Why does Paired t-test show not significant? I have a two-column and three-row paired data (data3)
138.2   13.64
64.00  12.76
81.36  14.90

I conducted the following analysis using R:
shapiro.test(data3[,1] - data3[,2])
#    Shapiro-Wilk normality test
#
#data:  data3[, 1] - data3[, 2]
#
#W = 0.89766, p-value = 0.3781

t.test(data3[,1], data3[,2], paired = T)

#   Paired t-test
#data:  data3[, 1] and data3[, 2]
#t = 3.6148, df = 2, p-value = 0.06873
#alternative hypothesis: true difference in means is not equal to 0
#95 percent confidence interval:
# -15.36643 176.87310
# 
#sample estimates:
#mean of the differences 
#           80.75333

I am wondering why the paired t-test shows no significant difference but when looking at the data we can see the first column > the second column.
 A: There are two parts to your analysis. Let's look at them one at a time.
(1) Shapiro-Wilk test for normality. Failure to reject does not mean that
the data are normal.
In particular, you look at three differences, but $n = 3$ is too small
a sample size for the test to have a useful probability of detecting
non-normality, if it exists. So there is no particular reason to believe
that the differences are from a normal population. Thus there is no
particular reason to suppose that results from a t test would be accurate.
An exponential population is very far from normal,
yet in 10,000 sets of $n = 3$ from $\mathsf{Exp}(1),$ the Shapiro-Wilk test
rejected the null hypothesis of normality only 748 times. That is to say,
the power of the test against the alternative that data are exponential
is only about 7.5%. (Simulation in R.)
set.seed(701);  m = 10^4
pv = replicate(m, shapiro.test(rexp(3))$p.val)
mean(pv < .05)
[1] 0.0748

(2) Paired test. It is true that the differences for the three pairs average about 81. This may seem a large value, but it is not a statistically significant value.
You have correctly shown that the (questionable) t test does not find significance at the 5% level.
Moreover, because there are only three differences, no reasonable rank-based, sign test, or permutation test can show a P-value for a two-sided
alternative smaller than 0.25. For a one-sided test (that column 1 is bigger than
column 2) no P-value can be smaller than 0.125.
To achieve a P-value below 5% with any of these tests (one-sided), you would need $n = 5$ differences (all positive or all negative).
d = c( -124.56,  -51.24,  -66.46)
wilcox.test(d)

        Wilcoxon signed rank test

data:  d
V = 0, p-value = 0.25
alternative hypothesis: true location is not equal to 0

A: Because your sample size is very small and the variance of the differences is very big. And because you evaluated the size of the difference using p values, which are heavily influenced by sample size. 
A: Yes there is a difference, but is the difference large enough to say that the population means are different?
As with every statistical test that relies on P-values for interpretation it is important to understand what a p-value is. Remember that we are trying to make inferences about the population based on a sample. 
a P-value is : "the chance that a given estimate would be found in the sample if the H0 would be true in the population".
The H0 for a t-test is that the means of the respective populations are the same.  
That paired t-test calculates that p-value. So the chance of finding these means in the sample, when the population means are the same is 0.068. 
Assuming that you decided upon a significance threshold of p<0.05 the conclusion of this test is that the difference found in your sample is coincidence. 
Think of it in this way. You want to test if younger siblings are systematically shorter in height than their older brother and sisters. To eliminate age you only question adults. 
You take a sample from three families and end up finding something like this: 
younger  older
192      198
175      176 
186      190 

Clearly in the sample the older siblings are taller right? But could they have come from the same population mean? Ie could the mean in the population of younger siblings and older siblings be the same number? 
Given the variance and means in the (very small) sample the t-test shows a non significant difference. If the t-test is non-significant you conclude that it could. 
