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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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    $\begingroup$ 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 right interpretation should contain the following information:

  1. There are two approaches to interpretating of p-values:

    • The Frequentist interpretation, which your answer correctly used: The p-value is the probability of observing a value (in your case, the association between y-intercept and response) as extreme or more ('extreme' implies a two-tailed test), if the null hypothesis is true (in your case that is, the association between y-intercept and response is truly absent in the population, i.e. y-intercept = 0. In some tests it can mean the difference is 0);

    • or, 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, you do not need to assume the other coefficients are correct when interpreting p-values in a regression model... just that the same approach was used. However, it does assume that those parameters are estimated. So, your definition lacks in saying that 'first coefficient is 0.514'. All you need to assert was that the first coefficient is being estimated, i.e. '...the true value of the y-intercept is zero, in the presence of x.'. The values of other coefficients are immaterial to the definition of the p-value of any coefficients.
  3. The y-intecept is referring to the value of y when all xs are zero. You correctly implied this point.

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

With 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 used 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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