A report states that 38% of home owners have a vegetable garden. How large a sample is needed to estimate the true proportion of home owners who have vegetable gardens to
Within 6 percentage points with 98% confidence?
A) 178
B) 276
C) 105
D) 356

Question

Quizplus · Accepted Answer

To estimate the sample size needed for a proportion with a given 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 (from Z-tables). Given $p = 0.38$ and $E = 0.06$, substituting these values gives $n = \frac{(2.33)^2 \cdot 0.38 \cdot (1-0.38)}{(0.06)^2}$, which calculates to approximately 355.8, rounding up to 356.

A Report States That 38% of Home Owners Have a Vegetable