Test standby time and compare significance statements

$\bar x=3040.8/90=33.7866666667$ and $\begin{aligned}&s^2\\ & =\frac{115773.66-3040.8^2/90}{89}\\ & =\frac{3258791}{22250}\\ & =146.4625168539.\end{aligned}$ Test $H_0:\mu=36$ against $H_1:\mu<36$. For independent large-sample observations the source uses $\bar X\approx N(36,s^2/90)$ by CLT with estimated variance, so its standardised statistic is approximately standard normal.

$z=(33.7866666667-36)/\sqrt{146.4625168539/90}=-1.735021388$. The lower-tail $p=0.04136850844\approx0.0414>0.01$: do not reject. Equivalently, $z>-2.326347874$, the 1% lower critical value. There is insufficient evidence at 1% that mean standby time is below 36 h. The source screen enters $\mu_0=36$, $\sigma=12.10206593934$ from rounded variance 146.46, $\bar x=33.787$, $n=90$, alternative <, producing $z\approx-1.734774996$, $p\approx0.0413902936$. Its hand calculation instead uses mean 33.78; these small premature-rounding differences leave the requested conclusion unchanged.

For talk-time, $H_0:\mu=5$, $H_1:\mu<5$ has $p\le0.05$ by the given result. Therefore $p\le0.05<0.10$, so the same data/model/alternative also reject at 10%.

For the source’s symmetric continuous normal reference, the observed lower-tail statistic is negative and the two-sided p-value is $2p$. The given $p\le0.05$ only gives $2p\le0.10$, not $2p\le0.05$. For example, one-sided $p=0.02$ gives two-sided 0.04 and rejects at 5%; one-sided 0.04 gives two-sided 0.08 and does not. Changing the alternative changes the p-value; the standby-time test does not supply this talk-time p-value.