A random sample of 1000 people was chosen for a survey. 80% of them had health insurance. Find the 95% confidence interval. Round answers to 4 significant digits.
To find the 95% confidence interval, we need to use the formula:
Confidence Interval = Sample Proportion ± (Z * Standard Error)
Step 1: Calculate the Sample Proportion
The sample proportion (p̂) is the percentage of people in the sample who had health insurance. In this case, it is given as 80%, which can be written as 0.8.
Sample Proportion (p̂) = 0.8
Step 2: Calculate the Standard Error
The standard error measures how accurately the sample proportion represents the population proportion. It is calculated using the formula:
Standard Error (SE) = √[(p̂ * (1 - p̂)) / n]
where:
- √ represents the square root
- p̂ is the sample proportion
- n is the sample size, which is 1000
Standard Error (SE) = √[(0.8 * (1 - 0.8)) / 1000]
Step 3: Determine the Z-score for a 95% confidence level
The Z-score corresponds to the desired confidence level. For a 95% confidence level, the Z-score is approximately 1.96.
Z = 1.96
Step 4: Calculate the Confidence Interval
Now we can calculate the confidence interval using the formula mentioned earlier:
Confidence Interval = Sample Proportion ± (Z * Standard Error)
Confidence Interval = 0.8 ± (1.96 * √[(0.8 * (1 - 0.8)) / 1000])
Simplify the expression and round the answer to 4 significant digits:
Confidence Interval = 0.8 ± (1.96 * 0.0137)
Confidence Interval = 0.8 ± 0.0268492
Rounded to 4 significant digits:
Confidence Interval ≈ 0.8 ± 0.0268
Therefore, the 95% confidence interval for the proportion of people with health insurance is approximately 0.7732 to 0.8268.