Using Decibels in Statistics

Clash Royale CLAN TAG#URR8PPP .everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty margin-bottom:0;
up vote
11
down vote
favorite
I'm working on a project that involves reading RFID Tags and comparing the signal strength the reader sees when you change the antenna configuration (number of antenna, position, etc...). As part of the project, I need to compare the setups to see which are most effective.
Ideally, I would be able to perform either an Unpaired t-Test or an ANOVA between two antenna positions (or MANOVA between multiple). However, as the response is in decibels which are logarithmic, I'm wondering what the best way to proceed with that is?
Would it be best to convert the results into a linear scale and then compare using one of the methods I mentioned, or should I use decibels as they are with a different statistical test to compare them?
data-transformation linear-model descriptive-statistics
add a comment |Â
up vote
11
down vote
favorite
I'm working on a project that involves reading RFID Tags and comparing the signal strength the reader sees when you change the antenna configuration (number of antenna, position, etc...). As part of the project, I need to compare the setups to see which are most effective.
Ideally, I would be able to perform either an Unpaired t-Test or an ANOVA between two antenna positions (or MANOVA between multiple). However, as the response is in decibels which are logarithmic, I'm wondering what the best way to proceed with that is?
Would it be best to convert the results into a linear scale and then compare using one of the methods I mentioned, or should I use decibels as they are with a different statistical test to compare them?
data-transformation linear-model descriptive-statistics
2
Taken the liberty of editing tags. Mathematical statistics is in practice a useless tag. Logarithmic series refers to something quite different with a discrete response.
â Nick Cox
Aug 7 at 16:27
1
As you are using tests assuming a Gaussian distribution, if the distribution of responses is "more Gaussian" in dB than in the linear scale (i.e. the original data is approximately log normal), it makes sense to stay in the logarithmic scale.
â Luca Citi
Aug 7 at 19:28
@NickCox, I thinkmathematical-statisticsworks just fine when requesting a proof, the corresponding tag being a synonym of the former tag.
â Richard Hardy
Aug 8 at 17:10
Perhaps I should have said "a useless tag for this kind of question".
â Nick Cox
Aug 8 at 17:12
add a comment |Â
up vote
11
down vote
favorite
up vote
11
down vote
favorite
I'm working on a project that involves reading RFID Tags and comparing the signal strength the reader sees when you change the antenna configuration (number of antenna, position, etc...). As part of the project, I need to compare the setups to see which are most effective.
Ideally, I would be able to perform either an Unpaired t-Test or an ANOVA between two antenna positions (or MANOVA between multiple). However, as the response is in decibels which are logarithmic, I'm wondering what the best way to proceed with that is?
Would it be best to convert the results into a linear scale and then compare using one of the methods I mentioned, or should I use decibels as they are with a different statistical test to compare them?
data-transformation linear-model descriptive-statistics
I'm working on a project that involves reading RFID Tags and comparing the signal strength the reader sees when you change the antenna configuration (number of antenna, position, etc...). As part of the project, I need to compare the setups to see which are most effective.
Ideally, I would be able to perform either an Unpaired t-Test or an ANOVA between two antenna positions (or MANOVA between multiple). However, as the response is in decibels which are logarithmic, I'm wondering what the best way to proceed with that is?
Would it be best to convert the results into a linear scale and then compare using one of the methods I mentioned, or should I use decibels as they are with a different statistical test to compare them?
data-transformation linear-model descriptive-statistics
edited Aug 7 at 16:25
Nick Cox
36.9k476123
36.9k476123
asked Aug 7 at 16:07
Brian Truman
585
585
2
Taken the liberty of editing tags. Mathematical statistics is in practice a useless tag. Logarithmic series refers to something quite different with a discrete response.
â Nick Cox
Aug 7 at 16:27
1
As you are using tests assuming a Gaussian distribution, if the distribution of responses is "more Gaussian" in dB than in the linear scale (i.e. the original data is approximately log normal), it makes sense to stay in the logarithmic scale.
â Luca Citi
Aug 7 at 19:28
@NickCox, I thinkmathematical-statisticsworks just fine when requesting a proof, the corresponding tag being a synonym of the former tag.
â Richard Hardy
Aug 8 at 17:10
Perhaps I should have said "a useless tag for this kind of question".
â Nick Cox
Aug 8 at 17:12
add a comment |Â
2
Taken the liberty of editing tags. Mathematical statistics is in practice a useless tag. Logarithmic series refers to something quite different with a discrete response.
â Nick Cox
Aug 7 at 16:27
1
As you are using tests assuming a Gaussian distribution, if the distribution of responses is "more Gaussian" in dB than in the linear scale (i.e. the original data is approximately log normal), it makes sense to stay in the logarithmic scale.
