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): $4$


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


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 $n = 189$

I have no idea how this is right. Please help me.


1 Answer 1


When the length of the confidence interval is 4, the value of E is 2.

  • $\begingroup$ ah so when they says length of confidence interval, it's -2 +2 from the test statistic? Thus a length of 4? So the overall error estimate is 2 for this reason? $\endgroup$
    – user848500
    Commented Nov 12, 2020 at 2:53
  • $\begingroup$ 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 $2\cdot z_{\alpha /2}\frac{\sigma}{\sqrt{n}}$. $\endgroup$ Commented Nov 12, 2020 at 15:11

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