Answer:
538 books should be tested.
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{1-0.99}{2} = 0.005](https://tex.z-dn.net/?f=%5Calpha%20%3D%20%5Cfrac%7B1-0.99%7D%7B2%7D%20%3D%200.005)
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 = 2.575](https://tex.z-dn.net/?f=z%20%3D%202.575)
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)
In which
is the standard deviation of the population and n is the size of the sample.
How many books should be tested to estimate the average force required to break the binding to within 0.08 lb with 99% confidence?
n books should be tested.
n is found when ![M = 0.08](https://tex.z-dn.net/?f=M%20%3D%200.08)
We have that ![\sigma = 0.72](https://tex.z-dn.net/?f=%5Csigma%20%3D%200.72)
![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)
![0.08 = 2.575*\frac{0.72}{\sqrt{n}}](https://tex.z-dn.net/?f=0.08%20%3D%202.575%2A%5Cfrac%7B0.72%7D%7B%5Csqrt%7Bn%7D%7D)
![0.08\sqrt{n} = 2.575*0.72](https://tex.z-dn.net/?f=0.08%5Csqrt%7Bn%7D%20%3D%202.575%2A0.72)
![\sqrt{n} = \frac{2.575*0.72}{0.08}](https://tex.z-dn.net/?f=%5Csqrt%7Bn%7D%20%3D%20%5Cfrac%7B2.575%2A0.72%7D%7B0.08%7D)
![(\sqrt{n})^{2} = (\frac{2.575*0.72}{0.08})^{2}](https://tex.z-dn.net/?f=%28%5Csqrt%7Bn%7D%29%5E%7B2%7D%20%3D%20%28%5Cfrac%7B2.575%2A0.72%7D%7B0.08%7D%29%5E%7B2%7D)
![n = 537.1](https://tex.z-dn.net/?f=n%20%3D%20537.1)
Rounding up
538 books should be tested.