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My question grew out of a discussion with @whuber in the comments of a different question.

Specifically, @whuber 's comment was as follows:

One reason it might surprise you is that the assumptions underlying a correlation test and a regression slope test are different--so even when we understand that the correlation and slope are really measuring the same thing, why should their p-values be the same? That shows how these issues go deeper than simply whether $r$ and $\beta$ should be numerically equal.

This got my thinking about it and I came across a variety of interesting answers. For example, I found this question "Assumptions of correlation coefficient" but can't see how this would clarify the comment above.

I found more interesting answers about the relationship of Pearson's $r$ and the slope $\beta$ in a simple linear regression (see here and here for example) but none of them seem to answer what @whuber was referring to in his comment (at least not apparent to me).

Question 1: What are the assumptions underlying a correlation test and a regression slope test?

For my 2nd question consider the following outputs in R:

model <- lm(Employed ~ Population, data = longley)
summary(model)

Call:
lm(formula = Employed ~ Population, data = longley)

Residuals:
    Min      1Q  Median      3Q     Max 
-1.4362 -0.9740  0.2021  0.5531  1.9048 

Coefficients:
            Estimate Std. Error t value Pr(>|t|)    
(Intercept)   8.3807     4.4224   1.895   0.0789 .  
Population    0.4849     0.0376  12.896 3.69e-09 ***
---
Signif. codes:  0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1

Residual standard error: 1.013 on 14 degrees of freedom
Multiple R-squared:  0.9224,    Adjusted R-squared:  0.9168 
F-statistic: 166.3 on 1 and 14 DF,  p-value: 3.693e-09

And the output of the cor.test() function:

with(longley, cor.test(Population, Employed))

    Pearson's product-moment correlation

data:  Population and Employed
t = 12.8956, df = 14, p-value = 3.693e-09
alternative hypothesis: true correlation is not equal to 0
95 percent confidence interval:
 0.8869236 0.9864676
sample estimates:
      cor 
0.9603906 

As can be seen by the lm() and cov.test() output, the Pearson's correlation coefficient $r$ and the slope estimate ($\beta_1$) are largely different, 0.96 vs. 0.485, respectively, but the t-value and the p-values are the same.

Then I also tried to see if I am able to calculate the t-value for $r$ and $\beta_1$, which are the same despite $r$ and $\beta_1$ being different. And that's where I get stuck, at least for $r$:

Calculate the the slope ($\beta_1$) in a simple linear regression using the total sums of squares of $x$ and $y$:

x <- longley$Population; y <- longley$Employed
xbar <- mean(x); ybar <- mean(y)
ss.x <- sum((x-xbar)^2)
ss.y <- sum((y-ybar)^2)
ss.xy <- sum((x-xbar)*(y-ybar))

Calculate the least-squares estimate of the regression slope, $\beta_{1}$ (there is a proof of this in Crawley's R Book 1st edition, page 393):

b1 <- ss.xy/ss.x                        
b1
# [1] 0.4848781

Calculate the standard error for $\beta_1$:

ss.residual <- sum((y-model$fitted)^2)
n <- length(x) # SAMPLE SIZE
k <- length(model$coef) # NUMBER OF MODEL PARAMETER (i.e. b0 and b1)
df.residual <- n-k
ms.residual <- ss.residual/df.residual # RESIDUAL MEAN SQUARE
se.b1 <- sqrt(ms.residual/ss.x)
se.b1
# [1] 0.03760029

And the t-value and p-value for $\beta_1$:

t.b1 <- b1/se.b1
p.b1 <- 2*pt(-abs(t.b1), df=n-2)
t.b1
# [1] 12.89559
p.b1
# [1] 3.693245e-09

What I don't know at this point, and this is Question 2, is, how to calculate the same t-value using $r$ instead of $\beta_1$ (perhaps in baby-steps)?

I assume that since cor.test()'s alternative hypothesis is whether the true correlation is not equal to 0 (see cor.test() output above), I would expect something like the Pearson correlation coefficient $r$ divided by the "standard error of the Pearson correlation coefficient" (similar to the b1/se.b1 above)?! But what would that standard error be and why?

Maybe this has something to do with the aforementioned assumptions underlying a correlation test and a regression slope test?!

