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' ± [tex]Z_{1-\alpha /2} * \sqrt{\frac{p'(1-p')}{n} }[/tex]]
Where
p' is the sample proportion/point estimator of the population proportion
[tex]Z_{1-\alpha /2} * \sqrt{\frac{p'(1-p')}{n} }[/tex] is the margin of error (d) of the confidence interval.
[tex]Z_{1-\alpha /2} = Z_{1-0.05} = Z_{0.95}= 1.648[/tex]
So
[tex]d= Z_{1-\alpha /2} * \sqrt{\frac{p'(1-p')}{n} }[/tex]
[tex]d *Z_{1-\alpha /2}= \sqrt{\frac{p'(1-p')}{n} }[/tex]
[tex](d*Z_{1-\alpha /2})^2= \frac{p'(1-p')}{n}[/tex]
[tex]n*(d*Z_{1-\alpha /2})^2= p'(1-p')[/tex]
[tex]n= \frac{p'(1-p')}{(d*Z_{1-\alpha /2})^2}[/tex]
[tex]n= \frac{0.2(1-0.2)}{(0.02*1.648)^2}[/tex]
n= 147.28 ≅ 148 boxes.
I hope it helps!
