3.3 Sample size

Size sample

Very frequently asked question in statistical consulting is, how large should the sample size be to estimate a mean?

The answer will depend on the variation associated with the random variable under observation. The statistician could correctly respond, only one item is needed, provided that the standard deviation of the distribution is zero. That is, if $\sigma$ is equal zero, then the value of that one item would necessarily equal the unknown mean of the distribution. This is the extreme case and one that is not met in practice. However, the smaller the variance, the smaller the sample size needed to achieve a given degree of accuracy.

Most likely, the person involved in this project would find this a more reasonable sample size. Of course, any sample size greater than 93 could be used. Then either the length of the confidence interval could be decreased from that of $\overline{x}\pm 2$ or the confidence coefficient could be increased from 80% or a combination of both. Also, since there might be some question of whether the standard deviation $\sigma$ actually equals 15, the sample standard deviations would no doubt be used in the construction of the interval.

For example , suppose that the sample characteristics observed are $$n=145,\overline{x}=77.2,s=13.2;$$ then, $\overline{x}\pm \frac{1.282s}{\sqrt{n}}$ or $77.2\pm 1.41$ provides an approximate 80% confidence interval for $\mu$ .

In general, if we want the $100\left(1-\alpha \right)\%$ confidence interval for $\mu$ , $\overline{x}\pm z_{\alpha /2}\left(\sigma /\sqrt{n}\right)$ , to be no longer than that given by $\overline{x}\pm \epsilon$ , the sample size n is the solution of $\epsilon =\frac{z_{\alpha /2}\sigma }{\sqrt{n}},$ where $\Phi \left(z_{\alpha /2}\right)=1-\frac{\alpha }{2}.$

That is, $$n=\frac{z_{\alpha /2}^2{\sigma }^2}{{\epsilon }^2},$$ where it is assumed that ${\sigma }^2$ is known.

Sometimes $$\epsilon =z_{\alpha /2}\sigma /\sqrt{n}$$ is called the maximum error of the estimate . If the experimenter has no ideas about the value of ${\sigma }^2$ , it may be necessary to first take a preliminary sample to estimate ${\sigma }^2$ .

The type of statistic we see most often in newspaper and magazines is an estimate of a proportion p . We might, for example, want to know the percentage of the labor force that is unemployed or the percentage of voters favoring a certain candidate. Sometimes extremely important decisions are made on the basis of these estimates. If this is the case, we would most certainly desire short confidence intervals for p with large confidence coefficients. We recognize that these conditions will require a large sample size. On the other hand, if the fraction p being estimated is not too important, an estimate associated with a longer confidence interval with a smaller confidence coefficients is satisfactory; and thus a smaller sample size can be used.

In general , to find the required sample size to estimate p , recall that the point estimate of p is $$\widehat{p}=z_{\alpha /2}\sqrt{\frac{\widehat{p}\left(1-\widehat{p}\right)}{n}}.$$

Suppose we want an estimate of p that is within $\epsilon$ of the unknown p with $100\left(1-\alpha \right)\%$ confidence where $\epsilon =z_{\alpha /2}\sqrt{\widehat{p}\left(1-\widehat{p}\right)/n}$ is the maximum error of the point estimate $\widehat{p}=y/n$ . Since $\widehat{p}$ is unknown before the experiment is run, we cannot use the value of $\widehat{p}$ in our determination of n . However, if it is known that p is about equal to $p^{*}$ , the necessary sample size n is the solution of $$\epsilon =\frac{z_{\alpha /2}\sqrt{p^{\ast }\left(1-p^{\ast }\right)}}{\sqrt{n}}.$$ That is, $$n=\frac{z_{\alpha /2}^2{p}^{\ast }\left(1-p^{\ast }\right)}{{\epsilon }^2}.$$