EDIT (27-Jul-2017): While @whuber provided a very detailed explanation for Question 1 (and partly Question 2, see comments under his answer), I did some further digging and found that these two posts (here and here) do show a specific standard error for $r$, which works well to answer Question 2, that is to reproduce the t-value given $r$:

r <- 0.9603906
n <- 16
r.se <- sqrt((1-r^2)/(n-2))
r/r.se
# [1] 12.8956
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    $\begingroup$ It is the same test or at least an equivalent test. If you reject the hypothesis that the correlation is not zero the test also reject the hypothesis that the slope is not zero. $\endgroup$ Commented Jan 19, 2017 at 22:42
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    $\begingroup$ @Michael Right--but there are many potential models here, and they are strikingly different. One of them is a standard model for correlation, of which the simplest is that the data are a sample from some unknown bivariate Normal distribution. Another is some version of an OLS model for regression of $Y$ against $X$--in two flavors, fixed regressors and random regressors. Another reverses the roles of $X$ and $Y$. If you have a feeling these should produce the same p-values for comparable hypothesis tests, that's probably only through extensive familiarity, but it's not intuitively obvious! $\endgroup$
    – whuber
    Commented Jan 20, 2017 at 0:53
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    $\begingroup$ @whuber Seeing that this Q is so well upvoted but lacks a satisfactory answer, I started a bounty that has ended earlier today; it's in the grace period now. One new answer was posted and it well explains the correlation-as-slope computations, but claims that there is no difference in assumptions, contrary to your quoted statement. My bounty is going to be automatically awarded to this new answer unless another one appears. I am letting you know in case you would consider posting your own answer as well. $\endgroup$
    – amoeba
    Commented Jul 25, 2017 at 14:19
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    $\begingroup$ @amoeba Thank you; I had not noticed the bounty. I have posted a partial account of what I had in mind when I wrote the remark that sparked this question. I hope it represents some progress in the direction you have suggested. $\endgroup$
    – whuber
    Commented Jul 25, 2017 at 15:27
  • $\begingroup$ For really beautiful comparison of correlation coefficient with regression (not only), I really recommend to check out this beautiful classic reference: Rodgers, J.L., Nicewander, W.A., 1988. Thirteen Ways to Look at the Correlation Coefficient. Am. Stat. 42, 59–66. doi.org/10.1080/00031305.1988.10475524 $\endgroup$
    – Tomas
    Commented Dec 1, 2021 at 18:43

4 Answers 4

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Introduction

This reply addresses the underlying motivation for this set of questions:

What are the assumptions underlying a correlation test and a regression slope test?

In light of the background provided in the question, though, I would like to suggest expanding this question a little: let us explore the different purposes and conceptions of correlation and regression.

Correlation typically is invoked in situations where

  • Data are bivariate: exactly two distinct values of interest are associated with each "subject" or "observation".

  • The data are observational: neither of the values was set by the experimenter. Both were observed or measured.

  • Interest lies in identifying, quantifying, and testing some kind of relationship between the variables.

Regression is used where

  • Data are bivariate or multivariate: there may be more than two distinct values of interest.

  • Interest focuses on understanding what can be said about a subset of the variables--the "dependent" variables or "responses"--based on what might be known about the other subset--the "independent" variables or "regressors."

  • Specific values of the regressors may have been set by the experimenter.

These differing aims and situations lead to distinct approaches. Because this thread is concerned about their similarities, let's focus on the case where they are most similar: bivariate data. In either case those data will typically be modeled as realizations of a random variable $(X,Y)$. Very generally, both forms of analysis seek relatively simple characterizations of this variable.

Correlation

I believe "correlation analysis" has never been generally defined. Should it be limited to computing correlation coefficients, or could it be considered more extensively as comprising PCA, cluster analysis, and other forms of analysis that relate two variables? Whether your point of view is narrowly circumscribed or broad, perhaps you would agree that the following description applies:

Correlation is an analysis that makes assumptions about the distribution of $(X,Y)$, without privileging either variable, and uses the data to draw more specific conclusions about that distribution.

For instance, you might begin by assuming $(X,Y)$ has a bivariate Normal distribution and use the Pearson correlation coefficient of the data to estimate one of the parameters of that distribution. This is one of the narrowest (and oldest) conceptions of correlation.

As another example, you might being by assuming $(X,Y)$ could have any distribution and use a cluster analysis to identify $k$ "centers." One might construe that as the beginnings of a resolution of the distribution of $(X,Y)$ into a mixture of unimodal bivariate distributions, one for each cluster.

One thing common to all these approaches is a symmetric treatment of $X$ and $Y$: neither is privileged over the other. Both play equivalent roles.

