Cfrgtkky

I have some data which is fit along a roughly linear line:

When I do a linear regression of these values, I get a linear equation:

$$y = 0.997x-0.0136$$

In an ideal world, the equation should be $y = x$.

Clearly, my linear values are close to that ideal, but not exactly. My question is, how can I determine whether this result is statistically significant?

Is the value of 0.997 significantly different from 1? Is -0.01 significantly different from 0? Or are they statistically the same and I can conclude that $y=x$ with some reasonable confidence level?

What is a good statistical test I can use?

Thanks

This type of situation can be handled by a standard F-test for nested models. Since you want to test both of the parameters against a null model with fixed parameters, your hypotheses are:

$$H_0: boldsymbol{beta} = begin{bmatrix} 0 \ 1 end{bmatrix} quad quad quad H_A: boldsymbol{beta} neq begin{bmatrix} 0 \ 1 end{bmatrix} .$$

The F-test involves fitting both models and comparing their residual sum-of-squares, which are:

$$SSE_0 = sum_{i=1}^n (y_i-x_i)^2 quad quad quad SSE_A = sum_{i=1}^n (y_i - hat{beta}_0 - hat{beta}_1 x_i)^2$$

The test statistic is:

$$F equiv F(mathbf{y}, mathbf{x}) = frac{n-2}{2} cdot frac{SSE_0 - SSE_A}{SSE_A}.$$

The corresponding p-value is:

$$p equiv p(mathbf{y}, mathbf{x}) = int limits_{F(mathbf{y}, mathbf{x}) }^infty text{F-Dist}(r | 2, n-2) dr.$$

Implementation in R: Suppose your data is in a data-frame called DATA with variables called y and x. The F-test can be performed manually with the following code. In the simulated mock data I have used, you can see that the estimated coefficients are close to the ones in the null hypothesis, and the p-value of the test shows no significant evidence to falsify the null hypothesis that the true regression function is the identity function.

#Generate mock data (you can substitute your data if you prefer)

set.seed(12345);

n    <- 1000;

x    <- rnorm(n, mean = 0, sd = 5);

e    <- rnorm(n, mean = 0, sd = 2/sqrt(1+abs(x)));

y    <- x + e;

DATA <- data.frame(y = y, x = x);



#Fit initial regression model

MODEL <- lm(y ~ x, data = DATA);



#Calculate test statistic

SSE0   <- sum((DATA$y-DATA$x)^2);

SSEA   <- sum(MODEL$residuals^2);

F_STAT <- ((n-2)/2)*((SSE0 - SSEA)/SSEA);

P_VAL  <- pf(q = F_STAT, df1 = 2, df2 = n-2, lower.tail = FALSE);



#Plot the data and show test outcome

plot(DATA$x, DATA$y,

     main = 'All Residuals',

     sub  = paste0('(Test against identity function - F-Stat = ',

            sprintf("%.4f", F_STAT), ', p-value = ', sprintf("%.4f", P_VAL), ')'),

     xlab = 'Dataset #1 Normalized residuals',

     ylab = 'Dataset #2 Normalized residuals');

abline(lm(y ~ x, DATA), col = 'red', lty = 2, lwd = 2);

The summary output and plot for this data look like this:

summary(MODEL);



Call:

lm(formula = y ~ x, data = DATA)



Residuals:

    Min      1Q  Median      3Q     Max 

-4.8276 -0.6742  0.0043  0.6703  5.1462 



Coefficients:

            Estimate Std. Error t value Pr(>|t|)    

(Intercept) -0.02784    0.03552  -0.784    0.433    

x            1.00507    0.00711 141.370   <2e-16 ***

---

Signif. codes:  0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1



Residual standard error: 1.122 on 998 degrees of freedom

Multiple R-squared:  0.9524,    Adjusted R-squared:  0.9524 

F-statistic: 1.999e+04 on 1 and 998 DF,  p-value: < 2.2e-16



F_STAT;

[1] 0.5370824



P_VAL;

[1] 0.5846198

Here is a cool graphical method which I cribbed from Julian Faraway's excellent book "Linear Models With R (Second Edition)". It's simultaneous 95% confidence intervals for the intercept and slope, plotted as an ellipse.

For illustration, I created 500 observations with a variable "x" having N(mean=10,sd=5) distribution and then a variable "y" whose distribution is N(mean=x,sd=2). That yields a correlation of a little over 0.9 which may not be quite as tight as your data.

You can check the ellipse to see if the point (intercept=0,slope=1) fall within or outside that simultaneous confidence interval.

library(tidyverse)

library(ellipse)

#> 

#> Attaching package: 'ellipse'

#> The following object is masked from 'package:graphics':

#> 

#>     pairs



set.seed(50)

dat <- data.frame(x=rnorm(500,10,5)) %>% mutate(y=rnorm(n(),x,2))



lmod1 <- lm(y~x,data=dat)

summary(lmod1)

#> 

#> Call:

#> lm(formula = y ~ x, data = dat)

#> 

#> Residuals:

#>     Min      1Q  Median      3Q     Max 

#> -6.9652 -1.1796 -0.0576  1.2802  6.0212 

#> 

#> Coefficients:

#>             Estimate Std. Error t value Pr(>|t|)    

#> (Intercept)  0.24171    0.20074   1.204    0.229    

#> x            0.97753    0.01802  54.246   <2e-16 ***

#> ---

#> Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1

#> 

#> Residual standard error: 2.057 on 498 degrees of freedom

#> Multiple R-squared:  0.8553, Adjusted R-squared:  0.855 

#> F-statistic:  2943 on 1 and 498 DF,  p-value: < 2.2e-16



cor(dat$y,dat$x)

#> [1] 0.9248032



plot(y~x,dat)

abline(0,1)

confint(lmod1)

#>                  2.5 %    97.5 %

#> (Intercept) -0.1526848 0.6361047

#> x            0.9421270 1.0129370



plot(ellipse(lmod1,c("(Intercept)","x")),type="l")

points(coef(lmod1)["(Intercept)"],coef(lmod1)["x"],pch=19)



abline(v=confint(lmod1)["(Intercept)",],lty=2)

abline(h=confint(lmod1)["x",],lty=2)



points(0,1,pch=1,size=3)

#> Warning in plot.xy(xy.coords(x, y), type = type, ...): "size" is not a

#> graphical parameter



abline(v=0,lty=10)

abline(h=0,lty=10)

^{Created on 2019-01-21 by the reprex package (v0.2.1)}

You could compute the coefficients with n bootstrapped samples. This will likely result in normal distributed coefficient values (Central limit theorem). With that you could then construct a (e.g. 95%) confidence interval with t-values (n-1 degrees of freedom) around the mean. If your CI does not include 1 (0), it is statistically significant different, or more precise: You can reject the null hypothesis of an equal slope.

You could perform a simple test of hypothesis, namely a t-test. For the intercept your null hypothesis is $beta_0=0$ (note that this is the significance test), and for the slope you have that under H0 $beta_1=1$.

You should fit a linear regression and check the 95% confidence intervals for the two parameters. If the CI of the slope includes 1 and the CI of the offset includes 0 the two sided test is insignificant approx. on the (95%)^2 level -- as we use two separate tests the typ-I risk increases.

Using R:

fit = lm(Y ~ X)

confint(fit)

or you use

summary(fit)

and calc the 2 sigma intervals by yourself.