A random sample of size 30 from a normal population yields = 32.8 with a population standard deviation of 4.51.Construct a 95 percent confidence interval for μ.
A) [23.96,41.64]
B) [32.04,33.56]
C) [31.45,34.15]
D) [31.19,34.41]

Question

Quizplus · Accepted Answer

The correct answer is D. The formula for a 95% confidence interval for the mean of a normal population is:x̄ ± z * (σ / √n)where x̄ is the sample mean, σ is the population standard deviation, n is the sample size, and z is the z-score corresponding to the desired confidence level.In this case, x̄ = 32.8, σ = 4.51, n = 30, and z = 1.96 (for a 95% confidence level).Plugging these values into the formula, we get:32.8 ± 1.96 * (4.51 / √30)= 32.8 ± 1.96 * (4.51 / 5.48)= 32.8 ± 1.61= [31.19, 34.41]Therefore, the 95% confidence interval for the mean of the population is [31.19, 34.41].

A Random Sample of Size 30 from a Normal Population