A data contains a random sample of 100 employees with eduation levels. Please construct a 90% confidence interval of the proportion of employees, which is 51 total, who do not have a college degree.
To construct a confidence interval for the proportion of employees who do not have a college degree, we can follow these steps:
1. Calculate the sample proportion: Count the number of employees in the sample who do not have a college degree (let's call it "x") and divide it by the total sample size (n). In this case, x = 51 and n = 100. So the sample proportion is: p̂ = x / n.
2. Calculate the margin of error: The margin of error indicates the range within which the true population proportion is likely to fall. It depends on the confidence level and the sample size. For a 90% confidence interval, we need to find the critical value associated with the desired confidence level. Since we have a large sample size (n > 30), we can use the Z-distribution. The critical value for a 90% confidence level is approximately 1.645. The formula to calculate the margin of error is: ME = Z * sqrt(p̂ * (1-p̂) / n).
3. Calculate the lower and upper bounds: Subtract the margin of error from the sample proportion to get the lower bound, and add it to the sample proportion to get the upper bound. The lower bound is p̂ - ME, and the upper bound is p̂ + ME.
4. Calculate the confidence interval: Combine the lower and upper bounds to form the confidence interval. The confidence interval can be expressed as: [lower bound, upper bound].
For your specific example:
1. Sample proportion (p̂) = 51 / 100 = 0.51
2. Margin of error (ME) = 1.645 * sqrt(0.51 * (1-0.51) / 100) ≈ 0.069
3. Lower bound = 0.51 - 0.069 ≈ 0.441
Upper bound = 0.51 + 0.069 ≈ 0.579
4. Confidence interval = [0.441, 0.579]
Therefore, we can say with 90% confidence that the proportion of employees who do not have a college degree is between 0.441 and 0.579.