1
$\begingroup$

I would like to model the functional relationship between two acoustic measurements of traffic noise exposure, e.g. noise during day (x) and noise during night (y), expressed in Decibel values, using some kind of linear regression. The purpose of the exercise is to predict the noise at night from the noise during day and vice versa in a sample of receiver points. Important: there is no variable assuming the role of "dependent" or "independent" variable and the measurement error is the same in both x and y variables. I am therefore pretty sure that the standard OLS regression is the wrong approach. Normally, the two noise measures are shifted up to 10 decibels and are highly correlated with a slope close to 1. Therefore the intercept of the regression line is basically the only parameter of interest when it comes to predicting one measure from the other (and vice versa). However, there are plenty of "different" intercepts possible, depending on a) whether y is regressed on x or x is regressed on y, and 2) which estimation method in the linear model is used. I used lmodel2 in R to carry out a orthogonal major axis regression (MA) and the result is close to what I want. However, if switching x with y I get non-symmetric (or non-mirrored) estimates for the intercept, but this does not make sense in the present case as I can not say e.g.: The noise at location i is estimated to be 5 decibels lower during night than during day but the noise at the same location during day is also estimated to be 7 decibels louder than during night. I'm looking for the best possible estimate for the intercept (let's call it Delta) that works in "both ways" (i.e. x=y+Delta; y=x-Delta).

My questions are:

  1. Do I need to force the slope to equal 1 to solve this?
  2. Is there a way to force the slope to the value 1 in a method that does orthogonal regression (TLS, or similar) instead of ordinary least squares regression (OLS)?
  3. Shall I just calculate the mean difference between y and x to estimate the intercept (Delta)? (seems to me the most straightforward option)

Thanks!

Improve this question
$\endgroup$
3
  • 1
    $\begingroup$ If you know the slope is 1, then isn't it good enough to just calculate the centroid of the data and draw a line of slope 1 through it? Alternatively, isn't this an ideal candidate for a matched-pairs test, where you just want the average difference between daytime and nighttime measurements from a single receiver? $\endgroup$
    – Jacob Socolar
    Commented Dec 9, 2016 at 19:09
  • $\begingroup$ In OLS you can fix the slope to 1 using what is called an offset. However your use case reminds me of the medical literature comparing two sorts of measurement.stats.stackexchange.com/questions/99835/… has some details. I am not sure if it fits your problem exactly but worth a look. $\endgroup$
    – mdewey
    Commented Dec 9, 2016 at 21:52
  • $\begingroup$ Actually, there are very clear roles for the variables: when you are predicting the night values, the day values are the regressors and the night values are the response. When you are predicting the day values, the night values are the regressors and the day values are the response. Just fit two regression models. $\endgroup$
    – whuber ♦
    Commented Dec 9, 2016 at 23:22

1 Answer 1

Reset to default
3
$\begingroup$

If slope is known, the regression line is straightforward: It's just the line with that slope going through the point $(\bar x, \bar y)$ - and that is true for orthogonal regression as well as OLS regression. Therefore, with slope=1, your regression line is just:

$$y-\bar y=x-\bar x$$

or

$$y=x+(\bar y-\bar x)=x+\delta$$

Where $\delta=\bar y-\bar x$.

Said that - that is strictly the answer to your question -, I agree with user43849's comment that you don't need regression tools to deal with such a simple relation and that a paired t-test would be more useful to estimate $\delta$.

Improve this answer
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.