â Luca Citi
Aug 7 at 19:28
@NickCox, I thinkmathematical-statisticsworks just fine when requesting a proof, the corresponding tag being a synonym of the former tag.
â Richard Hardy
Aug 8 at 17:10
Perhaps I should have said "a useless tag for this kind of question".
â Nick Cox
Aug 8 at 17:12
2
2
Taken the liberty of editing tags. Mathematical statistics is in practice a useless tag. Logarithmic series refers to something quite different with a discrete response.
â Nick Cox
Aug 7 at 16:27
Taken the liberty of editing tags. Mathematical statistics is in practice a useless tag. Logarithmic series refers to something quite different with a discrete response.
â Nick Cox
Aug 7 at 16:27
1
1
As you are using tests assuming a Gaussian distribution, if the distribution of responses is "more Gaussian" in dB than in the linear scale (i.e. the original data is approximately log normal), it makes sense to stay in the logarithmic scale.
â Luca Citi
Aug 7 at 19:28
As you are using tests assuming a Gaussian distribution, if the distribution of responses is "more Gaussian" in dB than in the linear scale (i.e. the original data is approximately log normal), it makes sense to stay in the logarithmic scale.
â Luca Citi
Aug 7 at 19:28
@NickCox, I think
mathematical-statistics works just fine when requesting a proof, the corresponding tag being a synonym of the former tag.â Richard Hardy
Aug 8 at 17:10
@NickCox, I think
mathematical-statistics works just fine when requesting a proof, the corresponding tag being a synonym of the former tag.â Richard Hardy
Aug 8 at 17:10
Perhaps I should have said "a useless tag for this kind of question".
â Nick Cox
Aug 8 at 17:12
Perhaps I should have said "a useless tag for this kind of question".
â Nick Cox
Aug 8 at 17:12
add a comment |Â
4 Answers
4
active
oldest
votes
up vote
5
down vote
accepted
Whether to transform depends on what scale you want your inference at.
Generally, the variance of a function of $x$ does not equal the function of the variance of $x$. Because $sigma^2_f(x) ne f(sigma^2_x)$ transforming $x$ with $f$, then performing statistical inference (hypothesis tests or confidence intervals) on $f(x)$, then back-transformingâÂÂ$f^-1$âÂÂthe results of that inference to apply to $x$ is invalid (since both test statistics and CIs require an estimate of the variance).
Basing CIs on transformed variables + back-transformation produces intervals without the nominal coverage probabilities, so back-transformed confidence about an estimate based on $f(x)$ is not confidence on an estimate based on $x$.
Likewise, inferences about untransformed variables based on hypothesis tests
on transformed variables means that any of the following can be true, for
example, when making inferences about $x$ based on some grouping variable $y$:
$x$ differs significantly across $y$, but $f(x)$ does not differ
significantly across $y$.$x$ differs significantly across $y$, and $f(x)$ differs significantly
across $y$.$x$ does not differ significantly across $y$, and $f(x)$ does not
differ significantly across $y$.$x$ does not differ significantly across $y$, but $f(x)$ differs
significantly across $y$.
In short, knowing whether $f(x)$ differs significantly across groups of $y$ does not tell you
whether $x$ differs across $y$.
So the question of whether to transform those dBs is answered by whether you care about dB or exponentiated dB.
add a comment |Â
up vote
14
down vote
Strictly we need to see your data to have any chance of giving definitive advice, but it is possible to guess.
As you say, decibels are already on a logarithmic scale. That is likely to mean, for a variety of physical and statistical reasons, that they are quite likely to behave well by being approximately additive, homoscedastic and symmetrically distributed conditional on predictors. But you may be able to give a physical or engineering argument of how the response should vary as you change your design variables.
I know no possible principle or theory that means that you are obliged to exponentiate them before applying a $t$ test or ANOVA. I would expect that to make statistical behaviour worse, not better.
The same kind of reasoning usually applies to other "pre-transformed" logarithmic scales such as pH or the Richter scale.
PS: No idea what RFID tags are.
4
RFID's tags are Radio Frequency ID tags... those things in your passport, library materials, chipped credit card, etc. that make token-based ID possible wirelessly.
â Alexis
Aug 7 at 18:02
2
Random-seeming downvote in there. I don't have much cause for complaint as I have several votes for little work and it's not a great answer. (I could have written a better one given sight of some data.) But the downvote is futile: without a reason given there is no scope for changing anyone's mind!
â Nick Cox
Aug 8 at 0:40
3
I know, right? I really wish down-voters would leave constructive feedback.
