Question

How large of a sample size is required to estimate the mean within 1.6 units with 90% confidence, given s=0.47.

Answer 1

The answer is exactly 1, but I don't understand how

Answer 2

didn't learn this sry cant help

Answer 3

use below :

Margin of error= \(\large Z^* \frac{s}{\sqrt{n}} \)

Answer 4

you're given :
Margin of error = \(1.6\)
\(s = 0.47\)

Answer 5

you need to find \(Z^*\) corresponding to 90% CI and plugin above

Answer 6

yes I get 0.611229, but its asking how large of a sample, my zscore table says 1.645

Answer 7

http://www.math.upenn.edu/~chhays/zscoretable.pdf

Answer 8

yes, for 90% CI, \(Z^* = 1.65\)

Answer 9

plugin that value and solve \(n\)

Answer 10

\(\large 1.6 = 1.65 \frac{0.47}{\sqrt{n}}\)

Answer 11

solve \(n\)

Answer 12

0.23350036035

Answer 13

yes, so the sample size needs to be a MINIMUM of 0.2335..

Answer 14

think a bit,
can u really sample 0.2335... persons ?

Answer 15

ohhh bless your heart, the problem is simpified!

Answer 16

i get it, thank you

Answer 17

np... u wlc :)