Posibly 760 but not really sure sorry
Answer:
The company should take a sample of 148 boxes.
Step-by-step explanation:
Hello!
The cable TV company whats to know what sample size to take to estimate the proportion/percentage of cable boxes in use during an evening hour.
They estimated a "pilot" proportion of p'=0.20
And using a 90% confidence level the CI should have a margin of error of 2% (0.02).
The CI for the population proportion is made using an approximation of the standard normal distribution, and its structure is "point estimation" ± "margin of error"
[p' ±
]
Where
p' is the sample proportion/point estimator of the population proportion
is the margin of error (d) of the confidence interval.
![Z_{1-\alpha /2} = Z_{1-0.05} = Z_{0.95}= 1.648](https://tex.z-dn.net/?f=Z_%7B1-%5Calpha%20%2F2%7D%20%3D%20Z_%7B1-0.05%7D%20%3D%20Z_%7B0.95%7D%3D%201.648)
So
![d= Z_{1-\alpha /2} * \sqrt{\frac{p'(1-p')}{n} }](https://tex.z-dn.net/?f=d%3D%20Z_%7B1-%5Calpha%20%2F2%7D%20%2A%20%5Csqrt%7B%5Cfrac%7Bp%27%281-p%27%29%7D%7Bn%7D%20%7D)
![d *Z_{1-\alpha /2}= \sqrt{\frac{p'(1-p')}{n} }](https://tex.z-dn.net/?f=d%20%2AZ_%7B1-%5Calpha%20%2F2%7D%3D%20%5Csqrt%7B%5Cfrac%7Bp%27%281-p%27%29%7D%7Bn%7D%20%7D)
![(d*Z_{1-\alpha /2})^2= \frac{p'(1-p')}{n}](https://tex.z-dn.net/?f=%28d%2AZ_%7B1-%5Calpha%20%2F2%7D%29%5E2%3D%20%5Cfrac%7Bp%27%281-p%27%29%7D%7Bn%7D)
![n*(d*Z_{1-\alpha /2})^2= p'(1-p')](https://tex.z-dn.net/?f=n%2A%28d%2AZ_%7B1-%5Calpha%20%2F2%7D%29%5E2%3D%20p%27%281-p%27%29)
![n= \frac{p'(1-p')}{(d*Z_{1-\alpha /2})^2}](https://tex.z-dn.net/?f=n%3D%20%5Cfrac%7Bp%27%281-p%27%29%7D%7B%28d%2AZ_%7B1-%5Calpha%20%2F2%7D%29%5E2%7D)
![n= \frac{0.2(1-0.2)}{(0.02*1.648)^2}](https://tex.z-dn.net/?f=n%3D%20%5Cfrac%7B0.2%281-0.2%29%7D%7B%280.02%2A1.648%29%5E2%7D)
n= 147.28 ≅ 148 boxes.
I hope it helps!
wouldn't this just be 2+2+2+2+10+10+10+10? or is it times I'm not sure but I hope this helps anyone to answer this fully :)