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 44 people, the mean body mass index (bmi) was 27.8 and the standard deviation was 6.05. 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, the population is normal, and σ is unknown. b. use a t - distribution because the sample is random, n≥30, and σ is unknown. c. use a normal distribution because the sample is random, the population is normal, and σ is known. d. use a normal distribution because the sample is random, n≥30, and σ is known. 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:

Step1: Check sample size and population standard deviation

We are given a sample size \(n = 44\) (\(n\geq30\)) and the sample standard deviation \(s = 6.05\) (population standard deviation \(\sigma\) is unknown).

Step2: Recall distribution - selection rules

When \(n\geq30\) and \(\sigma\) is unknown, we use the \(t -\)distribution. The \(t -\)distribution is used for confidence - interval construction for the population mean when the population standard deviation \(\sigma\) is unknown. For large samples (\(n\geq30\)), the \(t -\)distribution approximates the normal distribution well, but the formula for the confidence interval still uses the \(t -\)value because of the unknown \(\sigma\).

Answer:

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