â Alexis
Aug 8 at 1:17
add a comment |Â
up vote
3
down vote
Well, the only way to definitively answer this question is to look at some decibel data -- is there a simple distribution (e.g. Gaussian distribution) which is a good model for that? Or is the exponential of the data a better candidate? My guess is that the non-exponentialized data is more nearly Gaussian and therefore to make any ensuing analysis more straightforward, you should use that, but I'll let you be the judge of it.
I take issue with your proposed analysis, which is to apply a significance test to observed data from different experiments (namely different antenna positions). From considering the physics of this, there must be some difference, perhaps minuscule, perhaps substantial. But a priori there is some difference, therefore with a large enough data set, you must reject the null hypothesis of no difference. Thus the effect of a significance test is only to conclude "you have / you don't have a large data set". That doesn't seem very useful.
More useful would be to quantify the difference between different antenna positions, and perhaps also take into account costs and benefits to decide which position is to be selected. Quantified differences are sometimes called "effect size analysis"; a web search for that should turn up some resources. Costs and benefits come under the heading of utility theory and decision theory; again a search will find some resources.
add a comment |Â
up vote
2
down vote
The (logarithmic) Decibel scale is useful because the power of a signal can often be described by a (variable) series (or fluid range) of multiplications.
- E.g. if a 1cm thick wall reduces the signal to $frac110$ power (10 Decibel reduction).
- then a 2 cm thick wall reduces the signal to $frac1100$ power (20 Decibel redution).
- and a 3cm thick wall reduces the signal to $frac11000$ power (30 Decibel reduction)
- etc.
More generally, if you make the wall thickness non-discrete then
the signal (if you express it in untransformed units) could be expressed by an exponential function
$$P[mW] = P_0 left( frac110 right)^L[cm]$$
This is more simple, if you express the logarithm of the signal power, as a linear function (which, if you wish, requires some definition about the absolute scale, in this case 0dB relates to 1 mW)
$$P[dB] = 10 left(log(P_0[mW])-L[cm]right)$$
Whenever you have a process that is multiplicative like:
$$X propto e^Y $$
with the parameter $Y$ normal distributed: $$Y sim N(mu,sigma^2)$$
then $X$ has a log-normal distribution and $log(X)$ (or X expressed in any other logarithmic scale, such as the dB scale) has a normal distribution.
I expect that your error term will be multiplicative like that. That is: the signal strength will be a sum of many normal distributed error terms (e.g. amplifier temperature fluctuations, atmospheric conditions, etc.) that occur in the exponent of the expression for the signal strength.
$$y_i = e^x_i+epsilon_i$$
add a comment |Â
4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
5
down vote
accepted
Whether to transform depends on what scale you want your inference at.
Generally, the variance of a function of $x$ does not equal the function of the variance of $x$. Because $sigma^2_f(x) ne f(sigma^2_x)$ transforming $x$ with $f$, then performing statistical inference (hypothesis tests or confidence intervals) on $f(x)$, then back-transformingâÂÂ$f^-1$âÂÂthe results of that inference to apply to $x$ is invalid (since both test statistics and CIs require an estimate of the variance).
Basing CIs on transformed variables + back-transformation produces intervals without the nominal coverage probabilities, so back-transformed confidence about an estimate based on $f(x)$ is not confidence on an estimate based on $x$.
Likewise, inferences about untransformed variables based on hypothesis tests
on transformed variables means that any of the following can be true, for
example, when making inferences about $x$ based on some grouping variable $y$:
$x$ differs significantly across $y$, but $f(x)$ does not differ
significantly across $y$.$x$ differs significantly across $y$, and $f(x)$ differs significantly
across $y$.$x$ does not differ significantly across $y$, and $f(x)$ does not
differ significantly across $y$.$x$ does not differ significantly across $y$, but $f(x)$ differs
significantly across $y$.
In short, knowing whether $f(x)$ differs significantly across groups of $y$ does not tell you
whether $x$ differs across $y$.
So the question of whether to transform those dBs is answered by whether you care about dB or exponentiated dB.
add a comment |Â
up vote
5
down vote
accepted
Whether to transform depends on what scale you want your inference at.
Generally, the variance of a function of $x$ does not equal the function of the variance of $x$. Because $sigma^2_f(x) ne f(sigma^2_x)$ transforming $x$ with $f$, then performing statistical inference (hypothesis tests or confidence intervals) on $f(x)$, then back-transformingâÂÂ$f^-1$âÂÂthe results of that inference to apply to $x$ is invalid (since both test statistics and CIs require an estimate of the variance).
Basing CIs on transformed variables + back-transformation produces intervals without the nominal coverage probabilities, so back-transformed confidence about an estimate based on $f(x)$ is not confidence on an estimate based on $x$.
