I'm having trouble with a problem, where I can't understand how my textbook's answer makes sense:
An architect is measuring the distance between two points. Each value can be seen as an observation of a normal-distributed variable, with $\mu$ as expectation value and std $\sigma = 2.04$ meters. The architect performs $n=16$ measurements and finds the average $\bar{x} = 2316$ meter.
Create a 95% confidence-interval for the real distance $\mu$
Vi create a $Z$-interval.
$Z_{\alpha/2} = Z_{0.05/2} = 1.960$
$[\bar{X}-Z_{\alpha/2}*\frac{\sigma}{\sqrt{n}}, \bar{X}+Z_{\alpha/2}*\frac{\sigma}{\sqrt{n}} ]$
$\bar{X}\pm Z_{\alpha/2}*\frac{\sigma}{\sqrt{n}} = 2316 \pm 1.960 * \frac{2.04}{\sqrt{16}} = [2315,2317]$
How many measurements must the architect do so that the length of the confidence-interval is 0.2 meters?
Using Length rule for Z-intervals:
$n \ge (\frac{2*Z_{\alpha/2} * \sigma}{L})^2 (\frac{2*1.1960*2.04}{0.2})^{2} = 595$
I also tried:
$L = 2* Z_{\alpha/2} * \frac{\sigma}{\sqrt{n}} = 2\cdot1.196\cdot\frac{2.04}{\sqrt{16}} = 1.21$
$n \ge (\frac{2*Z_{\alpha/2} * \sigma}{L})^2 = n \ge (\frac{2*1.1960 * 2.04}{1.21})^2 = 16.263$
However, my textbooks answer is: Requires $L<0.2M$ Therefore: $n \ge 1599$.
Thus my question is what am I doing wrong?
I would highly appreciate any answer.
Solved: There was a misplaced number in $1.196$ which really should be $1.96$
Thanks to @Caran-d'Ache