Question

<sub>Near the city of Santo Domingo in Ecuador, there is a population indigenous of extreme poverty. The volunteer doctors who help these people suspect that this population suffers from diseases due to leukopenia (decreased white blood cells). The last 10 patients showed white blood cell levels of 4.64, 4.34, 5.32, 4.99, 3.98, 4.53, 4.82, 6.29, 4.35 and 4.12 (/ mm8).
1. If historical information assumes a standard deviation of 1.0 / mm8.Build a 95% confidence interval for the expected value of the current white blood cell levels in this population.
2. Explain in your own words what this confidence interval means?
3. To impact the health of the population, doctors recommend providing foods like beets, lettuce and carrots given that they help to rise white blood cells. If low levels are classified for values ​​less than 4.0 / mm8. Is it a good suggestion from doctors to modify the diet or can be a waste of money and effort?</sub>

Answer 1

1)
sample mean, xbar = 4.738
sample standard deviation, σ = 1
sample size, n = 10

Given CI level is 95%, hence α = 1 - 0.95 = 0.05
α/2 = 0.05/2 = 0.025, Zc = Z(α/2) = 1.96

ME = zc * σ/sqrt(n)
ME = 1.96 * 1/sqrt(10)
ME = 0.62

CI = (xbar - Zc * s/sqrt(n) , xbar + Zc * s/sqrt(n))
CI = (4.738 - 1.96 * 1/sqrt(10) , 4.738 + 1.96 * 1/sqrt(10))
CI = (4.12 , 5.36)

2)

we are 95% confident that the current white blood cell levels of this population. is between 4.12 and 5.36

3)

yes, it is a good suggestion from doctors because confidence interval does not contain 4