Question

A simple random sample with n=50 provided a sample mean of 23.0 and a sample standard deviation of 4.4 .

a. Develop a 90% confidence interval for the population mean (to 1 decimal).
b. Develop a 95% confidence interval for the population mean (to 1 decimal).
c. Develop a 99% confidence interval for the population mean (to 1 decimal).
d. What happens to the margin of error and the confidence interval as the confidence level is increased?

Answer 1

a. The 90% confidence interval is (21.3, 24.7). b. The 95% confidence interval is (20.9, 25.1). c. The 99% confidence interval is (19.9, 26.1). d. As the confidence level increases, the margin of error and the confidence interval also increase.

Step-by-step explanation:

a. To calculate the 90% confidence interval, we use the formula: CI = x ± Z * (σ/√n), where x is the sample mean, Z is the z-score for the confidence level, σ is the sample standard deviation, and n is the sample size. Plugging in the values, we get: CI = 23 ± 1.645 * (4.4/√50). Simplifying, we get a 90% confidence interval of (21.3, 24.7).

b. For a 95% confidence interval, we use the same formula but with a different z-score. Plugging in the values, we get: CI = 23 ± 1.96 * (4.4/√50). Simplifying, we get a 95% confidence interval of (20.9, 25.1).

c. For a 99% confidence interval, we use a different z-score. Plugging in the values, we get: CI = 23 ± 2.576 * (4.4/√50). Simplifying, we get a 99% confidence interval of (19.9, 26.1).

d. As the confidence level is increased, the margin of error and the confidence interval both become wider. This is because a higher confidence level requires a larger z-score, which increases the margin of error.