Question 317029
A randomly selected sample of n = 51 men in Brazil had an average lifespan of 59 years. The standard deviation was 10 years, and the standard error of the mean is 1.400. Calculate a 90% confidence interval for the average lifespan for all men in Brazil.
--------------
x-bar = 59
ME = (invT(0.05 with df = 50)*SE
ME = (1.6759)1.4 = 2.3463
---
90% CI: 59-2.3463 < u < 59+2.3463
90% CI: 56.65 < u < 61.35
==================================
Cheers,
Stan H.
===============================



1) (56.6, 61.4)
2)(42.2, 75.8)
3) (56.2, 61.8)
Kristal