Question

use the standard normal distribution or the t - distribution to construct a 90% confidence interval for the population mean. justify your decision. if neither distribution can be used, explain why. interpret the results.
in a random sample of 40 people, the mean body mass index (bmi) was 28.5 and the standard deviation was 6.21.
which distribution should be used to construct the confidence interval? choose the correct answer below.
○ a. use a t - distribution because the sample is random, n ≥ 30, and σ is unknown.
○ b. use a normal distribution because the sample is random, the population is normal, and σ is known.
○ c. use a normal distribution because the sample is random, n ≥ 30, and σ is known.
○ d. use a t - distribution because the sample is random, the population is normal, and σ is unknown.
○ e. neither a normal distribution nor a t - distribution can be used because either the sample is not random, or n < 30, and the population is not known to be normal.

Explanation:

• We are given a sample size \(n = 40\) (so \(n\geq30\)), the sample mean \(\bar{x}=28.5\), and the sample standard deviation \(s = 6.21\) (population standard deviation \(\sigma\) is unknown).
• For constructing a confidence interval for the population mean:
• If the population standard deviation \(\sigma\) is known, we use the standard - normal (\(z\)) distribution.
• If the population standard deviation \(\sigma\) is unknown, and either the population is normal or \(n\geq30\) (by the Central Limit Theorem, the sampling distribution of the sample mean \(\bar{X}\) is approximately normal), we use the \(t\) - distribution.

Answer:

A. Use a t - distribution because the sample is random, \(n\geq30\), and \(\sigma\) is unknown.