Assess Alice’s claim about mean mass

Let $X$ be a randomly selected man’s mass in kg and $\mu$ the population mean. Test $H_0:\mu=75$ against $H_1:\mu>75$. Under $H_0$, $X\sim N(75,80)$ and $\bar X\sim N(75,80/100)$. Normality and known variance make this sampling distribution exact; no CLT approximation is needed.

The standard error is $\sqrt{80/100}$ kg, not 80 or $80/100$. Thus $z=(77.2-75)/\sqrt{80/100}=2.459674775\ldots$ and $p=P(Z\ge z)=0.006953148\ldots\approx0.00695$. The source displays $z=2.46$ but retains the p-value from the unrounded input; keep full precision until the final result.

Reject $H_0$ if $p\le0.05$. Here $0.00695<0.05$, so reject $H_0$: there is sufficient evidence at the 5% level that the population mean mass exceeds 75 kg. Under the null, repeated random samples of 100 have probability about 0.00695 of a mean at least 77.2 kg. This is not a probability that the null is true.

The source’s calculator uses inverse normal with upper-tail area 0.05, mean 75 and standard deviation $\sqrt{80/100}$: $a=75+1.644853627\sqrt{0.8}=76.47120181\ldots$. The source writes the critical region $\bar x>76.471$; the observed 77.2 is inside it. Using the inclusive boundary convention $p\le\alpha$ gives the same decision; boundary equality has zero probability in this continuous model.