Find the significance levels that reject a claim

Let $\mu$ be the population mean examination mark. $\bar x=11872/250=47.488$ and $s^2=[646193-11872^2/250]/249=330.9857991968\ldots$. The source rounds this variance to 330.99. Under $H_0$, use $\bar X\approx N(49.5,s^2/250)$ by the large-sample approximation.

For $H_0:\mu=49.5$ against $H_1:\mu<49.5$, $z=(47.488-49.5)/\sqrt{s^2/250}=-1.748611379\ldots$ and $p=0.04017910900\ldots$. The source’s p-value strategy is appropriate because the unknown is the significance level. Its Z-Test Stats screen uses $\mu_0=49.5$, estimated SD 18.1931305713 from rounded variance 330.99, $\bar x=47.488$, $n=250$, lower alternative; it reports $z=-1.748600283$, $p=0.0401800257$, printed as 0.040180. These rounded-input/source-algorithm differences are retained separately.

Rejection requires $p\le\alpha/100$. Hence $100p\le\alpha<100$, with exact lower endpoint approximately $4.017910900\%$ for the full-precision normal approximation. The source’s final $4.02\le\alpha<100$ is a rounded display of that threshold; it is not the identical set of real significance percentages. If percentages are restricted to 0.01 increments, 4.02% is the least rejecting grid value. The source includes a stray 5% rejection-rule line before solving for $\alpha$; 5% is not supplied by the question.