Question

According to the Center for Disease Control and Prevention (CDC), the mean life expectancy in 2015 for Hispanic females was 84.3 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 females. Determine the interval centered at the population mean y 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

Answer 1

Solution :

Given that,

Point estimate = sample mean = $\bar x$ = 84.3

Population standard deviation =   $\sigma$ = 15

Sample size = n = 100

At 95% confidence level

$\alpha$ = 1 - 95%  

$\alpha$ = 1 - 0.95 =0.05

$\alpha$/2 = 0.025

Z$\alpha$/2 = Z_{0.025 =} 1.96

Margin of error = E = Z$\alpha$/2 * ( $\sigma$ /$\sqrt$n)

= 1.96 * ( 15 /  $\sqrt$100 )

= 2.9

At 95% confidence interval estimate of the population mean is,

$\bar x$  ± E

84.3 ± 2.9

( 81.4, 87.2 )

lower limit = 81.4 years

upper limit = 87.2 years