Interpret an upper-tail p-value in context

$\bar x=8+1400/60=94/3$. The unbiased variance is $\begin{aligned}&s^2\\ & =\frac{34800-1400^2/60}{59}\\ & =\frac{6400}{177}.\end{aligned}$

Let $\mu$ be the population mean daily case count. Test $H_0:\mu=30$ against $H_1:\mu>30$. Under $H_0$, $\bar X\approx N(30,(6400/177)/60)$ using the large sample. Then $\begin{aligned}&z\\ & =\frac{94/3-30}{\sqrt{(6400/177)/60}}\\ & =1.717556404.\end{aligned}$ The upper-tail p-value is $0.0429387790<0.05$, so reject $H_0$: there is sufficient evidence at 5% that the mean daily count exceeds 30. Source calculator input: Z-Test Stats, null 30, estimated standard deviation 6.013168224, sample mean 94/3,n 60, greater-than alternative; its displayed p≈.0429387435 agrees.

The 5% upper-tail critical value is $\Phi^{-1}(0.95)=1.644853627$. Since $z=1.717556404$ exceeds it, the same rejection decision follows.

If the population mean really is 30 cases per day, the approximate probability that a random sample of 60 days has a mean at least as large as the observed $94/3$ is 0.04294. This probability is conditional on the null model; it is not the probability that the increased-mean claim is false.