A randomly selected sample of n =51 men in Brazil had an average lifespan of 59 years. The standard deviation was 10 years. Calculate a 98% confidence interval for the average lifespan for all men in Brazil.
A) (55.6, 62.4)
B) (56.2, 61.8)
C) (35.0, 83.0)
D) None of the above

Question

Quizplus · Accepted Answer

The correct answer is A because the 98% confidence interval is (55.6, 62.4). This interval is calculated using the formula:CI = 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 = 59, σ = 10, n = 51, and z = 2.326 (for a 98% confidence level). Plugging these values into the formula, we get:CI = 59 ± 2.326 * (10 / √51)CI = 59 ± 2.326 * (10 / 7.14)CI = 59 ± 2.326 * 1.4CI = 59 ± 3.25CI = (55.6, 62.4)

A Randomly Selected Sample of N =51 Men in Brazil