
$begingroup$
Dear @DavidK, as far as I know, $pm 0.1$ is for one-sigma i.e. $68%$ confidence region. There are actually no other clues. I was thinking about t-test and F-test (F distribution) given one degree of freedom, but I'm not really sure!
$endgroup$
– Ehsan M. Kermani
Dec 8 '14 at 2:26
$begingroup$
Good, it sounds like you have sample standard deviations. If you don't know sample sizes, that's a bit of a hindrance.
$endgroup$
– David K
Dec 8 '14 at 5:24
$begingroup$
We don't know what $pm 0.1$ means. It seem likely to be a $95$% confidence region under an assumption of normal distribution, but it could be a $90$% confidence region; and if these measurements are from a "six-sigma" method then the confidence is much higher than $95$%. Are there really no other clues as to what was meant? (And, of course, what are your thoughts so far on how to do this?)
$endgroup$
– David K
Dec 7 '14 at 21:15
$begingroup$
We don't know what $pm 0.1$ means. It seem likely to be a $95$% confidence region under an assumption of normal distribution, but it could be a $90$% confidence region; and if these measurements are from a "six-sigma" method then the confidence is much higher than $95$%. Are there really no other clues as to what was meant? (And, of course, what are your thoughts so far on how to do this?)
$endgroup$
– David K
Dec 7 '14 at 21:15
$begingroup$
Dear @DavidK, as far as I know, $pm 0.1$ is for one-sigma i.e. $68%$ confidence region. There are actually no other clues. I was thinking about t-test and F-test (F distribution) given one degree of freedom, but I'm not really sure!
$endgroup$
– Ehsan M. Kermani
Dec 8 '14 at 2:26
$begingroup$
Dear @DavidK, as far as I know, $pm 0.1$ is for one-sigma i.e. $68%$ confidence region. There are actually no other clues. I was thinking about t-test and F-test (F distribution) given one degree of freedom, but I'm not really sure!
$endgroup$
– Ehsan M. Kermani
Dec 8 '14 at 2:26
$begingroup$
Good, it sounds like you have sample standard deviations. If you don't know sample sizes, that's a bit of a hindrance.
$endgroup$
– David K
Dec 8 '14 at 5:24
$begingroup$
Good, it sounds like you have sample standard deviations. If you don't know sample sizes, that's a bit of a hindrance.
$endgroup$
– David K
Dec 8 '14 at 5:24
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
If we assume that the $pm$ denotes a $x$ confidence interval and the confidence intervals are symmetrical and that there is a single true value $y$ then you can say:
$$p=begin{cases}
frac{(1-x)^2}{4},&ylt0.7\
frac{x(1-x)}{2},&0.7le yle0.9\\
frac{(1-x)^2}{4},&0.9lt ylt1.0\\
frac{x(1-x)}{2},&1.0le yle1.4\\
frac{(1-x)^2}{4},&1.4lt y\\
end{cases}$$
Now if you add all this up, you get a value which is (surprisingly) $in[0,1]$. You might be tempted to say that this is the probability that there is a single true value but you would be wrong!
And this is why - consider $y$ and $y+Delta$. For $Delta$ sufficiently large so that you are happy to say that they are distinct, the probabilities that they each fall into their respective intervals would be exactly the same - this is the definition of a confidence interval! So the probability that they are identical is equal to the probability that they are not identical for $Delta$, and $-Delta$ and $2Delta$ and an infinite number of other $Delta$ variants. Add up an infinite number of numbers $gt0$ and you will soon have a number $gt 1$ so it cannot represent a probability.
Without more information on the methodology, you can only state that the probability of a single true value is $in[0,1]$, but then, what isn't?
answered Dec 8 '14 at 2:34
Dale MDale M
2,5621822
$endgroup$
$begingroup$
Thanks Dale. Let me ask What those probability functions are coming from?
$endgroup$
– Ehsan M. Kermani
Dec 8 '14 at 18:34
$begingroup$
If you construct a confidence interval for a mean, say 95%, then there is a 95% chance that the value lies within it. Therefore there is a 2.5% it lies above and a 2.5% it lies below. Since the experiments (and therefore the confidence intervals) are independent, the probabilities follow. To be totally accurate, you should partition the low and high probabilities but the maths has no real meaning and was provided to illustrate the point.
$endgroup$
– Dale M
Dec 8 '14 at 22:33
add a comment |
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1 Answer
1
active
oldest
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1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
If we assume that the $pm$ denotes a $x$ confidence interval and the confidence intervals are symmetrical and that there is a single true value $y$ then you can say:
$$p=begin{cases}
frac{(1-x)^2}{4},&ylt0.7\
frac{x(1-x)}{2},&0.7le yle0.9\\
frac{(1-x)^2}{4},&0.9lt ylt1.0\\
frac{x(1-x)}{2},&1.0le yle1.4\\
frac{(1-x)^2}{4},&1.4lt y\\
end{cases}$$
Now if you add all this up, you get a value which is (surprisingly) $in[0,1]$. You might be tempted to say that this is the probability that there is a single true value but you would be wrong!
