I've come across a discrepancy in one of my practice questions.
I'm given the following: $\sigma=150$ units. For a 98% CI, how many observations do I need to be accurate within 15 units.
Now, the 15 refers to the width of my CI: $$w=2\cdot Z_{\frac{\alpha}{2}}\cdot \frac{\sigma}{\sqrt{n}}$$
This comes from $$w=(\bar{x}+Z_{\frac{\alpha}{2}}\cdot \frac{\sigma}{\sqrt{n}})-(\bar{x}-Z_{\frac{\alpha}{2}}\cdot \frac{\sigma}{\sqrt{n}})$$
For a 98% CI, 2-tailed, I need the Z-score for 0.99. That is 2.33 appoximately (qnorm(.99)=2.3263).
so when I plug it in and solve for n: $$n=(2\cdot 2.33\cdot 150/30)^2$$ $$=23.3^2=543$$.
However, the answer given is 644.... if someone could help me out I would appreciate it.