Test for a change after servicing a machine

Let $X$ denote a post-service bar length in cm and $\mu$ the population mean length. Test $H_0:\mu=420$ against $H_1:\mu\ne420$. “A change” requires both directions even though the observed mean is higher. Under $H_0$, $X\sim N(420,12^2)$ and $\bar X\sim N(420,12^2/100)$.

The standard error is $12/\sqrt{100}=1.2$ cm, so $z=(423-420)/1.2=2.5$. By symmetry, $p=P(|Z|\ge2.5)=2P(Z\ge2.5)=0.01241933065\ldots\approx0.0124$. The source’s Z-Test screen uses $\mu_0=420$, $\sigma=12$, $\bar x=423$, $n=100$ and the $\mu\ne\mu_0$ alternative; it reports $z=2.5$ and $p=0.0124193597$. Its last digits differ slightly from the independent CDF calculation but both round to 0.0124.

At 5%, reject $H_0$ when $p\le0.05$. Since $0.0124\le0.05$, reject $H_0$: there is sufficient evidence that the population mean bar length has changed after servicing, assuming the standard deviation remains 12 cm. This is evidence for a change in the mean, not for a change in variance.