Regression

Regression enjoys a clear, universally understood definition:

Regression characterizes the conditional distribution of $Y$ (the response) given $X$ (the regressor).

Historically, regression traces its roots to Galton's discovery (c. 1885) that bivariate Normal data $(X,Y)$ enjoy a linear regression: the conditional expectation of $Y$ is a linear function of $X$. At one pole of the special-general spectrum is Ordinary Least Squares (OLS) regression where the conditional distribution of $Y$ is assumed to be Normal$(\beta_0+\beta_1 X, \sigma^2)$ for fixed parameters $\beta_0, \beta_1,$ and $\sigma$ to be estimated from the data.

At the extremely general end of this spectrum are generalized linear models, generalized additive models, and others of their ilk that relax all aspects of OLS: the expectation, variance, and even the shape of the conditional distribution of $Y$ may be allowed to vary nonlinearly with $X$. The concept that survives all this generalization is that interest remains focused on understanding how $Y$ depends on $X$. That fundamental asymmetry is still there.

Correlation and Regression

One very special situation is common to both approaches and is frequently encountered: the bivariate Normal model. In this model, a scatterplot of data will assume a classic "football," oval, or cigar shape: the data are spread elliptically around an orthogonal pair of axes.

  • A correlation analysis focuses on the "strength" of this relationship, in the sense that a relatively small spread around the major axis is "strong."

  • As remarked above, the regression of $Y$ on $X$ (and, equally, the regression of $X$ on $Y$) is linear: the conditional expectation of the response is a linear function of the regressor.

(It is worthwhile pondering the clear geometric differences between these two descriptions: they illuminate the underlying statistical differences.)

Of the five bivariate Normal parameters (two means, two spreads, and one more that measures the dependence between the two variables), one is of common interest: the fifth parameter, $\rho$. It is directly (and simply) related to

  1. The coefficient of $X$ in the regression of $Y$ on $X$.

  2. The coefficient of $Y$ in the regression of $X$ on $Y$.

  3. The conditional variances in either of the regressions $(1)$ and $(2)$.

  4. The spreads of $(X,Y)$ around the axes of an ellipse (measured as variances).

A correlation analysis focuses on $(4)$, without distinguishing the roles of $X$ and $Y$.

A regression analysis focuses on the versions of $(1)$ through $(3)$ appropriate to the choice of regressor and response variables.

In both cases, the hypothesis $H_0: \rho=0$ enjoys a special role: it indicates no correlation as well as no variation of $Y$ with respect to $X$. Because (in this simplest situation) both the probability model and the null hypothesis are common to correlation and regression, it should be no surprise that both methods share an interest in the same statistics (whether called "$r$" or "$\hat\beta$"); that the null sampling distributions of those statistics are the same; and (therefore) that hypothesis tests can produce identical p-values.

This common application, which is the first one anybody learns, can make it difficult to recognize just how different correlation and regression are in their concepts and aims. It is only when we learn about their generalizations that the underlying differences are exposed. It would be difficult to construe a GAM as giving much information about "correlation," just as it would be hard to frame a cluster analysis as a form of "regression." The two are different families of procedures with different objectives, each useful in its own right when applied appropriately.


I hope that this rather general and somewhat vague review has illuminated some of the ways in which "these issues go deeper than simply whether $r$ and $\hat\beta$ should be numerically equal." An appreciation of these differences has helped me understand what various techniques are attempting to accomplish, as well as to make better use of them in solving statistical problems.

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  • $\begingroup$ Thank you whuber for this insightful answer! As mentioned in the comments to @matt-barstead 's answer, I did came across a standard error for $r$, regarding my 2nd question. What I don't quite understand though is how it is derived and why (similar to the question here) $\endgroup$
    – Stefan
    Commented Jul 25, 2017 at 21:26
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    $\begingroup$ The SE for $r$ can be derived only by making specific distributional assumptions, such as that $(X,Y)$ is bivariate Normal. At that point it's an exercise in integral Calculus--which for this question is not an illuminating thing to pursue. The distribution of $r$ is quoted by Wikipedia and is derived (geometrically) in my post at stats.stackexchange.com/a/85977/919. $\endgroup$
    – whuber
    Commented Jul 25, 2017 at 21:44
  • $\begingroup$ I will leave this can of worms for some other time then :) Thanks for your comment @whuber ! $\endgroup$
    – Stefan
    Commented Jul 25, 2017 at 21:50
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As @whuber's answer suggests there are a number of models and techniques that may fall under the correlation umbrella that do not have clear analogues in a regression world and vice versa. However, by and large when people think about, compare, and contrast regression and correlation they are in fact considering two sides of the same mathematical coin (typically a linear regression and a Pearson' correlation). Whether they should take a broader view of both families of analyses is something of a separate debate, and one that researchers should wrestle with at least minimally.

