Reverse a known-variance test for mean time and sample size

Let $\mu$ be the population mean installation time in minutes after music is introduced. Test $H_0:\mu=38.0$ against $H_1:\mu<38.0$. The known standard deviation remains 5.0 minutes. Under $H_0$, $T\sim N(38,25)$ and $\bar T\sim N(38,25/n)$ exactly for independent normal sampling.

At 5%, the lower critical value is $c=\Phi^{-1}(0.05)=-1.644853627\ldots$. For $n=50$, reject when $(\bar t-38)/(5/\sqrt{50})\le c$. Thus $\bar t\le38+5c/\sqrt{50}=36.8369128463\ldots$ minutes. In the positive-time context this gives $0<\bar t\le36.8369128463\ldots$. The source gives $0<\bar t\le36.8$ after rounding; 36.8 is a boundary label, not the exact rejection set for real-valued means.

With $\bar t=37.1$, the observed statistic is $(37.1-38)/(5/\sqrt n)=-0.18\sqrt n$. Since rejection includes equality, non-rejection means $-0.18\sqrt n>c$, hence $\sqrt n<-c/0.18$ and $n<(c/0.18)^2=83.5044275955\ldots$. As $n$ is a positive integer, $n\in\{1,2,\ldots,83\}$. The source uses rounded $c=-1.6449$, $\sqrt n<9.1383$, $n<83.5$ and the same integer set. Here normality and known variance justify the test even for small n; do not impose an unnecessary n≥30 restriction.