Question

According to the Center for Disease Control and Prevention (CDC), the mean life expectancy in 2015 for non‑Hispanic white males was 76.3 years. Assume that the standard deviation was 15 years, as suggested by the Bureau of Economic Research.The distribution of age at death, ????, is not normal because it is skewed to the left. Nevertheless, the distribution of the mean, x⎯⎯⎯, in all possible samples of size ???? is approximately normal if ???? is large enough, by the central limit theorem.Let x⎯⎯⎯ be the mean life expectancy in a sample of 100 non‑Hispanic white males. Determine the interval centered at the population mean ???? such that 95% of sample means x⎯⎯⎯ will fall in the interval. Give your answers precise to one decimal.

Answer 1

Answer:

The 95% confidence interval for the mean life expectancy of non-hispanic white males is (73.3 years, 79.3 years).

Step-by-step explanation:

This is the 95% confidence interval for the mean life expectancy of non-hispanic white males.

Our sample size is 100

The first step to solve this problem is finding our degrees of freedom, that is, the sample size subtracted by 1. So

$df = 100-1 = 99$

Then, we need to subtract one by the confidence level $\alpha$ and divide by 2. So:

$\frac{1-0.95}{2} = \frac{0.05}{2} = 0.025$

Now, we need our answers from both steps above to find a value T in the t-distribution table. So, with 99 and 0.025 in the two-sided t-distribution table, we have $T = 1.984$

Now, we find the standard deviation of the sample. This is the division of the standard deviation by the square root of the sample size. So

$s = \frac{15}{\sqrt{100}} = 1.5$

Now, we multiply T and s

$M = T*s = 1.984*1.5 = 2.976$

Then

The lower end of the confidence interval is the mean subtracted by M. So:

$L = 76.3 - 2.976 = 73.3$

The upper end of the confidence interval is the mean added to M. So:

$L = 76.3 + 2.976 = 79.3$

The 95% confidence interval for the mean life expectancy of non-hispanic white males is (73.3 years, 79.3 years).