Answer:
The smallest sample size required to obtain the desired margin of error is 44.
Step-by-step explanation:
I think there was a small typing error, we have that
is the standard deviation of these weighs.
We have that to find our
level, that is the subtraction of 1 by the confidence interval divided by 2. So:
![\alpha = \frac{1-0.9}{2} = 0.05](https://tex.z-dn.net/?f=%5Calpha%20%3D%20%5Cfrac%7B1-0.9%7D%7B2%7D%20%3D%200.05)
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 ![z = 1.645](https://tex.z-dn.net/?f=z%20%3D%201.645)
Now, find the margin of error M as such
![M = z*\frac{\sigma}{\sqrt{n}}](https://tex.z-dn.net/?f=M%20%3D%20z%2A%5Cfrac%7B%5Csigma%7D%7B%5Csqrt%7Bn%7D%7D)
Which of these is the smallest approximate sample size required to obtain the desired margin of error?
This sample size is n.
n is found when ![M = 15](https://tex.z-dn.net/?f=M%20%3D%2015)
So
![M = z*\frac{\sigma}{\sqrt{n}}](https://tex.z-dn.net/?f=M%20%3D%20z%2A%5Cfrac%7B%5Csigma%7D%7B%5Csqrt%7Bn%7D%7D)
![15 = 1.645*\frac{60}{\sqrt{n}}](https://tex.z-dn.net/?f=15%20%3D%201.645%2A%5Cfrac%7B60%7D%7B%5Csqrt%7Bn%7D%7D)
![15\sqrt{n} = 1.645*60](https://tex.z-dn.net/?f=15%5Csqrt%7Bn%7D%20%3D%201.645%2A60)
Simplifying by 15
![\sqrt{n} = 1.645*4](https://tex.z-dn.net/?f=%5Csqrt%7Bn%7D%20%3D%201.645%2A4)
![(\sqrt{n})^{2} = (1.645*4)^{2}](https://tex.z-dn.net/?f=%28%5Csqrt%7Bn%7D%29%5E%7B2%7D%20%3D%20%281.645%2A4%29%5E%7B2%7D)
![n = 43.3](https://tex.z-dn.net/?f=n%20%3D%2043.3)
Rounding up
The smallest sample size required to obtain the desired margin of error is 44.