Likewise, inferences about untransformed variables based on hypothesis tests
on transformed variables means that any of the following can be true, for
example, when making inferences about $x$ based on some grouping variable $y$:
$x$ differs significantly across $y$, but $f(x)$ does not differ
significantly across $y$.$x$ differs significantly across $y$, and $f(x)$ differs significantly
across $y$.$x$ does not differ significantly across $y$, and $f(x)$ does not
differ significantly across $y$.$x$ does not differ significantly across $y$, but $f(x)$ differs
significantly across $y$.
In short, knowing whether $f(x)$ differs significantly across groups of $y$ does not tell you
whether $x$ differs across $y$.
So the question of whether to transform those dBs is answered by whether you care about dB or exponentiated dB.
add a comment |Â
up vote
5
down vote
accepted
up vote
5
down vote
accepted
Whether to transform depends on what scale you want your inference at.
Generally, the variance of a function of $x$ does not equal the function of the variance of $x$. Because $sigma^2_f(x) ne f(sigma^2_x)$ transforming $x$ with $f$, then performing statistical inference (hypothesis tests or confidence intervals) on $f(x)$, then back-transformingâÂÂ$f^-1$âÂÂthe results of that inference to apply to $x$ is invalid (since both test statistics and CIs require an estimate of the variance).
Basing CIs on transformed variables + back-transformation produces intervals without the nominal coverage probabilities, so back-transformed confidence about an estimate based on $f(x)$ is not confidence on an estimate based on $x$.
Likewise, inferences about untransformed variables based on hypothesis tests
on transformed variables means that any of the following can be true, for
example, when making inferences about $x$ based on some grouping variable $y$:
$x$ differs significantly across $y$, but $f(x)$ does not differ
significantly across $y$.$x$ differs significantly across $y$, and $f(x)$ differs significantly
across $y$.$x$ does not differ significantly across $y$, and $f(x)$ does not
differ significantly across $y$.$x$ does not differ significantly across $y$, but $f(x)$ differs
significantly across $y$.
In short, knowing whether $f(x)$ differs significantly across groups of $y$ does not tell you
whether $x$ differs across $y$.
So the question of whether to transform those dBs is answered by whether you care about dB or exponentiated dB.
Whether to transform depends on what scale you want your inference at.
Generally, the variance of a function of $x$ does not equal the function of the variance of $x$. Because $sigma^2_f(x) ne f(sigma^2_x)$ transforming $x$ with $f$, then performing statistical inference (hypothesis tests or confidence intervals) on $f(x)$, then back-transformingâÂÂ$f^-1$âÂÂthe results of that inference to apply to $x$ is invalid (since both test statistics and CIs require an estimate of the variance).
Basing CIs on transformed variables + back-transformation produces intervals without the nominal coverage probabilities, so back-transformed confidence about an estimate based on $f(x)$ is not confidence on an estimate based on $x$.
Likewise, inferences about untransformed variables based on hypothesis tests
on transformed variables means that any of the following can be true, for
example, when making inferences about $x$ based on some grouping variable $y$:
$x$ differs significantly across $y$, but $f(x)$ does not differ
significantly across $y$.$x$ differs significantly across $y$, and $f(x)$ differs significantly
across $y$.$x$ does not differ significantly across $y$, and $f(x)$ does not
differ significantly across $y$.$x$ does not differ significantly across $y$, but $f(x)$ differs
significantly across $y$.
In short, knowing whether $f(x)$ differs significantly across groups of $y$ does not tell you
whether $x$ differs across $y$.
So the question of whether to transform those dBs is answered by whether you care about dB or exponentiated dB.
edited Aug 8 at 19:44
answered Aug 7 at 16:29
Alexis
14.6k34389
14.6k34389
add a comment |Â
add a comment |Â
up vote
14
down vote
Strictly we need to see your data to have any chance of giving definitive advice, but it is possible to guess.
As you say, decibels are already on a logarithmic scale. That is likely to mean, for a variety of physical and statistical reasons, that they are quite likely to behave well by being approximately additive, homoscedastic and symmetrically distributed conditional on predictors. But you may be able to give a physical or engineering argument of how the response should vary as you change your design variables.
I know no possible principle or theory that means that you are obliged to exponentiate them before applying a $t$ test or ANOVA. I would expect that to make statistical behaviour worse, not better.
The same kind of reasoning usually applies to other "pre-transformed" logarithmic scales such as pH or the Richter scale.
PS: No idea what RFID tags are.
4
RFID's tags are Radio Frequency ID tags... those things in your passport, library materials, chipped credit card, etc. that make token-based ID possible wirelessly.
â Alexis
Aug 7 at 18:02
2
Random-seeming downvote in there. I don't have much cause for complaint as I have several votes for little work and it's not a great answer. (I could have written a better one given sight of some data.) But the downvote is futile: without a reason given there is no scope for changing anyone's mind!
