A lower-tail test at the 1% level

Test $H_0:\mu=22$ against $H_1:\mu<22$, where $\mu$ is the population mean. Under $H_0$, $X\sim N(22,36)$ and $\bar X\sim N(22,36/50)$. The known standard deviation is 6. The teacher strategy recommends the p-value approach because all required summary data are available.

$z=(20.2-22)/(6/\sqrt{50})=-2.121320344\ldots$. The lower-tail probability is $P(\bar X\le20.2)=P(Z\le-2.121320344\ldots)=0.01694742676\ldots\approx0.0169$. The source’s alternative calculator route is STAT → TESTS → Z-Test → Stats: enter $\mu_0=22$, $\sigma=6$, $\bar x=20.2$, $n=50$, select $\mu<\mu_0$, then Calculate. Its screen reports $z=-2.121320344$, $p=0.0169473661$; that slightly different last-digit value is preserved here, while the independent CDF calculation agrees to the printed 0.0169.

Reject $H_0$ when $p\le0.01$. Since $0.0169>0.01$, do not reject $H_0$. There is insufficient evidence at the 1% level that the population mean is below 22. A result that would be significant at 5% need not be significant at 1%.