Test a revision-time claim and find a minimum sample size

$\bar x=120-1500/150=110$ minutes. The unbiased variance estimate is $\begin{aligned}&s^2\\ & =\frac{462000-(-1500)^2/150}{149}\\ & =3000\ \mathrm{min}^2.\end{aligned}$

Let $\mu$ be the population mean daily revision time in minutes. Test $H_0:\mu=120$ against $H_1:\mu<120$. With the large random sample, under $H_0$, $\bar X\approx N(120,3000/150)$ by the CLT, using the estimated variance. Thus $\begin{aligned}&z\\ & =\frac{110-120}{\sqrt{3000/150}}\\ & =-2.236067977.\end{aligned}$ The left-tail p-value is $0.0126736593\approx 0.0127>0.01$. Do not reject $H_0$: there is insufficient evidence at 1% that the teacher overstated the mean. The source calculator workflow is Z-Test, Stats, null mean 120, estimated standard deviation $\sqrt{3000}=54.77225575$, sample mean 110, n 150, alternative less-than; it displays z≈−2.236067977 and p≈.0126736174.

The 1% lower-tail critical value is $-2.326347874$. Reject when $z\le-2.326347874$. Since $-2.236067977$ is to the right of this boundary, the same non-rejection conclusion follows. The comparison diagram shades the 1% rejection tail.

The printed “110 hours” conflicts with all the minute-based data and the source working. To reproduce the intended question, take the new mean to be 110 minutes and retain the 1% level and $s^2=3000$. Rejection requires $\frac{-10}{\sqrt{3000/n}}\le-2.326347874.$ Therefore $n\ge 30(2.326347874)^2=162.3568329$, so the least integer is $163$. At n 162, p≈.01007; at n 163, p≈.00988, straddling .01. If 110 hours were literal, the mean would be 6600 minutes and could never reject in this lower-tail direction;163 is conditional on the intended minute unit.