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The first output uses the equal-variance, pooled Student's t-test, as shown by its seven statistical degrees of freedom. Its exact model assumes two independent samples from normal distributions, with means ${\mu }_X,{\mu }_Y$ and a common unknown variance ${\sigma }^2$. The hypotheses are $H_0:{\mu }_X={\mu }_Y$ against $H_1:{\mu }_X<{\mu }_Y$. The sample means are $\bar{x}=4.35$ and $\bar{y}=9.88$, and the unbiased within-sample sample variances areThe pooled sample variance and its resulting standard error areHenceTo justify its reference distribution, under the null $(\bar{X}-\bar{Y})/(\sigma \sqrt{1/4+1/5})$ is standard normal, while $(3s_X^2+4s_Y^2)/{\sigma }^2$ is ${\chi }_7^2$, independently of the two sample means. Dividing the first variable by the square root of the second divided by seven gives the exact pooled two-sample t statistic, with Student t-distribution $t_7$. Its one-sided p-value is the lower tailThe associated one-sided 95% confidence interval for ${\mu }_X-{\mu }_Y$ isThe printed missing lower bound means that this is a one-sided upper confidence limit, not that a finite lower endpoint should be estimated. With such small samples the normality and common-variance assumptions matter; a Welch analysis would use a different variance estimate and generally different degrees of freedom.

The second output is the Wilcoxon rank sum test for two independent continuous samples. Under a common-distribution null hypothesis, labels are exchangeable conditional on the pooled values, so the four ranks assigned to the first sample form a uniformly chosen four-element subset of $\{1,\dots ,9\}$. The first sample receives ranks $2,1,3,5$, giving raw rank sumEquivalently the Mann–Whitney U test statistic is $U_X=W_X-4\cdot 5/2=1$, the number of cross-sample pairs with an $X$ observation larger than a $Y$ observation. This output reports the raw rank sum, rather than the shifted statistic used by some software.

A first sample tending to be smaller has a small rank sum. Under the null, all $\left(\genfrac{}{}{0ex}{}{9}{4}\right)=126$ rank allocations have equal probability. Only $\{1,2,3,4\}$ and $\{1,2,3,5\}$ have rank sum at most 11. By the exact rank-sum tail one above the minimum, the exact lower-tail p-value isNo large-sample normal approximation or continuity correction is involved; the absence of ties makes this simple enumeration valid. For a common-shape location-shift family, the test's alternative is a negative location shift of $X$ relative to $Y$. Without that model, the null is equality of distributions and the rank comparison is not simply a test of equality of means or medians.

Both tests reject at the 5% level in the direction of smaller $X$ observations; only the pooled t-test rejects at 1%. The t-test uses the numerical spacing and targets a mean difference under its normal model. The rank-sum test uses ordering and an exact exchangeability calculation, avoiding the normality and common-normal-variance assumptions. Their different p-values therefore reflect both different statistics and different modelling assumptions, not a numerical contradiction.

Write the common variance as ${\sigma }^2$. Independence and the normal distribution giveMoreover $\bar{X}$ and $S_{XX}$ are independent. To see both assertions, standardize the observations into an isotropic Gaussian vector and choose an orthonormal basis with first vector $(1,\dots ,1)/\sqrt{m}$. Its first coordinate is $\sqrt{m}(\bar{X}-{\mu }_X)/\sigma$; the remaining independent standard normal coordinates have squared sum $S_{XX}/{\sigma }^2$. The same argument applies to the $Y$ sample, independently of the first sample.

Under the null hypothesis, the difference of sample means is $N(0,{\sigma }^2(1/m+1/n))$. The residual sums of squares add to an independent chi-squared distribution with $\nu =m+n-2$ degrees of freedom. Define the pooled sample variance and test statistic byFor $\nu >0$, under $H_0$ this is $Z/\sqrt{U/\nu }$ with independent $Z\sim N(0,1)$ and $U\sim {\chi }_{\nu }^2$. Hence $T\sim t_{m+n-2}$ under the null, the Student t-distribution.

Let $t_{\nu ,0.99}$ be its 99th percentile. For the specified one-sided alternative, reject $H_0$ precisely whenBecause this distribution is continuous, the rejection probability under $H_0$ is exactly $0.01$, irrespective of the nuisance mean and variance. Large positive values supply evidence for ${\mu }_X>{\mu }_Y$; a two-sided or 0.995 cutoff would test a different alternative. The common-variance and independence assumptions are essential to this exact Student t-test. If both samples have size one, there are no residual degrees of freedom and the proposed unknown-variance test cannot be formed.