Answer:
A sample of 601 bricks will be needed.
Step-by-step explanation:
In a sample with a number n of people surveyed with a probability of a success of
, and a confidence level of
, we have the following confidence interval of proportions.
![\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}](https://tex.z-dn.net/?f=%5Cpi%20%5Cpm%20z%5Csqrt%7B%5Cfrac%7B%5Cpi%281-%5Cpi%29%7D%7Bn%7D%7D)
In which
z is the zscore that has a pvalue of
.
The margin of error is:
![M = z\sqrt{\frac{\pi(1-\pi)}{n}}](https://tex.z-dn.net/?f=M%20%3D%20z%5Csqrt%7B%5Cfrac%7B%5Cpi%281-%5Cpi%29%7D%7Bn%7D%7D)
95% confidence level
So
, z is the value of Z that has a pvalue of
, so
.
How large a sample will we need?
A sample of n bricks will be needed.
n is found when M = 0.04.
We have no estimate for the proportion of defective bricks, so we work with the worst case scenario, which is ![M = 0.04](https://tex.z-dn.net/?f=M%20%3D%200.04)
![M = z\sqrt{\frac{\pi(1-\pi)}{n}}](https://tex.z-dn.net/?f=M%20%3D%20z%5Csqrt%7B%5Cfrac%7B%5Cpi%281-%5Cpi%29%7D%7Bn%7D%7D)
![0.04 = 1.96\sqrt{\frac{0.5*0.5}{n}}](https://tex.z-dn.net/?f=0.04%20%3D%201.96%5Csqrt%7B%5Cfrac%7B0.5%2A0.5%7D%7Bn%7D%7D)
![0.04\sqrt{n} = 1.96*0.5](https://tex.z-dn.net/?f=0.04%5Csqrt%7Bn%7D%20%3D%201.96%2A0.5)
![\sqrt{n} = \frac{1.96*0.5}{0.04}](https://tex.z-dn.net/?f=%5Csqrt%7Bn%7D%20%3D%20%5Cfrac%7B1.96%2A0.5%7D%7B0.04%7D)
![(\sqrt{n})^{2} = (\frac{1.96*0.5}{0.04})^{2}](https://tex.z-dn.net/?f=%28%5Csqrt%7Bn%7D%29%5E%7B2%7D%20%3D%20%28%5Cfrac%7B1.96%2A0.5%7D%7B0.04%7D%29%5E%7B2%7D)
![n = 600.25](https://tex.z-dn.net/?f=n%20%3D%20600.25)
Rounding up
A sample of 601 bricks will be needed.