Answer:
At least 832 teenargers must be interviewed.
Step-by-step explanation:
We have that to find our
level, that is the subtraction of 1 by the confidence interval divided by 2. So:
Now, we have to find z in the Ztable as such z has a pvalue of
.
So it is z with a pvalue of
, so
Now, find the margin of error M as such
In which
is the standard deviation of the population and n is the size of the sample.
How many teenagers must the firm interview in order to have a margin of error of at most 0.1 liter when constructing a 99% confidence interval
At least n teenargers must be interviewed.
n is found when M = 0.1.
We have that ![\sigma = 1.12](https://tex.z-dn.net/?f=%5Csigma%20%3D%201.12)
So
![0.1\sqrt{n} = 2.575*1.12](https://tex.z-dn.net/?f=0.1%5Csqrt%7Bn%7D%20%3D%202.575%2A1.12)
![\sqrt{n} = \frac{2.575*1.12}{0.1}](https://tex.z-dn.net/?f=%5Csqrt%7Bn%7D%20%3D%20%5Cfrac%7B2.575%2A1.12%7D%7B0.1%7D)
![(\sqrt{n})^{2} = (\frac{2.575*1.12}{0.1})^{2}](https://tex.z-dn.net/?f=%28%5Csqrt%7Bn%7D%29%5E%7B2%7D%20%3D%20%28%5Cfrac%7B2.575%2A1.12%7D%7B0.1%7D%29%5E%7B2%7D)
![n = 831.7](https://tex.z-dn.net/?f=n%20%3D%20831.7)
Rounding up
At least 832 teenargers must be interviewed.