Question

Consider the distribution of serum cholesterol levels for all 20 to 74 year old males living in the United States. The mean of this population is 211 mg/100 ml, and the standard deviation is 46 mg/100 ml. How large would the samples need to be for 95% of their means to lie within ±5 mg/100 ml of the population mean?

Answer 1

The corresponding critical z score for a 95% confidence interval = 1.96

Mean, $\mu$ = 211 mg/100 ml

Standard deviation, $\sigma$ = 46 mg/100 ml

Let the required sample size be n

Given that the allowable margin of error = 5 mg/100 ml

Margin of error = $z*\frac{\sigma}{\sqrt{n}}$

= $1.96*\frac{46}{\sqrt{n}}$ <= 5

-> n >= 325.15

Thus, the sample of size 326 or higher is needed to be 95% of their means to lie within (+-)5 mg/100 ml of the population mean