Find the reference means rejected by an upper-tail test

$\bar x=511.5/75=6.82$ and $s^2=[4027.89-511.5^2/75]/74=7.29$. Test $H_0:\mu=\mu_0$ against $H_1:\mu>\mu_0$. Using the source’s independent large-sample normal approximation, $\bar X\approx N(\mu_0,7.29/75)$ under $H_0$. Reject for $(6.82-\mu_0)/(2.7/\sqrt{75})\ge1.281551566$. Therefore $\begin{aligned}&\mu_0\le6.82\\ &{}-1.281551566\frac{2.7}{\sqrt{75}}\\ & =6.420451763672\ldots.\end{aligned}$ This is rejection in favour of the upper alternative, not proof of it.

The printed set is $\{\mu_0\in\mathbb R:0<\mu_0\le6.42\}$. Its positive lower bound is an extra restriction on reference means, not a consequence of the test inequality. Without that restriction, the real-valued rejection set is $(-\infty,6.420451763672\ldots]$; with positive reference means, intersect it with $(0,\infty)$. The bound 6.42 is rounded; if reference means are explicitly restricted to a 0.01 grid, it is the largest rejecting grid value.

Weekend-only sampling does not cover all times during the week and need not represent the whole-week target. Randomness within weekends does not repair the missing weekday frame. The source rejects the whole-week inference for possible time-pattern bias; that does not establish any specific dietary or clinical effect. Even random times over the week need a suitable dependence model: temporal correlation can invalidate the iid standard error used here.