mjhjmtu

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:

$$H_0: boldsymbolbeta = beginbmatrix 0 \ 1 endbmatrix quad quad quad H_A: boldsymbolbeta neq beginbmatrix 0 \ 1 endbmatrix .$$

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

$$SSE_0 = sum_i=1^n (y_i-x_i)^2 quad quad quad SSE_A = sum_i=1^n (y_i - hatbeta_0 - hatbeta_1 x_i)^2$$

The test statistic is:

$$F equiv F(mathbfy, mathbfx) = fracn-22 cdot fracSSE_0 - SSE_ASSE_A.$$

The corresponding p-value is:

$$p equiv p(mathbfy, mathbfx) = int limits_F(mathbfy, mathbfx) ^infty textF-Dist(r | 2, n-2) dr.$$

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.