Test a minimum string length and interpret a strict boundary

$\bar x=300-60/100=299.4$ m and $\begin{aligned}&s^2\\ & =\frac{1240-(-60)^2/100}{99}\\ & =\frac{1204}{99}\\ & =12.1616161616\ldots\approx12.2\text{ m}^2.\end{aligned}$ The source truncates this to 12.161 before testing; retain the full estimate.

Use $H_0:\mu=300$ versus $H_1:\mu<300$, with 300 the boundary of the composite claim $\mu\ge300$. For independent large-sample observations, $\bar X\approx N(300,(1204/99)/100)$ under the boundary null. Then $z=-1.720503673$ and lower-tail $p=0.04267046326\approx0.0427\le0.05$. Reject: sufficient evidence at 5% that mean length is below 300 m. Equivalently $z<-1.644853627$. The source Z-Test screen enters SD 3.4872625367185 from truncated variance 12.161, mean 299.4,n 100 and alternative <, giving $z\approx-1.720547259$, $p\approx0.0426664692$; the conclusion agrees.

For the improved process, use the known new variance 12.1, not the old estimate. Under the boundary null, $\bar X\approx N(300,12.1/100)$ by CLT; the individual population is not stated normal. The source interprets “claim is valid” as do not reject the lower-tail null at 10%. Since rejection includes $p=0.10$, non-rejection requires $p>0.10$, or $z>-1.281551566$. Thus $\begin{aligned}&k>300\\ &{}+\Phi^{-1}(0.10)\sqrt{12.1/100}\\ & =299.554211592532\ldots.\end{aligned}$ This is only insufficient evidence against the claim, not proof that $\mu\ge300$.

Over real-valued $k$, this strict set has no least member; its infimum is 299.554211592532…. The source prints $k>299.55$ after rounding, which is not the exact set:299.551 would satisfy that printed inequality but fail the test condition. If the sample mean itself is required to lie on a 0.01 m grid, its least admissible value is 299.56 m. Rounding the infimum to 299.55 is not the same as a feasible minimum.