Distinguish unbiased variance, estimated SD and significance

$\bar x=12290/200-10=51.45$ and $\begin{aligned}&s^2\\ & =\frac{811212-12290^2/200}{199}\\ & =\frac{111983}{398}\\ & =281.3643216080.\end{aligned}$ The usual estimated SD is $s=16.7739178968\approx16.8$. The source calls that an unbiased SD estimate, matching its question, but generally $\sqrt{S^2}$ is not unbiased for population SD even when $S^2$ is unbiased for variance. No distribution-free correction making this square root unbiased follows from these summaries alone.

Test $H_0:\mu=50$, $H_1:\mu>50$. With representative independent observations and finite variance, use $\bar X\approx N(50,s^2/200)$ under $H_0$ by CLT, with estimated variance. No individual normality assumption is needed for that approximation; sampling and finite-moment assumptions still are. The statistic is $z=(51.45-50)/\sqrt{281.3643216080/200}=1.222498928$. Its upper p-value is 0.1107595044, printed 0.111. Do not reject at 5%; there is insufficient evidence of a mean above 50. Equivalently $1.2225<1.644853627$. The source prints 1.6445 as its critical value: this is a typo, not the 5% normal quantile. Its calculator uses rounded variance 281.36 (SD 16.773789077009), giving $z\approx1.222508317$, $p\approx0.110757794$, with the same decision.

To reject under the source convention $p\le\alpha/100$, require $11.0759504434\ldots\le\alpha<100.$ The printed $11.1\le\alpha<100$ uses a rounded p-value and is not the exact real-valued set. If significance percentages themselves are given to one decimal place,11.1% is the least rejecting grid value. “Accepting the alternative” here means rejecting the null in its favour, not establishing it as certainly true.

When the actual population mean score is 50, this rule has approximately $\alpha/100$ probability of wrongly concluding that the population mean exceeds 50, under the stated large-sample model. This is not a probability that the alternative hypothesis is true; variance estimation and CLT make the nominal error rate approximate.