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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.

PopeP
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  • I'm not sure what you did in the second part, but the first part seems correct. Try $n=1599$ instead of $n=16$ in that part, and see what interval you get. – Joe May 29 '21 at 17:56
  • That gave me a nonsensical answer. The result from the last equation is supposed to be 1599. – PopeP May 29 '21 at 18:06
  • why do you plug in $L=1.21$ in to the last equation , when you want to get $L=0.2$? – Caran-d'Ache May 29 '21 at 18:10
  • That is an edit, I originally attempted 0.2 But that didn't give me the right answer, then I saw in my text book that there was a formula for calculating $L$, so I tried that. I will include my previous answer as well in a new edit. – PopeP May 29 '21 at 18:12
  • If you plug $L=0.2$ and $Z=1.96$ (and not $1.196$ as in you calculations above) you will get 1599. – Caran-d'Ache May 29 '21 at 18:14
  • Oh lord.. How embarrassing. Thank you, Sir. – PopeP May 29 '21 at 18:15
  • I'm not sure why it gave you a nonsensical answer. Using $n=16$ gave you an interval of width $2$, so using $n=1599$ should have given you an interval of width $\approx 0.2$ – Joe May 29 '21 at 19:00

1 Answers1

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Solved: There was a misplaced number in $1.196$ which really should be $1.96$

Thanks to @Caran-d'Ache

PopeP
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