Inconsistent x-y relationship from paired t-test and linear regression

Question

I am analyzing an experiment comparing the effect of treatment A vs. B on the matched subject. Here are the measurements on 34 subjects:

 A     B
-1.15 -1.16
-1.13 -0.94
-0.16 -1.18
-0.37 -1.20
-1.09 -1.20
-1.20 -1.20
-0.94 -1.20
-0.84 -1.16
-1.18 -1.17
-1.20 -1.11
-0.78 -0.68
-0.83 -0.73
-1.05 -1.20
-0.71 -1.20
 0.07  0.12
-1.20 -0.98
-1.20 -1.20
-1.02 -1.17
-0.28 -0.84
 1.33  1.47
-1.19 -1.20
-1.20 -1.17
-0.40 -1.20
 0.66 -0.21
-0.63  0.21
-0.88 -1.16
-0.46 -1.20
-0.76 -1.20
-0.38 -1.20
-0.67 -0.97
-0.90 -1.20
-0.90 -1.20
-1.20 -1.15
-1.01 -0.79

The differences between the two treatment (dat[,"A"]-dat[,"B"]) looks normally distributed. I first applied a paired t-test:

t.test(dat$A, dat$B, alternative = c("two.sided"), mu = 0, paired = TRUE)

Paired t-test
data:  dat$A and dat$B 
t = 2.894, df = 33, p-value = 0.006692
alternative hypothesis: true difference in means is not equal to 0 
95 percent confidence interval:
 0.05870022 0.33659390 
sample estimates:
mean of the differences 
              0.1976471 

The paired t-test indicates that on average treatment A has a significantly higher measurement than treatment B.

On the other hand, I applied a linear model on A~B:

mod1 <- lm(A ~ B, data=dat)

Coefficients:
        Estimate Std. Error t value Pr(>|t|)    
(Intercept) -0.03345    0.12739  -0.263    0.795    
B            0.75111    0.11801   6.365 3.79e-07 ***

The 95% confidence interval of the coefficient for B (0.51-0.99) does not cover 1. This result indicates that treatment A has on average a smaller measurement than treatment B, which is contradictory to the findings from the t-test.

Can anyone help me to explain these contradictory findings?

and to expand my question: Does a paired t-test (test mean of difference against 0) equal to a linear regression without intercept (test the coefficient against 1)? I mean in terms of testing against the null hypothesis, rather than the estimate or the meaning of the coefficient. Because both tests are testing against the null hypothesis that $B_{i}-A_{i}=\epsilon_{i}$.

Answer 1

The linear regression tests whether B "predicts" A. The t-test tests whether the means are different. If you wanted to run a regression that was equivalent to the t-test you would have to structure the data so that A and B were two levels of another variable, say "group" that took on the values of your data, and, as @MichaelMayer points out, you would need a random intercept to account for the pairing.

As an example of why the two are not the same, consider:

set.seed(123)
x <- rnorm(100)
y <- x + 0.0001 + rnorm(100, 0, .01)

mod1 <- lm(y ~ x)
summary(mod1)                 #pvalue has 15 0's
t.test(x, y, paired = TRUE)   #p value = 0.32

If you know x, you can give a very good estimate of y. But the means are equivalent.

Answer 2

In addition to the excellent answer by @Peter Flom: This data shouts out for some plotting, something is going on here which we cannot interpret without knowing some context. First a simple plot with a smooth:

Then a Tukey mean-difference plot:

For reference and ease of further experimentation, the R code:

mydata <- 
"A     B 
-1.15 -1.16
-1.13 -0.94
-0.16 -1.18
-0.37 -1.20
-1.09 -1.20
-1.20 -1.20
-0.94 -1.20
-0.84 -1.16
-1.18 -1.17
-1.20 -1.11
-0.78 -0.68
-0.83 -0.73
-1.05 -1.20
-0.71 -1.20
0.07  0.12
-1.20 -0.98
-1.20 -1.20
-1.02 -1.17
-0.28 -0.84
1.33  1.47
-1.19 -1.20
-1.20 -1.17
-0.40 -1.20
0.66 -0.21
-0.63  0.21
-0.88 -1.16
-0.46 -1.20
-0.76 -1.20
-0.38 -1.20
-0.67 -0.97
-0.90 -1.20
-0.90 -1.20
-1.20 -1.15
-1.01 -0.79 "
md <- read.table(textConnection(mydata), header=TRUE)

library(ggplot2)

ggplot(md, aes(A,B)) + geom_point() + geom_smooth()

ggplot(md, aes( (A+B)/2, A-B)) + geom_point()