= Solution
For a sample of size $n\ge2$, define
$$
\overline X=\frac1n\sum_{i=1}^nX_i,\qquad
S^2=\frac1{n-1}\sum_{i=1}^n(X_i-\overline X)^2.
$$
For the zero-mean Gaussian sample, the standardized <vector> $(X_1/\sigma,\ldots,X_n/\sigma)$ has independent <standard normal> coordinates. Apply an <orthogonal transformation> whose first row is $(1,\ldots,1)/\sqrt n$. The resulting coordinates $Z_1,\ldots,Z_n$ remain independent <standard normals>: their <joint density> depends only on squared Euclidean length and the transformation has unit absolute Jacobian. Its first coordinate is $Z_1=\sqrt n\,\overline X/\sigma$, while <orthogonal decomposition> gives
$$
\sum_{j=2}^nZ_j^2=\frac1{\sigma^2}\sum_i(X_i-\overline X)^2
=\frac{(n-1)S^2}{\sigma^2}.
$$
Thus
$$
\boxed{\frac{\sqrt n}{\sigma}\overline X\sim N(0,1),\quad
\frac{n-1}{\sigma^2}S^2\sim\chi^2_{n-1},\quad\overline X\text{ and }S^2\text{ independent}.}
$$
The ratio of the first independent <standard normal> and the <square root> of the normalized chi-square variable has a Student <probability distribution>, giving
$$
\boxed{\frac{\sqrt n\,\overline X}{S}\sim t_{n-1}.}
$$
For the paired height observations, form one cross-minus-self difference per pot. The sampling units for this test are the fifteen differences, not thirty separate plant heights. Under the usual independent normal model for differences, a null mean of zero gives a $t_{14}$ statistic. The observed statistic is
$$
t_{\rm obs}=\frac{2.607}{4.713/\sqrt{15}}.
$$
For the two-sided alternative, reject at <significance level> $\delta$ if $|t_{\rm obs}|>t_{14,1-\delta/2}$. Its two-sided $p$-value is $2[1-F_{t_{14}}(|t_{\rm obs}|)]$. This uses the recorded values without performing the numerical calculation forbidden by the question. The normal difference model is what makes the small-sample Student calibration exact.
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