90% confidence interval

Question

The distribution of blood cholesterol level in the population of young men aged 20 to 34 years is close to Normal with standard deviation 𝜎 = 45 milligrams per deciliter (mg/dL). You measure the blood cholesterol of 13 cross-country runners. The mean level is x = 174 mg/dL. Assume that 𝜎 is the same as in the general population.

How large a sample is needed to cut the margin-of-error in half for a 90% confidence interval of the mean level 𝜇 among the cross-country runners?

Answer 1

1. Figure out the margin of error provided by the current sample:
2. Assuming a standard Normal distribution, the critical value of a 90% CI is z = 1.645
3. Margin of Error = Z_crit*(σ/√n) = 1.645*(45/√13) = 20.53
4. The target margin of error is half of the current MOE: 20.53/2 = 10.265
5. Set the target value equal to the MOE formula, then solve for n.
6. Target MOE = Z_crit*(σ/√n)
7. 10.265 = 1.645*(45/√n)
8. 6.240 = (45/√n)
9. 6.24(√n) = 45
10. √n = (45/6.24)
11. √n = 7.212
12. n = (7.212)² ≈ 52.01
13. You need at least 52.01 people to achieve the desired margin of error, so let's say n = 53 is what you want. My calculations are subject to rounding error, so your results may vary slightly.