Question

A marketing firm is asked to estimate the percent of existing customers who would purchase a "digital upgrade" to their basic cable TV service. The firm wants 99 percent confidence and an error of ± 5 percent. What is the required sample size (to the next higher integer)?A. 664B. 625C. 801D. 957

Answer 1

Answer: A. 664

Step-by-step explanation:

Given : A marketing firm is asked to estimate the percent of existing customers who would purchase a "digital upgrade" to their basic cable TV service.

But there is no information regarding the population proportion is mentioned.

Formula to find the samples size , if the prior estimate to the population proportion is unknown :

$n=0.25(\dfrac{z*}{E})^2$

, where E = Margin of error.

z* = Two -tailed critical z-value

We know that critical value for 99% confidence interval = $z*=2.576$  [By z-table]

Margin of error = 0.05

Then, the minimum sample size would become :

$n=0.25(\dfrac{2.576}{0.05})^2$

Simplify,

$n=0.25\times2654.3104=663.5776\approx664$

Thus, the required sample size= 664

Hence, the correct answer is A. 664.