Question

How many randomly selected adults must be surveyed to ask me the percentage of adults in the country who now use the internet assume that you want to be 99% confident that the sample percentage is within 2 percentage points of the true population percentage and that you know nothing about the percentage of adults using the internet at a local high school football game 150 randomly selected people were so surveyed about the location of the concession stand 108 of the people feel the same should be closer to the bleachers and the best point estimate of the proportion of all people attending the game who feel the concession stand should be closer to the bleachers

Answer 1

Two answers:

1) Sample size needed for national Internet-use percentage (99% confidence, ±2% margin, p unknown)
- Use p = 0.5 for maximum variability. z for 99% confidence = 2.576.
- n = (z^2 p(1−p)) / E^2 = (2.576^2 × 0.25) / 0.02^2
- 2.576^2 × 0.25 = 1.658944 → n = 1.658944 / 0.0004 = 4147.36
- Always round up → n = 4,148

So you need to survey 4,148 adults.

2) Point estimate from the football-game survey
- Sample proportion p̂ = 108 / 150 = 0.72 = 72%.

So the best point estimate is 0.72 (72%).