A researcher wishes to estimate the number of households with two computers. How large a sample is neededin order to be 98% confident that the sample proportion will not differ from the true proportion by more than2%? A previous study indicates that the proportion of households with two computers is 20%.
A) 2172
B) 1537
C) 2715
D) 3

Question

Quizplus · Accepted Answer

To determine the sample size needed for a proportion with a specific confidence level and margin of error, the formula is $n = \frac{Z^2 \cdot p \cdot (1-p)}{E^2}$, where $Z$ is the Z-score corresponding to the confidence level, $p$ is the estimated proportion, and $E$ is the margin of error. For a 98% confidence level, the Z-score is approximately 2.33. Given $p = 0.20$ and $E = 0.02$, the calculation is $n = \frac{(2.33)^2 \cdot 0.20 \cdot (1-0.20)}{(0.02)^2}$, which approximates to 2167.36. Rounding up, the closest provided option is 2172.

A Researcher Wishes to Estimate the Number of Households with Two