Question

Assume that a sample is used to estimate a population mean μ . Find the 98% confidence interval for a sample of size 809 with a mean of 23.3 and a standard deviation of 12.5. Enter your answer as a tri-linear inequality accurate to one decimal place (because the sample statistics are reported accurate to one decimal place).

Answer 1

To find the 98% confidence interval, we can use the formula: CI = x ± z * (σ / √n). Given the sample mean (x) of 23.3, the standard deviation (σ) of 12.5, and the sample size (n) of 809, the confidence interval is (22.4, 24.2).

Step-by-step explanation:

To find the 98% confidence interval, we can use the formula:

CI = x ± z * (σ / √n)

Where:

• CI is the confidence interval
• x is the sample mean
• z is the z-score corresponding to the desired level of confidence
• σ is the population standard deviation
• n is the sample size

Given that the sample mean (x) is 23.3, the standard deviation (σ) is 12.5, and the sample size (n) is 809, we can find the z-score for a 98% confidence level using a standard normal distribution table or a calculator. The z-score for a 98% confidence level is approximately 2.33.

Plugging in the values, the confidence interval is:

(23.3 - 2.33 * (12.5 / √809), 23.3 + 2.33 * (12.5 / √809))

Simplifying the expression, we get:

(22.4, 24.2)