The diameters of 25 steel rods have a sample mean of 0.980 cm and a standard deviation of 0.015 cm. Assuming the diameters are normally distributed with this standard deviation, find the 99% confidence interval for the population mean diameter.

The 99% CI is $\bar{x} \pm z_{0.005}\cdot\frac{s}{\sqrt n}$ with $z_{0.005}=2.576$, $s=0.015$, $n=25$. The margin is $2.576\times\frac{0.015}{\sqrt{25}}=2.576\times0.003=0.00773$, giving $(0.980-0.00773,\,0.980+0.00773)=(0.9723,\,0.9877)$ cm. Using the 95% multiplier $z=1.96$ instead of the 99% value gives the first option. Forgetting to divide the standard deviation by $\sqrt n$ before multiplying by $z$ gives the much wider third option. Using the one-tailed 99% value $z=2.326$ instead of the two-tailed $z_{0.005}=2.576$ gives the fourth option.