Answer:
She needs to sample 189 trees.
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 - 0.95}{2} = 0.025](https://tex.z-dn.net/?f=%5Calpha%20%3D%20%5Cfrac%7B1%20-%200.95%7D%7B2%7D%20%3D%200.025)
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 = 1.96.
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 standard deviation of the population is 70 peaches per tree.
This means that ![\sigma = 70](https://tex.z-dn.net/?f=%5Csigma%20%3D%2070)
How many trees does she need to sample to obtain an average accurate to within 10 peaches per tree?
She needs to sample n trees.
n is found when M = 10. 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)
![10 = 1.96\frac{70}{\sqrt{n}}](https://tex.z-dn.net/?f=10%20%3D%201.96%5Cfrac%7B70%7D%7B%5Csqrt%7Bn%7D%7D)
![10\sqrt{n} = 1.96*70](https://tex.z-dn.net/?f=10%5Csqrt%7Bn%7D%20%3D%201.96%2A70)
Dividing both sides by 10:
![\sqrt{n} = 1.96*7](https://tex.z-dn.net/?f=%5Csqrt%7Bn%7D%20%3D%201.96%2A7)
![(\sqrt{n})^2 = (1.96*7)^2](https://tex.z-dn.net/?f=%28%5Csqrt%7Bn%7D%29%5E2%20%3D%20%281.96%2A7%29%5E2)
![n = 188.2](https://tex.z-dn.net/?f=n%20%3D%20188.2)
Rounding up:
She needs to sample 189 trees.