# Find the required sample size for the investigator.

An investigator interested in estimating a population mean wants to be 95% certain that the length of the confidence interval does not exceed 4. Find the required sample size for his study if the population standard deviation is 14.

My work:

Confidence interval: $$95$$%

Error estimate(E): $$4$$

$$\sigma=14$$

$$\alpha=0.05$$ so $$\alpha/2=0.025$$

$$Z_{0.025}=1.96$$

so using $$n =(\cfrac{Z_{0.025} \cdot \sigma}{E})^2 = (\cfrac{1.96 \cdot 14}{4})^2 = 47.0596$$

The sample size cannot be a decimal so round up to $$48$$ so $$n = 48$$

So he needs a sample size of $$48$$ but the book's answer says it's $$n = 189$$

• The interval estimation is $\bar{x}\pm z_{\alpha /2}\frac{\sigma}{\sqrt{n}}$. The second component $z_{\alpha /2}\frac{\sigma}{\sqrt{n}}$ is called the bound on the error of estimation, which is $E$ in your formula. Now let $z_{\alpha /2}\frac{\sigma}{\sqrt{n}}=E$, you will know why you have the formula for the sample size. The length of the interval estimation is $2\cdot z_{\alpha /2}\frac{\sigma}{\sqrt{n}}$. Commented Nov 12, 2020 at 15:11