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This question already has an answer here:

I am trying to interpret one of the p-values in a one variable linear regression. Some of the answers I've seen for similar questions were not worded as thoroughly as I would have liked. My interpretation is deliberately verbose because it will aid my understanding if faults are found within it.

From Microsoft Excel the linear regression formula from 90 samples of (x,y) pairs is

y = 0.514x + 0.00087

and the p-value of the first coefficient is 4e-16 (scientific notation) and for the second it is 0.0027.

Would it be correct to say that the interpretation of the p-value of the 0.00087 term is:

Under the assumption that the true value of the y-intercept is zero and the first coefficient is 0.514, random sampling of the same number of (x,y) pairs, specifically 90, would result in a least squares best fit line with a y-intercept at least as extreme as 0.00087, with a probability of 0.0027.

If not, then what would be the correct interpretation?

Not so importantly, but just to be complete, I am also inquiring if it would be more accurate and complete to put the relevant phrase as

"at least as extreme as 0.00087 in the same direction, that is, positive".

Edit: The Excel funcion is Tools > Data Analysis > Regression in Office 2003 with service pack 2. Excel regression p-values on coefficients are 2 sided.

Edit: Regarding differentiation from this question here: The most up voted answer there discusses the p-value of a hypothesis, which seems ill defined or at least not specific. I am not interested in that. I am interested in the p-value of a coefficient that is not the coefficient of an independent variable. I am being very specific.

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marked as duplicate by whuber Oct 13 '13 at 16:31

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

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    $\begingroup$ Yes, you are being specific, but your specific question is covered by all the answers to the duplicate: you are testing the hypothesis that the intercept equals zero. That is well defined and specific. $\endgroup$ – whuber Oct 13 '13 at 17:10
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    $\begingroup$ I think there's an overlooked distinction between the most up voted answer in the purported duplicate, and the answer sought here. The "dup" provides an understanding of the p-value, whereas this question seeks an answer that is an example of precise wording of the interpretation of the p-value. A concise answer, such as that in the comments here might be valuable to some practitioners. $\endgroup$ – H2ONaCl Oct 17 '13 at 8:35
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Your interpretation is almost right.

A correct interpretation requires the following components:

  1. The interpretation of p-values is in reference to either a) [frequentist] the probability of obtaining a value if you ran the same experiment (and statistic) many times over that you would obtain values that were as extreme or more extreme (the 'extreme' language implies a two-tailed test) if the true population value were 0 (or, in some tests that the difference is 0), i.e. if the null hypothesis were true; or b) the probability of obtaining that estimate of the parameter (e.g. intercept; using this statistical approach), or a more extreme value, if the population value for that parameter is 0. Your definition correctly uses the frequentist form.
  2. As you can see from point 1 above, the interpretation of p-values in a regression equation are not dependent on assumptions that the rest of the model is correct... just that the same approach was used. However, it does assume that those parameters are estimated. So, your definition lacks in that you say 'first coefficient is 0.514'. All you need to assert is that the first parameter is estimated... and that overall the model used has the same parameters estimated. The values obtained for the estimate are immaterial to the definition of the p-value for any given parameter.
  3. That we are referring to the value of y when all xs are at the value of zero. You correctly imply this point.

You should also note that your example, in using the frequentist form, is not free from your wants and subjective beliefs. Specifically, the p-value is tied to the design of the experiment you ran. You acknowledge this when you mention using the same number of sampling pairs.

In regards to your second question, the typical p-value reported for a regression equation is implicitly two-tailed. So, it refers to the absolute value of the parameters obtained. You didn't provide the Excel function you uses to calaculate the p-value, but I'd check there to see if Excel is calculating one-tailed (in the same direction) or two-tailed (extreme or more extreme) p-values.

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  • $\begingroup$ Ok Thanks. I can simply remove the mention of the 0.514 but I will not be offended if anybody offers to do a, superior in several respects, rewrite. $\endgroup$ – H2ONaCl Oct 13 '13 at 14:17
  • $\begingroup$ You understand this answer and not the one I gave, Wow! Yep I think I got your question completely wrong then. My apologies. $\endgroup$ – htrahdis Oct 13 '13 at 17:55
  • $\begingroup$ My suggested interpretation, informed by the above, becomes: Under the assumption that the true value of the y-intercept is zero, random sampling of the same number of (x,y) pairs, specifically 90, produced by the same process, would result in a least squares best fit line with a y-intercept at least as extreme as +0.00087, with a probability of 0.0027, and equal to or greater than +0.00087, with a probability of 0.00135. $\endgroup$ – H2ONaCl Oct 14 '13 at 4:16
  • $\begingroup$ @broiyan: Sounds close. I think technically the precise value obtained, if re-obtained, would be included in the probability. Therefore, I would go to the expense of extra words to make that clear by saying "as extreme or more extreme". Also, because you are phrasing it in a frequentist framework, I wouldn't talk about the probability of that event, but the proportion of times you would expect it to occur if the experiment were repeated over and over. $\endgroup$ – russellpierce Oct 15 '13 at 15:08
  • $\begingroup$ Informed by the above, it becomes: Under the assumption that the true value of the y-intercept is zero, random sampling of the same number of (x,y) pairs, specifically 90, produced by the same process, would result in a least squares best fit line with a y-intercept as extreme as or more extreme than +0.00087, with an expectation of 27 times out of 10000 trials, and equal to or greater than +0.00087, with an expectation of 135 times out of 100000 trials. $\endgroup$ – H2ONaCl Oct 17 '13 at 8:25
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Assuming the coefficients to be normally distributed with the mean of 0 and an estimated standard error which you have not mentioned, the p value tells the quantile of how far the calculated value is from the mean. In the given case if you think that the value is significant at 99.73% level, even then the coefficient is different from 0. If the confidence level that you want is higher than this, then you fail to reject the hypothesis that the coefficient is different from 0.

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  • $\begingroup$ You wrote "if you think that" and "if the confidence level that you want". I believe the interpretation is easier to understand if it is stated objectively; independent of what I think. I believe the interpretation will be easier to understand if it is independent of my wants. Please see that my example is free of my wants and my subjective beliefs. $\endgroup$ – H2ONaCl Oct 13 '13 at 12:51
  • $\begingroup$ a 90% confidence level means that you are fine if the estimate is right 9 out of 10 times. 99.73% confidence level means that you are fine if the estimate is right 9973 out of 10000 times. Generally used values are 90%, 95% and 99%. But it is a subjective decision that has to be decided by the researcher. $\endgroup$ – htrahdis Oct 13 '13 at 13:08
  • $\begingroup$ That's sweet but you probably don't know me well enough to say "you are fine". Please rephrase your answer if you care to. $\endgroup$ – H2ONaCl Oct 13 '13 at 13:25

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