Central limit theorem problem with sample size.

Question

Suppose an anthropologist wants the difference between the sample mean and the population mean to be less than 0.4 inches, with a probability of 0.95. How many men would have to select to reach the goal?

I know that:

$$P(|\bar{X}-\mu|<0.4)=0.95$$ $$P\left(\sqrt{n}\frac{-0.4}{2.5}<Z<\sqrt{n}\frac{0.4}{2.5}\right)=P(-0.16\sqrt{n}<Z<0.16\sqrt{n})$$

then:

$$2P(0<Z<0.16\sqrt{n})=0.95$$

But the $n$ that I'm getting from this is very small. There's something wrong with this?

Answer 1

For a 95% probability, you want to compute a 95% confidence interval in which the top and bottom of the interval are 0.4 inches more and less than the mean value. In other words, you want the "margin of error" to be 0.4. I'm assuming from your formula above that the standard deviation of an individual value is 2.5.

Then the margin of error = 0.4 = 1.96 * sd / (sqrt(n)

(1.96 is the z-score corresponding to a two-tailed 95% confidence interval)

Solving (with sd= 2.5) gives n = 150.