And this is why - consider $y$ and $y+Delta$. For $Delta$ sufficiently large so that you are happy to say that they are distinct, the probabilities that they each fall into their respective intervals would be exactly the same - this is the definition of a confidence interval! So the probability that they are identical is equal to the probability that they are not identical for $Delta$, and $-Delta$ and $2Delta$ and an infinite number of other $Delta$ variants. Add up an infinite number of numbers $gt0$ and you will soon have a number $gt 1$ so it cannot represent a probability.
Without more information on the methodology, you can only state that the probability of a single true value is $in[0,1]$, but then, what isn't?
answered Dec 8 '14 at 2:34
Dale MDale M
2,5621822
$endgroup$
$begingroup$
Thanks Dale. Let me ask What those probability functions are coming from?
$endgroup$
– Ehsan M. Kermani
Dec 8 '14 at 18:34
$begingroup$
If you construct a confidence interval for a mean, say 95%, then there is a 95% chance that the value lies within it. Therefore there is a 2.5% it lies above and a 2.5% it lies below. Since the experiments (and therefore the confidence intervals) are independent, the probabilities follow. To be totally accurate, you should partition the low and high probabilities but the maths has no real meaning and was provided to illustrate the point.
$endgroup$
– Dale M
Dec 8 '14 at 22:33
add a comment |
$begingroup$
If we assume that the $pm$ denotes a $x$ confidence interval and the confidence intervals are symmetrical and that there is a single true value $y$ then you can say:
$$p=begin{cases}
frac{(1-x)^2}{4},&ylt0.7\
frac{x(1-x)}{2},&0.7le yle0.9\\
frac{(1-x)^2}{4},&0.9lt ylt1.0\\
frac{x(1-x)}{2},&1.0le yle1.4\\
frac{(1-x)^2}{4},&1.4lt y\\
end{cases}$$
Now if you add all this up, you get a value which is (surprisingly) $in[0,1]$. You might be tempted to say that this is the probability that there is a single true value but you would be wrong!
And this is why - consider $y$ and $y+Delta$. For $Delta$ sufficiently large so that you are happy to say that they are distinct, the probabilities that they each fall into their respective intervals would be exactly the same - this is the definition of a confidence interval! So the probability that they are identical is equal to the probability that they are not identical for $Delta$, and $-Delta$ and $2Delta$ and an infinite number of other $Delta$ variants. Add up an infinite number of numbers $gt0$ and you will soon have a number $gt 1$ so it cannot represent a probability.
Without more information on the methodology, you can only state that the probability of a single true value is $in[0,1]$, but then, what isn't?
answered Dec 8 '14 at 2:34
Dale MDale M
2,5621822
$endgroup$
$begingroup$
Thanks Dale. Let me ask What those probability functions are coming from?
$endgroup$
– Ehsan M. Kermani
Dec 8 '14 at 18:34
$begingroup$
If you construct a confidence interval for a mean, say 95%, then there is a 95% chance that the value lies within it. Therefore there is a 2.5% it lies above and a 2.5% it lies below. Since the experiments (and therefore the confidence intervals) are independent, the probabilities follow. To be totally accurate, you should partition the low and high probabilities but the maths has no real meaning and was provided to illustrate the point.
$endgroup$
– Dale M
Dec 8 '14 at 22:33
add a comment |
$begingroup$
If we assume that the $pm$ denotes a $x$ confidence interval and the confidence intervals are symmetrical and that there is a single true value $y$ then you can say:
$$p=begin{cases}
frac{(1-x)^2}{4},&ylt0.7\
frac{x(1-x)}{2},&0.7le yle0.9\\
frac{(1-x)^2}{4},&0.9lt ylt1.0\\
frac{x(1-x)}{2},&1.0le yle1.4\\
frac{(1-x)^2}{4},&1.4lt y\\
end{cases}$$
Now if you add all this up, you get a value which is (surprisingly) $in[0,1]$. You might be tempted to say that this is the probability that there is a single true value but you would be wrong!
And this is why - consider $y$ and $y+Delta$. For $Delta$ sufficiently large so that you are happy to say that they are distinct, the probabilities that they each fall into their respective intervals would be exactly the same - this is the definition of a confidence interval! So the probability that they are identical is equal to the probability that they are not identical for $Delta$, and $-Delta$ and $2Delta$ and an infinite number of other $Delta$ variants. Add up an infinite number of numbers $gt0$ and you will soon have a number $gt 1$ so it cannot represent a probability.
