A marketing firm is asked to estimate the percentage 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)664
B)625
C)801
D)957

Question

Quizplus · Accepted Answer

To achieve a 99% confidence level with a margin of error of ±5%, the formula for sample size is $n = \frac{Z^2 \times p(1-p)}{E^2}$, where $Z$ is the Z-score corresponding to the confidence level (2.576 for 99%), $p$ is the estimated proportion (0.5 for maximum variability), and $E$ is the margin of error (0.05). Plugging in the values: $n = \frac{2.576^2 \times 0.5(1-0.5)}{0.05^2} \approx 664.3$. Since we round up to the next higher integer, the required sample size is 665, but since that option is not available, the closest provided option is A) 664.

A Marketing Firm Is Asked to Estimate the Percentage of Existing