Using Decibels in Statistics

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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?







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  • 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.
    – 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.
    – 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.
    – Richard Hardy
    Aug 8 at 17:10










  • Perhaps I should have said "a useless tag for this kind of question".
    – Nick Cox
    Aug 8 at 17:12
















up vote
11
down vote

favorite
3












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?







share|cite|improve this question

















  • 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.
    – 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.
    – 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.
    – Richard Hardy
    Aug 8 at 17:10










  • Perhaps I should have said "a useless tag for this kind of question".
    – Nick Cox
    Aug 8 at 17:12












up vote
11
down vote

favorite
3









up vote
11
down vote

favorite
3






3





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?







share|cite|improve this question













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?









share|cite|improve this question












share|cite|improve this question




share|cite|improve this question








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.
    – 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.
    – 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.
    – Richard Hardy
    Aug 8 at 17:10










  • Perhaps I should have said "a useless tag for this kind of question".
    – Nick Cox
    Aug 8 at 17:12












  • 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.
    – 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.
    – 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.
    – Richard Hardy
    Aug 8 at 17:10










  • Perhaps I should have said "a useless tag for this kind of question".
    – 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.
– 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.
– 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.
– 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.
– 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.
– 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.
– Richard Hardy
Aug 8 at 17:10












Perhaps I should have said "a useless tag for this kind of question".
– Nick Cox
Aug 8 at 17:12




Perhaps I should have said "a useless tag for this kind of question".
– Nick Cox
Aug 8 at 17:12










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-transforming—$f^-1$—the 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 $y$:



  1. $x$ differs significantly across $y$, but $f(x)$ does not differ
    significantly across $y$.


  2. $x$ differs significantly across $y$, and $f(x)$ differs significantly
    across $y$.


  3. $x$ does not differ significantly across $y$, and $f(x)$ does not
    differ significantly across $y$.


  4. $x$ 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.






share|cite|improve this answer






























    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.






    share|cite|improve this answer

















    • 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.
      – 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!
      – Nick Cox
      Aug 8 at 0:40






    • 3




      I know, right? I really wish down-voters would leave constructive feedback.
      – Alexis
      Aug 8 at 1:17

















    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.






    share|cite|improve this answer




























      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.



      $$y_i = e^x_i+epsilon_i$$






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        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-transforming—$f^-1$—the 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 $y$:



        1. $x$ differs significantly across $y$, but $f(x)$ does not differ
          significantly across $y$.


        2. $x$ differs significantly across $y$, and $f(x)$ differs significantly
          across $y$.


        3. $x$ does not differ significantly across $y$, and $f(x)$ does not
          differ significantly across $y$.


        4. $x$ 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.






        share|cite|improve this answer



























          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-transforming—$f^-1$—the 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 $y$:



          1. $x$ differs significantly across $y$, but $f(x)$ does not differ
            significantly across $y$.


          2. $x$ differs significantly across $y$, and $f(x)$ differs significantly
            across $y$.


          3. $x$ does not differ significantly across $y$, and $f(x)$ does not
            differ significantly across $y$.


          4. $x$ 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.






          share|cite|improve this answer

























            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-transforming—$f^-1$—the 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 $y$:



            1. $x$ differs significantly across $y$, but $f(x)$ does not differ
              significantly across $y$.


            2. $x$ differs significantly across $y$, and $f(x)$ differs significantly
              across $y$.


            3. $x$ does not differ significantly across $y$, and $f(x)$ does not
              differ significantly across $y$.


            4. $x$ 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.






            share|cite|improve this answer















            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-transforming—$f^-1$—the 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 $y$:



            1. $x$ differs significantly across $y$, but $f(x)$ does not differ
              significantly across $y$.


            2. $x$ differs significantly across $y$, and $f(x)$ differs significantly
              across $y$.


            3. $x$ does not differ significantly across $y$, and $f(x)$ does not
              differ significantly across $y$.


            4. $x$ 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.







            share|cite|improve this answer















            share|cite|improve this answer



            share|cite|improve this answer








            edited Aug 8 at 19:44


























            answered Aug 7 at 16:29









            Alexis

            14.6k34389




            14.6k34389






















                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.






                share|cite|improve this answer

















                • 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.
                  – 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!
                  – Nick Cox
                  Aug 8 at 0:40






                • 3




                  I know, right? I really wish down-voters would leave constructive feedback.
                  – Alexis
                  Aug 8 at 1:17














                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.






                share|cite|improve this answer

















                • 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.
                  – 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!
                  – Nick Cox
                  Aug 8 at 0:40






                • 3




                  I know, right? I really wish down-voters would leave constructive feedback.
                  – Alexis
                  Aug 8 at 1:17












                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.






                share|cite|improve this answer













                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.







                share|cite|improve this answer













                share|cite|improve this answer



                share|cite|improve this answer











                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.
                  – 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!
                  – Nick Cox
                  Aug 8 at 0:40






                • 3




                  I know, right? I really wish down-voters would leave constructive feedback.
                  – Alexis
                  Aug 8 at 1:17












                • 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.
                  – 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!
                  – Nick Cox
                  Aug 8 at 0:40






                • 3




                  I know, right? I really wish down-voters would leave constructive feedback.
                  – 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.
                – 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.
                – 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!
                – 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!
                – Nick Cox
                Aug 8 at 0:40




                3




                3




                I know, right? I really wish down-voters would leave constructive feedback.
                – Alexis
                Aug 8 at 1:17




                I know, right? I really wish down-voters would leave constructive feedback.
                – Alexis
                Aug 8 at 1:17










                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.






                share|cite|improve this answer

























                  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.






                  share|cite|improve this answer























                    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.






                    share|cite|improve this answer













                    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.







                    share|cite|improve this answer













                    share|cite|improve this answer



                    share|cite|improve this answer











                    answered Aug 8 at 0:20









                    Robert Dodier

                    47229




                    47229




















                        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.



                        $$y_i = e^x_i+epsilon_i$$






                        share|cite|improve this answer



























                          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.



                          $$y_i = e^x_i+epsilon_i$$






                          share|cite|improve this answer

























                            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.



                            $$y_i = e^x_i+epsilon_i$$






                            share|cite|improve this answer















                            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.



                            $$y_i = e^x_i+epsilon_i$$







                            share|cite|improve this answer















                            share|cite|improve this answer



                            share|cite|improve this answer








                            edited Aug 8 at 11:25


























                            answered Aug 8 at 11:08









                            Martijn Weterings

                            8,1561047




                            8,1561047






















                                 

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