Question 3. a. Suppose you want to estimate the average concentration of vitamin

Question

Question 3. a. Suppose you want to estimate the average concentration of vitamin D in the blood stream of individuals within a given population. If you want to calculate a 95% confidence interval with a margin of error of +-2.5 nM/L, what sample size would you recommend? Assume for this exercise that the population standard deviation is 15.0 nM/L. b. To assess whether health promotion is called for, the government would like to know what proportion of people have low levels of vitamin D concentration. This population proportion is not known. To estimate this population proportion with a margin of error of +-0.02, what sample size would you recommend for 90% confidence interval or one with a CC of 90%?

Explanation / Answer

(a) Given a=0.05, Z(0.025) = 1.96 (from standard normal table)

So n=(Z*s/E)^2

= (1.96*15/2.5)^2

=138.2976

Take n=139

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(b) Given a=0.1, Z(0.05)= 1.645 (from standard normal table)

We use p=0.5 as estimated.

So n=(Z/E)^2*p*(1-p)

=(1.645/0.02)^2*0.5*0.5

=1691.266

Take n=1692

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