Answer:
She needs a sample size of 25.
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:
![\alpha = \frac{1 - 0.68}{2} = 0.16](https://tex.z-dn.net/?f=%5Calpha%20%3D%20%5Cfrac%7B1%20-%200.68%7D%7B2%7D%20%3D%200.16)
Now, we have to find z in the Ztable as such z has a pvalue of
.
That is z with a pvalue of
, so Z = 0.995.
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%5Cfrac%7B%5Csigma%7D%7B%5Csqrt%7Bn%7D%7D)
In which
is the standard deviation of the population and n is the size of the sample.
The population SD is 2 grams.
This means that ![\sigma = 2](https://tex.z-dn.net/?f=%5Csigma%20%3D%202)
What is the minimum sample size she needs to create a confidence interval that has a width of 0.4 grams?
She needs a sample size of n.
n is found when M = 0.4. So
![M = z\frac{\sigma}{\sqrt{n}}](https://tex.z-dn.net/?f=M%20%3D%20z%5Cfrac%7B%5Csigma%7D%7B%5Csqrt%7Bn%7D%7D)
![0.4 = 0.995\frac{2}{\sqrt{n}}](https://tex.z-dn.net/?f=0.4%20%3D%200.995%5Cfrac%7B2%7D%7B%5Csqrt%7Bn%7D%7D)
![\sqrt{n} = \frac{0.995*2}{0.4}](https://tex.z-dn.net/?f=%5Csqrt%7Bn%7D%20%3D%20%5Cfrac%7B0.995%2A2%7D%7B0.4%7D)
![(\sqrt{n})^2 = (\frac{0.995*2}{0.4})^2](https://tex.z-dn.net/?f=%28%5Csqrt%7Bn%7D%29%5E2%20%3D%20%28%5Cfrac%7B0.995%2A2%7D%7B0.4%7D%29%5E2)
![n = 24.8](https://tex.z-dn.net/?f=n%20%3D%2024.8)
Rounding up:
She needs a sample size of 25.