one tail hypothesis testing, Gaussian versus t-distribution Since the t-distribution represents more uncertainty than the Gaussian distribution I am accustomed to expecting a greater p-value when hypothesis testing. Consider autumn and winter temperatures as illustrated as input to my program for the apache commons math library.
autumn +1.0, +4.5, -1.0, +3.5, -2.0, +2.0, +1.5, +2.0, +0.0, -1.0, -2.0, +3.0, +4.5, +6.5, +6.0, -1.0, +2.0
winter -7.0, +1.0, -2.5, -2.0, +1.0, +2.0, -3.5, -4.0, -1.0, +2.5, -1.0, -5.5, -2.0, +1.0, +0.0, -3.0, +1.5

The test of the hypothesis that the true mean is zero. The results are as illustrated.
hypothesis is that the true mean equals 0
autumn observed mean  +1.7, gaussian dist p-value 0.008, t-dist p-value 0.017
winter observed mean  -1.3, gaussian dist p-value 0.045, t-dist p-value 0.062

Equal to zero, Autumn
The above agrees with the intuition that the t-dist p-value should be greater because the assumed distribution represents more uncertainty.
Equal to zero, Winter
The above agrees with the intuition that the t-dist p-value should be greater because the assumed distribution represents more uncertainty.
The test of the hypothesis that the true mean is less-than-or-equal to zero. The results are as illustrated.
hypothesis is that the true mean is LTE 0
autumn observed mean  +1.7, gaussian dist p-value 0.004, t-dist p-value 0.008
winter observed mean  -1.3, gaussian dist p-value 0.978, t-dist p-value 0.969

LTE zero, Autumn
When the observed mean is positive (and not supportive of the hypothesis) the usual effect is observed that the t-dist p-value is greater. From the Gaussian p-value there is a strong rejection of the hypothesis and from the t-dist p-value there is a weaker rejection. This aligns with intuition because the t-dist represents more uncertainty so the rejection ought to be weaker.
LTE zero, Winter
When the observed mean is negative (and supportive of the hypothesis) the t-dist p-value is actually smaller. From the Gaussian p-value there is strong support. From the t-dist p-value there is weaker support. This aligns with intuition because the t-dist represents more uncertainty so the support ought to be weaker.
Note that in the last case (LTE zero, Winter), the uncertainty of the t-dist makes the p-value go in the direction opposite to what was observed in the other cases.
Is this description of an intuitive understanding correct despite the last case being different? A good answer would explain why the last case is or is not an anomaly.
 A: In the last case, where the question is whether the data supports the claim that the winter mean is less than or equal to zero, it appears that you are doing a one sided test and reporting $1 - p$, where $p = Pr(\bar{X}_{winter} \le -1.3)$ assuming than $\mu_{winter}=0$. When reporting one-sided $p$ values, $p$ is the value that is typically reported.
winter <- c(-7, 1, -2.5, -2, 1, 2, -3.5, -4, -1, 2.5, -1, -5.5, -2, 1, 0, -3, 1.5)
1 - pnorm(mean(winter), 0, sd(winter)/sqrt(n))
[1] 0.9775428
1 - t.test(x=winter, alternative = "less")$p.value
[1] 0.968934

That $p$ is larger ($1-p$ smaller) for the $t$ version  is expected because $t$ distributions have "fatter tails". The plot below illustrates this; the normal distribution is more concentrated around its mode, whereas the $t$ distribution has more mass out in the tails of the distribution.
R code:
winter <- c(-7, 1, -2.5, -2, 1, 2, -3.5, -4, -1, 2.5, -1, -5.5, -2, 1, 0, -3, 1.5)
n <- length(winter)
x <- seq(-2,2,0.01)
df <- n-1
colors <- c("red", "black")
labels <- c("t(df=16)", "normal")
se <- sd(winter)/sqrt(n)
plot(x, dnorm(x, 0, se), type="l", lty=2, xlab="sample mean",
     ylab="", yaxt='n', main="comparison of null sampling distributions")
lines(x, 1/se * dt(x/se,df), lwd=2, col=colors[i])
legend("topright", title="",
       labels, lwd=2, lty=c(1, 2), col=colors)
abline(v=mean(winter))


A: There is no anomaly.
A p-value of 1.0 is extreme and it means support for the hypothesis being tested.
A p-value of 0.0 is extreme and it means rejection of the hypothesis being tested.
A p-value of 0.5 is moderate and it is consistent with uncertainty.
Greater uncertainty doesn't necessarily mean a greater p-value; it means a p-value closer to 0.5. Since the t-distribution is consistent with greater uncertainty compared to the Gaussian distribution, the p-value of the t-distribution will tend to be closer to 0.5 under the same procedural and computational circumstances.
In all 4 of the described cases, the p-value of the t-distribution is closer to 0.5. Under this intuition there is no anomaly.