â Nick Cox
Aug 8 at 0:40
3
I know, right? I really wish down-voters would leave constructive feedback.
â Alexis
Aug 8 at 1:17
add a comment |Â
up vote
14
down vote
Strictly we need to see your data to have any chance of giving definitive advice, but it is possible to guess.
As you say, decibels are already on a logarithmic scale. That is likely to mean, for a variety of physical and statistical reasons, that they are quite likely to behave well by being approximately additive, homoscedastic and symmetrically distributed conditional on predictors. But you may be able to give a physical or engineering argument of how the response should vary as you change your design variables.
I know no possible principle or theory that means that you are obliged to exponentiate them before applying a $t$ test or ANOVA. I would expect that to make statistical behaviour worse, not better.
The same kind of reasoning usually applies to other "pre-transformed" logarithmic scales such as pH or the Richter scale.
PS: No idea what RFID tags are.
4
RFID's tags are Radio Frequency ID tags... those things in your passport, library materials, chipped credit card, etc. that make token-based ID possible wirelessly.
â Alexis
Aug 7 at 18:02
2
Random-seeming downvote in there. I don't have much cause for complaint as I have several votes for little work and it's not a great answer. (I could have written a better one given sight of some data.) But the downvote is futile: without a reason given there is no scope for changing anyone's mind!
â Nick Cox
Aug 8 at 0:40
3
I know, right? I really wish down-voters would leave constructive feedback.
â Alexis
Aug 8 at 1:17
add a comment |Â
up vote
14
down vote
up vote
14
down vote
Strictly we need to see your data to have any chance of giving definitive advice, but it is possible to guess.
As you say, decibels are already on a logarithmic scale. That is likely to mean, for a variety of physical and statistical reasons, that they are quite likely to behave well by being approximately additive, homoscedastic and symmetrically distributed conditional on predictors. But you may be able to give a physical or engineering argument of how the response should vary as you change your design variables.
I know no possible principle or theory that means that you are obliged to exponentiate them before applying a $t$ test or ANOVA. I would expect that to make statistical behaviour worse, not better.
The same kind of reasoning usually applies to other "pre-transformed" logarithmic scales such as pH or the Richter scale.
PS: No idea what RFID tags are.
Strictly we need to see your data to have any chance of giving definitive advice, but it is possible to guess.
As you say, decibels are already on a logarithmic scale. That is likely to mean, for a variety of physical and statistical reasons, that they are quite likely to behave well by being approximately additive, homoscedastic and symmetrically distributed conditional on predictors. But you may be able to give a physical or engineering argument of how the response should vary as you change your design variables.
I know no possible principle or theory that means that you are obliged to exponentiate them before applying a $t$ test or ANOVA. I would expect that to make statistical behaviour worse, not better.
The same kind of reasoning usually applies to other "pre-transformed" logarithmic scales such as pH or the Richter scale.
PS: No idea what RFID tags are.
answered Aug 7 at 16:22
Nick Cox
36.9k476123
36.9k476123
4
RFID's tags are Radio Frequency ID tags... those things in your passport, library materials, chipped credit card, etc. that make token-based ID possible wirelessly.
â Alexis
Aug 7 at 18:02
2
Random-seeming downvote in there. I don't have much cause for complaint as I have several votes for little work and it's not a great answer. (I could have written a better one given sight of some data.) But the downvote is futile: without a reason given there is no scope for changing anyone's mind!
â Nick Cox
Aug 8 at 0:40
3
I know, right? I really wish down-voters would leave constructive feedback.
â Alexis
Aug 8 at 1:17
add a comment |Â
4
RFID's tags are Radio Frequency ID tags... those things in your passport, library materials, chipped credit card, etc. that make token-based ID possible wirelessly.
â Alexis
Aug 7 at 18:02
2
Random-seeming downvote in there. I don't have much cause for complaint as I have several votes for little work and it's not a great answer. (I could have written a better one given sight of some data.) But the downvote is futile: without a reason given there is no scope for changing anyone's mind!
â Nick Cox
Aug 8 at 0:40
3
I know, right? I really wish down-voters would leave constructive feedback.
â Alexis
Aug 8 at 1:17
4
4
RFID's tags are Radio Frequency ID tags... those things in your passport, library materials, chipped credit card, etc. that make token-based ID possible wirelessly.
â Alexis
Aug 7 at 18:02
RFID's tags are Radio Frequency ID tags... those things in your passport, library materials, chipped credit card, etc. that make token-based ID possible wirelessly.
