Test cereal mass and invert a two-sided rejection

$\bar x=79.53/80=0.994125$ kg and $s^2=[100.4621-79.53^2/80]/79=0.270877705696\ldots\ \mathrm{kg}^2$. Test $H_0:\mu=1.10$ against $H_1:\mu<1.10$, with $\mu$ the population mean content mass. Under $H_0$, use $\bar X\approx N(1.10,s^2/80)$ by the large-sample approximation. Individual normality is not given.

$z=(0.994125-1.10)/\sqrt{s^2/80}=-1.819498851\ldots$ gives $p=0.03441767898\ldots<0.04$. Reject $H_0$: sufficient evidence at 4% that mean content mass is below 1.10 kg. The source prints variance 0.27088 and uses estimated SD 0.52046133381837 in Z-Test Stats with $\mu_0=1.1$, $\bar x=0.994125$, $n=80$, lower alternative. Its screen shows $z=-1.819491146$ and $p=0.0344182092$, printed as 0.034418. This source result is distinct from the full-precision calculation but gives the same decision.

For $H_0:\mu=\mu_0$ against $H_1:\mu\ne\mu_0$, the 4% test assigns 0.02 to each tail. With $c=\Phi^{-1}(0.98)=2.053748911\ldots$, reject when $|(\bar x-\mu_0)/\sqrt{s^2/80}|\ge c$. The original PDF visibly includes the absolute-value bars, although extraction loses them. Equivalently, $\mu_0\le\bar x-c\sqrt{s^2/80}$ or $\mu_0\ge\bar x+c\sqrt{s^2/80}$.

The full-precision boundaries are $\mu_0\le0.874619199928\ldots$ or $\mu_0\ge1.113630800072\ldots$. The source’s intermediate values are 0.87462 and 1.1136; its final real-valued set is $0<\mu_0\le0.875$ or $\mu_0\ge1.11$. Those final rounded endpoints admit some means that do not reject: for example, 1.11 is inside the non-rejection interval. Preserve the exact expressions for classification. The additional positive restriction reflects the mass context; intersect with $\mu_0>0$ if positive reference means are required. The teacher note also calls $z_{\mathrm{test}}$ by $z_{\mathrm{calc}}$ and $z_{\mathrm{critical}}$ by $z_\alpha$; these are alternate labels for the same roles, not different statistics.