Question

If you wish to estimate a population mean with a sampling error of SE = .3 using a 95% confidence interval, and you know from prior sampling that ${\mathbf{\sigma }}^{\mathbf{2}}$is approximately equal to 7.2, how many observations would have to be included in your sample?

Short answer

Sample of size n=18 observations Is required.

Step-by-step solution

Given information

The sampling error is 0.3, level of significance is 5% and ${\mathrm{\sigma }}^2=7.2$.

Finding the sample size

$\mathrm{SE}=0.3,\mathrm{\alpha }=0.05,\mathrm{and}{\mathrm{\sigma }}^2=7.2$

The standard deviation is,

$\begin{array}{rcl}\mathrm{\sigma }& =& \sqrt{7.2}\\ & =& 2.6833\end{array}$

The sample size required obtained by,

$\begin{array}{rcl}{\mathrm{z}}_{\raisebox{1ex}{$\mathrm{\alpha }$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\left(\frac{\mathrm{\sigma }}{\sqrt{\mathrm{n}}}\right)& =& \mathrm{SE}\\ \mathrm{n}& =& {\left[\frac{\left({\mathrm{z}}_{\raisebox{1ex}{$\mathrm{\alpha }$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\right)\mathrm{\sigma }}{\mathrm{SE}}\right]}^2\\ \mathrm{n}& =& \left[\frac{{\mathrm{z}}_{0.025}\times 2.6833}{0.3}\right]\\ \mathrm{n}& =& \left[\frac{1.960\times 2.6833}{0.3}\right]\left[\mathrm{From}\mathrm{Standard}\mathrm{Normal}\mathrm{Table}\right]\\ \mathrm{n}& =& \frac{5.2593}{0.3}\\ \mathrm{n}& =& 17.531\\ \mathrm{n}& \approx & 18\end{array}$

Therefore, 18 observations have to be included in the sample.