â Alexis
Aug 7 at 18:02
2
2
Random-seeming downvote in there. I don't have much cause for complaint as I have several votes for little work and it's not a great answer. (I could have written a better one given sight of some data.) But the downvote is futile: without a reason given there is no scope for changing anyone's mind!
â Nick Cox
Aug 8 at 0:40
Random-seeming downvote in there. I don't have much cause for complaint as I have several votes for little work and it's not a great answer. (I could have written a better one given sight of some data.) But the downvote is futile: without a reason given there is no scope for changing anyone's mind!
â Nick Cox
Aug 8 at 0:40
3
3
I know, right? I really wish down-voters would leave constructive feedback.
â Alexis
Aug 8 at 1:17
I know, right? I really wish down-voters would leave constructive feedback.
â Alexis
Aug 8 at 1:17
add a comment |Â
up vote
3
down vote
Well, the only way to definitively answer this question is to look at some decibel data -- is there a simple distribution (e.g. Gaussian distribution) which is a good model for that? Or is the exponential of the data a better candidate? My guess is that the non-exponentialized data is more nearly Gaussian and therefore to make any ensuing analysis more straightforward, you should use that, but I'll let you be the judge of it.
I take issue with your proposed analysis, which is to apply a significance test to observed data from different experiments (namely different antenna positions). From considering the physics of this, there must be some difference, perhaps minuscule, perhaps substantial. But a priori there is some difference, therefore with a large enough data set, you must reject the null hypothesis of no difference. Thus the effect of a significance test is only to conclude "you have / you don't have a large data set". That doesn't seem very useful.
More useful would be to quantify the difference between different antenna positions, and perhaps also take into account costs and benefits to decide which position is to be selected. Quantified differences are sometimes called "effect size analysis"; a web search for that should turn up some resources. Costs and benefits come under the heading of utility theory and decision theory; again a search will find some resources.
add a comment |Â
up vote
3
down vote
Well, the only way to definitively answer this question is to look at some decibel data -- is there a simple distribution (e.g. Gaussian distribution) which is a good model for that? Or is the exponential of the data a better candidate? My guess is that the non-exponentialized data is more nearly Gaussian and therefore to make any ensuing analysis more straightforward, you should use that, but I'll let you be the judge of it.
I take issue with your proposed analysis, which is to apply a significance test to observed data from different experiments (namely different antenna positions). From considering the physics of this, there must be some difference, perhaps minuscule, perhaps substantial. But a priori there is some difference, therefore with a large enough data set, you must reject the null hypothesis of no difference. Thus the effect of a significance test is only to conclude "you have / you don't have a large data set". That doesn't seem very useful.
More useful would be to quantify the difference between different antenna positions, and perhaps also take into account costs and benefits to decide which position is to be selected. Quantified differences are sometimes called "effect size analysis"; a web search for that should turn up some resources. Costs and benefits come under the heading of utility theory and decision theory; again a search will find some resources.
add a comment |Â
up vote
3
down vote
up vote
3
down vote
Well, the only way to definitively answer this question is to look at some decibel data -- is there a simple distribution (e.g. Gaussian distribution) which is a good model for that? Or is the exponential of the data a better candidate? My guess is that the non-exponentialized data is more nearly Gaussian and therefore to make any ensuing analysis more straightforward, you should use that, but I'll let you be the judge of it.
I take issue with your proposed analysis, which is to apply a significance test to observed data from different experiments (namely different antenna positions). From considering the physics of this, there must be some difference, perhaps minuscule, perhaps substantial. But a priori there is some difference, therefore with a large enough data set, you must reject the null hypothesis of no difference. Thus the effect of a significance test is only to conclude "you have / you don't have a large data set". That doesn't seem very useful.
More useful would be to quantify the difference between different antenna positions, and perhaps also take into account costs and benefits to decide which position is to be selected. Quantified differences are sometimes called "effect size analysis"; a web search for that should turn up some resources. Costs and benefits come under the heading of utility theory and decision theory; again a search will find some resources.
Well, the only way to definitively answer this question is to look at some decibel data -- is there a simple distribution (e.g. Gaussian distribution) which is a good model for that? Or is the exponential of the data a better candidate? My guess is that the non-exponentialized data is more nearly Gaussian and therefore to make any ensuing analysis more straightforward, you should use that, but I'll let you be the judge of it.
I take issue with your proposed analysis, which is to apply a significance test to observed data from different experiments (namely different antenna positions). From considering the physics of this, there must be some difference, perhaps minuscule, perhaps substantial. But a priori there is some difference, therefore with a large enough data set, you must reject the null hypothesis of no difference. Thus the effect of a significance test is only to conclude "you have / you don't have a large data set". That doesn't seem very useful.