Without more information on the methodology, you can only state that the probability of a single true value is $in[0,1]$, but then, what isn't?
answered Dec 8 '14 at 2:34
Dale MDale M
2,5621822
$endgroup$
If we assume that the $pm$ denotes a $x$ confidence interval and the confidence intervals are symmetrical and that there is a single true value $y$ then you can say:
$$p=begin{cases}
frac{(1-x)^2}{4},&ylt0.7\
frac{x(1-x)}{2},&0.7le yle0.9\\
frac{(1-x)^2}{4},&0.9lt ylt1.0\\
frac{x(1-x)}{2},&1.0le yle1.4\\
frac{(1-x)^2}{4},&1.4lt y\\
end{cases}$$
Now if you add all this up, you get a value which is (surprisingly) $in[0,1]$. You might be tempted to say that this is the probability that there is a single true value but you would be wrong!
And this is why - consider $y$ and $y+Delta$. For $Delta$ sufficiently large so that you are happy to say that they are distinct, the probabilities that they each fall into their respective intervals would be exactly the same - this is the definition of a confidence interval! So the probability that they are identical is equal to the probability that they are not identical for $Delta$, and $-Delta$ and $2Delta$ and an infinite number of other $Delta$ variants. Add up an infinite number of numbers $gt0$ and you will soon have a number $gt 1$ so it cannot represent a probability.
Without more information on the methodology, you can only state that the probability of a single true value is $in[0,1]$, but then, what isn't?
answered Dec 8 '14 at 2:34
Dale MDale M
2,5621822
answered Dec 8 '14 at 2:34
Dale MDale M
2,5621822
answered Dec 8 '14 at 2:34
Dale MDale M
2,5621822
answered Dec 8 '14 at 2:34
Dale MDale M
2,5621822
2,5621822
$begingroup$
Thanks Dale. Let me ask What those probability functions are coming from?
$endgroup$
– Ehsan M. Kermani
Dec 8 '14 at 18:34
$begingroup$
If you construct a confidence interval for a mean, say 95%, then there is a 95% chance that the value lies within it. Therefore there is a 2.5% it lies above and a 2.5% it lies below. Since the experiments (and therefore the confidence intervals) are independent, the probabilities follow. To be totally accurate, you should partition the low and high probabilities but the maths has no real meaning and was provided to illustrate the point.
$endgroup$
– Dale M
Dec 8 '14 at 22:33
add a comment |
$begingroup$
Thanks Dale. Let me ask What those probability functions are coming from?
$endgroup$
– Ehsan M. Kermani
Dec 8 '14 at 18:34
$begingroup$
If you construct a confidence interval for a mean, say 95%, then there is a 95% chance that the value lies within it. Therefore there is a 2.5% it lies above and a 2.5% it lies below. Since the experiments (and therefore the confidence intervals) are independent, the probabilities follow. To be totally accurate, you should partition the low and high probabilities but the maths has no real meaning and was provided to illustrate the point.
$endgroup$
– Dale M
Dec 8 '14 at 22:33
$begingroup$
Thanks Dale. Let me ask What those probability functions are coming from?
$endgroup$
– Ehsan M. Kermani
Dec 8 '14 at 18:34
$begingroup$
Thanks Dale. Let me ask What those probability functions are coming from?
$endgroup$
– Ehsan M. Kermani
Dec 8 '14 at 18:34
$begingroup$
If you construct a confidence interval for a mean, say 95%, then there is a 95% chance that the value lies within it. Therefore there is a 2.5% it lies above and a 2.5% it lies below. Since the experiments (and therefore the confidence intervals) are independent, the probabilities follow. To be totally accurate, you should partition the low and high probabilities but the maths has no real meaning and was provided to illustrate the point.
$endgroup$
– Dale M
Dec 8 '14 at 22:33
$begingroup$
If you construct a confidence interval for a mean, say 95%, then there is a 95% chance that the value lies within it. Therefore there is a 2.5% it lies above and a 2.5% it lies below. Since the experiments (and therefore the confidence intervals) are independent, the probabilities follow. To be totally accurate, you should partition the low and high probabilities but the maths has no real meaning and was provided to illustrate the point.
$endgroup$
– Dale M
Dec 8 '14 at 22:33
add a comment |
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$begingroup$
We don't know what $pm 0.1$ means. It seem likely to be a $95$% confidence region under an assumption of normal distribution, but it could be a $90$% confidence region; and if these measurements are from a "six-sigma" method then the confidence is much higher than $95$%. Are there really no other clues as to what was meant? (And, of course, what are your thoughts so far on how to do this?)
$endgroup$
– David K
Dec 7 '14 at 21:15
$begingroup$
Dear @DavidK, as far as I know, $pm 0.1$ is for one-sigma i.e. $68%$ confidence region. There are actually no other clues. I was thinking about t-test and F-test (F distribution) given one degree of freedom, but I'm not really sure!
$endgroup$
– Ehsan M. Kermani
Dec 8 '14 at 2:26
$begingroup$
Good, it sounds like you have sample standard deviations. If you don't know sample sizes, that's a bit of a hindrance.
$endgroup$
– David K
Dec 8 '14 at 5:24