Ultimately, when evaluating correlation and regression in their most common applications, there are conceptual distinctions to be made between these two, but not not mathematical ones, aside from a linear transformation of $x$ and $y$ to specify certain distributional properties of $(x,y)$.

In this narrow view of both regression and correlation the following explanations should help elucidate how and why their estimates, standard errors and p values are essentially variants of one another.

With the dataframe dat being the longley data set referenced above we get the following for the cor.test. (There is nothing new here unless you skipped over the question above and went straight to reading the answers):

> cor.test(dat$Employed, dat$Population)

    Pearson's product-moment correlation

data:  dat$Employed and dat$Population
t = 12.896, df = 14, p-value = 3.693e-09
alternative hypothesis: true correlation is not equal to 0
95 percent confidence interval:
 0.8869236 0.9864676
sample estimates:
      cor 
0.9603906 

And the following for the linear model (also same as above):

> summary(lm(Employed~Population, data=dat))

Call:
lm(formula = Employed ~ Population, data = dat)

Residuals:
    Min      1Q  Median      3Q     Max 
-1.4362 -0.9740  0.2021  0.5531  1.9048 

Coefficients:
            Estimate Std. Error t value Pr(>|t|)    
(Intercept)   8.3807     4.4224   1.895   0.0789 .  
Population    0.4849     0.0376  12.896 3.69e-09 ***
---
Signif. codes:  
0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1

Residual standard error: 1.013 on 14 degrees of freedom
Multiple R-squared:  0.9224,    Adjusted R-squared:  0.9168 
F-statistic: 166.3 on 1 and 14 DF,  p-value: 3.693e-09

Now for the new component to this answer. First, create two new standardized versions of the Employed and Population variables:

> dat$zEmployed<-scale(dat$Employed)
> dat$zPopulation<-scale(dat$Population)

Second re-run the regression:

> summary(lm(zEmployed~zPopulation, data=dat))

Call:
lm(formula = zEmployed ~ zPopulation, data = dat)

Residuals:
     Min       1Q   Median       3Q      Max 
-0.40894 -0.27733  0.05755  0.15748  0.54238 

Coefficients:
              Estimate Std. Error t value Pr(>|t|)    
(Intercept) -2.956e-15  7.211e-02     0.0        1    
zPopulation  9.604e-01  7.447e-02    12.9 3.69e-09 ***
---
Signif. codes:  
0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1

Residual standard error: 0.2884 on 14 degrees of freedom
Multiple R-squared:  0.9224,    Adjusted R-squared:  0.9168 
F-statistic: 166.3 on 1 and 14 DF,  p-value: 3.693e-09

Voila! The regression slope equals the correlation coefficient from above. The answer to Question 1 then is that the assumptions for both tests are essentially the same:

  1. Independence of observations
  2. A linear relation between $x$ and $y$
  3. Normally distributed residuals with a mean of zero, $e\backsim N(0,\sigma_e^2)$
  4. Error terms are similarly distributed at each predicted value of the regression line (i.e., homogeneity of error variance)

Should any of these assumptions not be met, a researcher should interpret with caution results from either a correlation or a simple linear regression. After all, the only difference between a simple linear regression and a correlation (specifically Pearson's) is the linear transformation of both the $x$ and $y$ variables in which both variables are mean-centered and assigned a variance of 1 (sometimes called z-scoring or standardizing).

For Question 2, let's start with the standard error of the regression slope formula used above (implied in the R code - but stated outright below):

$$ b=\frac{\sum(X_i-\bar{X})(Y_i-\bar{Y})}{\sum(X_i-\bar{X})^2} $$

Therefore if we want to know the standard error of $b$ we need to be able to calculate its variance (or $Var(b)$). To make the notation simpler we can say $\mathbf{X_i}=(X_i-\bar{X})$ and $\mathbf{Y_i}=(Y_i-\bar{Y})$, which means that...

$$ Var(b)=Var(\frac{\sum(\mathbf{X_i}\mathbf{Y_i})}{\sum(\mathbf{X_i}^2)}) $$

From that formula you can get to the following, condensed and more useful expression (see this link for step-by-step):

$$ Var(b)=\frac{\sigma_e^2}{\sum(X_i-\bar{X})^2} $$ $$ SE(b) =\sqrt{Var(b)}=\sqrt{\frac{\sigma_e^2}{\sum(X_i-\bar{X})^2}} $$

where $\sigma_e^2$ represents the variance of the residuals.