More useful would be to quantify the difference between different antenna positions, and perhaps also take into account costs and benefits to decide which position is to be selected. Quantified differences are sometimes called "effect size analysis"; a web search for that should turn up some resources. Costs and benefits come under the heading of utility theory and decision theory; again a search will find some resources.
answered Aug 8 at 0:20
Robert Dodier
47229
47229
add a comment |Â
add a comment |Â
up vote
2
down vote
The (logarithmic) Decibel scale is useful because the power of a signal can often be described by a (variable) series (or fluid range) of multiplications.
- E.g. if a 1cm thick wall reduces the signal to $frac110$ power (10 Decibel reduction).
- then a 2 cm thick wall reduces the signal to $frac1100$ power (20 Decibel redution).
- and a 3cm thick wall reduces the signal to $frac11000$ power (30 Decibel reduction)
- etc.
More generally, if you make the wall thickness non-discrete then
the signal (if you express it in untransformed units) could be expressed by an exponential function
$$P[mW] = P_0 left( frac110 right)^L[cm]$$
This is more simple, if you express the logarithm of the signal power, as a linear function (which, if you wish, requires some definition about the absolute scale, in this case 0dB relates to 1 mW)
$$P[dB] = 10 left(log(P_0[mW])-L[cm]right)$$
Whenever you have a process that is multiplicative like:
$$X propto e^Y $$
with the parameter $Y$ normal distributed: $$Y sim N(mu,sigma^2)$$
then $X$ has a log-normal distribution and $log(X)$ (or X expressed in any other logarithmic scale, such as the dB scale) has a normal distribution.
I expect that your error term will be multiplicative like that. That is: the signal strength will be a sum of many normal distributed error terms (e.g. amplifier temperature fluctuations, atmospheric conditions, etc.) that occur in the exponent of the expression for the signal strength.
$$y_i = e^x_i+epsilon_i$$
add a comment |Â
up vote
2
down vote
The (logarithmic) Decibel scale is useful because the power of a signal can often be described by a (variable) series (or fluid range) of multiplications.
- E.g. if a 1cm thick wall reduces the signal to $frac110$ power (10 Decibel reduction).
- then a 2 cm thick wall reduces the signal to $frac1100$ power (20 Decibel redution).
- and a 3cm thick wall reduces the signal to $frac11000$ power (30 Decibel reduction)
- etc.
More generally, if you make the wall thickness non-discrete then
the signal (if you express it in untransformed units) could be expressed by an exponential function
$$P[mW] = P_0 left( frac110 right)^L[cm]$$
This is more simple, if you express the logarithm of the signal power, as a linear function (which, if you wish, requires some definition about the absolute scale, in this case 0dB relates to 1 mW)
$$P[dB] = 10 left(log(P_0[mW])-L[cm]right)$$
Whenever you have a process that is multiplicative like:
$$X propto e^Y $$
with the parameter $Y$ normal distributed: $$Y sim N(mu,sigma^2)$$
then $X$ has a log-normal distribution and $log(X)$ (or X expressed in any other logarithmic scale, such as the dB scale) has a normal distribution.
I expect that your error term will be multiplicative like that. That is: the signal strength will be a sum of many normal distributed error terms (e.g. amplifier temperature fluctuations, atmospheric conditions, etc.) that occur in the exponent of the expression for the signal strength.
$$y_i = e^x_i+epsilon_i$$
add a comment |Â
up vote
2
down vote
up vote
2
down vote
The (logarithmic) Decibel scale is useful because the power of a signal can often be described by a (variable) series (or fluid range) of multiplications.
- E.g. if a 1cm thick wall reduces the signal to $frac110$ power (10 Decibel reduction).
- then a 2 cm thick wall reduces the signal to $frac1100$ power (20 Decibel redution).
- and a 3cm thick wall reduces the signal to $frac11000$ power (30 Decibel reduction)
- etc.
More generally, if you make the wall thickness non-discrete then
the signal (if you express it in untransformed units) could be expressed by an exponential function
$$P[mW] = P_0 left( frac110 right)^L[cm]$$
This is more simple, if you express the logarithm of the signal power, as a linear function (which, if you wish, requires some definition about the absolute scale, in this case 0dB relates to 1 mW)
$$P[dB] = 10 left(log(P_0[mW])-L[cm]right)$$
Whenever you have a process that is multiplicative like:
$$X propto e^Y $$
with the parameter $Y$ normal distributed: $$Y sim N(mu,sigma^2)$$
then $X$ has a log-normal distribution and $log(X)$ (or X expressed in any other logarithmic scale, such as the dB scale) has a normal distribution.
