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just olya [345]
3 years ago
14

Suppose 250 randomly selected people are surveyed to determine if they own a tablet. Of the 250 surveyed, 98 reported owning a t

ablet. Using a 95% confidence level, compute a confidence interval estimate for the true proportion of people who own tablets.
Mathematics
1 answer:
SashulF [63]3 years ago
6 0

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.

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Answer:

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Step-by-step explanation:

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Pie
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pshichka [43]

Answer:

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Step-by-step explanation:

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