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: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 ( [tex]\frac{98}{250}=0.392[/tex])ME is the margin of Error from the meanand margin of error (ME) can be found using the formulaME=[tex]\frac{z*\sqrt{p*(1-p)}}{\sqrt{N} }[/tex] where z is the corresponding statistic in 95% confidence level (1.96)p is the sample proportion ( [tex]\frac{98}{250}=0.392[/tex])N is the sample size (250) Using the numbers ME=[tex]\frac{1.96*\sqrt{0.392*0.608}}{\sqrt{250} }[/tex] ≈ 0.06Then the true proportion for the population would be p±ME = 0.392±0.06 in 95% confidence level.