Question

According to the Center for Disease Control and Prevention (CDC), the mean life expectancy in 2015 for Hispanic males was 79.3 years. Assume that the standard deviation was 15 years, as suggested by the Bureau of Economic Research. The distribution of age at death, X, is not normal because it is skewed to the left. Nevertheless, the distribution of the mean, x, in all possible samples of size n is approximately normal if n is large enough, by the central limit theorem. Let x be the mean life expectancy in a sample of 100 Hispanic males. Determine the interval centered at the population mean μ such that 95% of sample means x will fall in the interval. Give your answers precise to one decimal. You may need to use software or a table of z-critical values. lower limit years upper limit years Murphy. Sherry L, et. al. Deaths: Final Data for 2015, National Vital Statisrics Reports, Volume 66, Number 6. November 27, 2017 https://www.ode.gownchsu/datamvsrinvsr66/nvsr66 06.pdf. Accessed 24 April 2018. Ryan D. Edwards, The Cost of Uncertain Life Span, NBER Working Paper No 14093 December, 2011. hup//www.nber.org/papensw 14093. Accessed 24 April 2018.

Answer 1

a 95% confince interval for population mean $\mu$ is

$=(\bar{x}-\frac{\sigma }{\sqrt{n}}*z_{0.025},\bar{x}+\frac{\sigma }{\sqrt{n}}*z_{0.025})$

$=(79.3-\frac{15}{\sqrt{100}}*z_{0.025},79.3+\frac{15}{\sqrt{100}}*z_{0.025})$

$=(79.3-\frac{15}{\sqrt{100}}*1.95996,79.3+\frac{15}{\sqrt{100}}*1.95996)$

Lower limit = 76.3601 years

Upper Limit = 82.2399 years