Question

Please explain this problem to me, it has me lost. Assume I know nothing at all. Thank you.

A cable TV company wants to estimate the percentage of cable boxes in use during an evening hour. An approximation based on previous surveys is 25 percent. The company wants the new estimate to be at the 90 percent confidence level and within 2 percent of the actual proportion. What sample size is needed?

Answer 1

Given,

Estimated proportion (p) = 0.25

So,

q = 1 - p = 1 - 0.25 = 0.75

Margin of error (E) = 2% = 0.02

For 90% confidence level, critical z value = 1.645

Hence,

Sample size needed

$n = p*q*(\frac{z}{E})^2$

$n = 0.25*0.75*(\frac{1.645}{0.02})^2$

n = 1268.4

Now n can't be in decimals so rounding off to next whole number:

n = 1269