Question

Suppose 250 randomly selected people are surveyed to determine if they own a tablet. Of the 250 surveyed, 98 reported owning a tablet. Using a 95% confidence level, compute a confidence interval estimate for the true proportion of people who own tablets.

Answer 1

Answer:

The true proportion for the population would be between 0.386 and 0,398 (0.392±0.06) in 95% confidence level.

Step-by-step explanation:

Confidence interval for true population proportion can be computed as p±ME where

• p is the sample proportion ( $\frac{98}{250}=0.392$)
• ME is the margin of Error from the mean

and margin of error (ME) can be found using the formula

ME=$\frac{z*\sqrt{p*(1-p)}}{\sqrt{N} }$ where

• z is the corresponding statistic in 95% confidence level (1.96)
• p is the sample proportion ( $\frac{98}{250}=0.392$)

• N is the sample size (250)

Using the numbers

ME=$\frac{1.96*\sqrt{0.392*0.608}}{\sqrt{250} }$ ≈ 0.06

Then the true proportion for the population would be p±ME = 0.392±0.06 in 95% confidence level.