I think you'll find if you solve this equation for the unstandardized and standardized (i.e., correlation) linear models you'll get the same p and t values for your slopes. Both tests are relying on ordinary least squares estimation and make the same assumptions. In practice, many researchers skip over assumption checking for both simple linear regression models and correlations, though I think it is even more prevalent to do so for correlations as many people do not recognize them as special cases of simple linear regressions. (Note: this is not a good practice to adopt)

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    $\begingroup$ This answer does not address the quote from @whuber reproduced in the question, where he claims that the assumptions are different. Do you mean to say that this statement was wrong? $\endgroup$
    – amoeba
    Commented Jul 25, 2017 at 6:46
  • $\begingroup$ If you follow out these equations, a Pearson's correlation has the same basic assumptions of a simple linear regression. I can amend my response to more clearly state this. $\endgroup$ Commented Jul 25, 2017 at 12:18
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    $\begingroup$ Thank for your answer! I was aware that the correlation coefficient equals the regression slope when standardized. This was shown in link 3 and 4 in my question. I was also aware of the general assumptions you listed and that's why @whuber 's comment got me thinking hence leading to this question. I should have explicitly stated which assumptions I am aware of - my apologies. $\endgroup$
    – Stefan
    Commented Jul 25, 2017 at 21:10
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    $\begingroup$ I actually did some further digging and found that these two posts (here and here) do show a specific standard error for $r$, which works well to answer my second question that is to reproduce the t-value given $r$: r <- 0.9603906; n <- 16; r/(sqrt((1-r^2)/(n-2))) # 12.8956. $\endgroup$
    – Stefan
    Commented Jul 25, 2017 at 21:13
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Here is an explanation of the equivalence of the test, also showing how r and b are related.

http://www.real-statistics.com/regression/hypothesis-testing-significance-regression-line-slope/

In order to perform OLS, you have to make https://en.wikipedia.org/wiki/Ordinary_least_squares#Assumptions

Additionally, OLS and corr require assumption of random sampling.

Construction of a corr test assumes:

We have a "random and large enough sample" from the population of (x,y).

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Regarding question 2

how to calculate the same t-value using r instead of β1

I do not think it is possible to calculate the $t$ statistic from the $r$ value, however the same statistical inference can be derived from the $F$ statistic, where the alternative hypothesis is that the model does not explain the data, and this can be calculated from $r$. $$ F = \frac{r^2/k}{(1-r^2)/(n-k)} $$

With $k=2$ parameters in the model and $n=datapoints$

With the restriction that

...the F ratio cannot be used when the model does not have intercept

Source: Hypothesis testing in the multiple regression model

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    $\begingroup$ I looked back at the original post to identify what question you might be responding to. I found two, numbered 1 (about assumptions) and 2 (about calculating a t-value), but neither seems to be addressed by this answer. Could you tell us more explicitly what question you are answering? $\endgroup$
    – whuber
    Commented May 16, 2017 at 13:16
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    $\begingroup$ Thank you for the clarification: the connection to the question is now apparent. I interpret the question differently, though. I take it to be asking how the p-value for the correlation analysis (that is, as based on the sample correlation coefficient $r$ and the model it implies) is computed (and implicitly to show explicitly why it ought to yield the same value for the regression analysis). Your answer, although correct, is also based on regression, so it still leaves us wondering. $\endgroup$
    – whuber
    Commented May 16, 2017 at 13:48
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    $\begingroup$ I think I understand, perhaps I was answering the question in the specific case rather than the general. I think it would be useful to be able to state the question in terms of a general null and alternative hypothesis to be able to consider this general case, as I am struggling to so. $\endgroup$ Commented May 17, 2017 at 12:53
  • $\begingroup$ I agree: exhibiting clear models and decision criteria for the correlation and regression analyses would be of great help in distinguishing them. Sometimes a good answer consists of little more than reframing or clarifying the question, and often the best answers begin with effective restatements of the question, so don't be afraid to go in that direction. $\endgroup$
    – whuber
    Commented May 17, 2017 at 12:59

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