a random sample of 1000 people was chosen for a survey. 80% of them had health insurance. Find the 95% confidence interval
To find the 95% confidence interval for the percentage of people with health insurance in the population, we can use the formula:
CI = p̂ ± Z * √((p̂ * (1 - p̂)) / n)
Where:
- p̂ is the sample proportion (the percentage of people in the sample with health insurance, which is given as 80% or 0.80).
- Z is the z-score for the desired confidence level. For a 95% confidence level, we can use Z = 1.96 (which corresponds to the standard normal distribution).
- n is the sample size (which is given as 1000 in this case).
Plugging in these values into the formula, we get:
CI = 0.80 ± 1.96 * √((0.80 * (1 - 0.80)) / 1000)
Simplifying the expression inside the square root:
CI = 0.80 ± 1.96 * √(0.16 / 1000)
CI = 0.80 ± 1.96 * 0.01265
CI ≈ 0.80 ± 0.02477
Therefore, the 95% confidence interval for the percentage of people with health insurance in the population is approximately:
CI ≈ (0.77523, 0.82477)