I expect that your error term will be multiplicative like that. That is: the signal strength will be a sum of many normal distributed error terms (e.g. amplifier temperature fluctuations, atmospheric conditions, etc.) that occur in the exponent of the expression for the signal strength.
$$y_i = e^x_i+epsilon_i$$
The (logarithmic) Decibel scale is useful because the power of a signal can often be described by a (variable) series (or fluid range) of multiplications.
- E.g. if a 1cm thick wall reduces the signal to $frac110$ power (10 Decibel reduction).
- then a 2 cm thick wall reduces the signal to $frac1100$ power (20 Decibel redution).
- and a 3cm thick wall reduces the signal to $frac11000$ power (30 Decibel reduction)
- etc.
More generally, if you make the wall thickness non-discrete then
the signal (if you express it in untransformed units) could be expressed by an exponential function
$$P[mW] = P_0 left( frac110 right)^L[cm]$$
This is more simple, if you express the logarithm of the signal power, as a linear function (which, if you wish, requires some definition about the absolute scale, in this case 0dB relates to 1 mW)
$$P[dB] = 10 left(log(P_0[mW])-L[cm]right)$$
Whenever you have a process that is multiplicative like:
$$X propto e^Y $$
with the parameter $Y$ normal distributed: $$Y sim N(mu,sigma^2)$$
then $X$ has a log-normal distribution and $log(X)$ (or X expressed in any other logarithmic scale, such as the dB scale) has a normal distribution.
I expect that your error term will be multiplicative like that. That is: the signal strength will be a sum of many normal distributed error terms (e.g. amplifier temperature fluctuations, atmospheric conditions, etc.) that occur in the exponent of the expression for the signal strength.
$$y_i = e^x_i+epsilon_i$$
edited Aug 8 at 11:25
answered Aug 8 at 11:08
Martijn Weterings
8,1561047
8,1561047
add a comment |Â
add a comment |Â
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
var $window = $(window),
onScroll = function(e)
var $elem = $('.new-login-left'),
docViewTop = $window.scrollTop(),
docViewBottom = docViewTop + $window.height(),
elemTop = $elem.offset().top,
elemBottom = elemTop + $elem.height();
if ((docViewTop elemBottom))
StackExchange.using('gps', function() StackExchange.gps.track('embedded_signup_form.view', location: 'question_page' ); );
$window.unbind('scroll', onScroll);
;
$window.on('scroll', onScroll);
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fstats.stackexchange.com%2fquestions%2f361150%2fusing-decibels-in-statistics%23new-answer', 'question_page');
);
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
var $window = $(window),
onScroll = function(e)
var $elem = $('.new-login-left'),
docViewTop = $window.scrollTop(),
docViewBottom = docViewTop + $window.height(),
elemTop = $elem.offset().top,
elemBottom = elemTop + $elem.height();
if ((docViewTop elemBottom))
StackExchange.using('gps', function() StackExchange.gps.track('embedded_signup_form.view', location: 'question_page' ); );
$window.unbind('scroll', onScroll);
;
$window.on('scroll', onScroll);
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
var $window = $(window),
onScroll = function(e)
var $elem = $('.new-login-left'),
docViewTop = $window.scrollTop(),
docViewBottom = docViewTop + $window.height(),
elemTop = $elem.offset().top,
elemBottom = elemTop + $elem.height();
if ((docViewTop elemBottom))
StackExchange.using('gps', function() StackExchange.gps.track('embedded_signup_form.view', location: 'question_page' ); );
$window.unbind('scroll', onScroll);
;
$window.on('scroll', onScroll);
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
var $window = $(window),
onScroll = function(e)
var $elem = $('.new-login-left'),
docViewTop = $window.scrollTop(),
docViewBottom = docViewTop + $window.height(),
elemTop = $elem.offset().top,
elemBottom = elemTop + $elem.height();
if ((docViewTop elemBottom))
StackExchange.using('gps', function() StackExchange.gps.track('embedded_signup_form.view', location: 'question_page' ); );
$window.unbind('scroll', onScroll);
;
$window.on('scroll', onScroll);
);
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
2
Taken the liberty of editing tags. Mathematical statistics is in practice a useless tag. Logarithmic series refers to something quite different with a discrete response.
â Nick Cox
Aug 7 at 16:27
1
As you are using tests assuming a Gaussian distribution, if the distribution of responses is "more Gaussian" in dB than in the linear scale (i.e. the original data is approximately log normal), it makes sense to stay in the logarithmic scale.
â Luca Citi
Aug 7 at 19:28
@NickCox, I think
mathematical-statisticsworks just fine when requesting a proof, the corresponding tag being a synonym of the former tag.â Richard Hardy
Aug 8 at 17:10
Perhaps I should have said "a useless tag for this kind of question".
â Nick Cox
Aug 8 at 17:12