Suppose you wish to estimate a population mean correct to within .20 with probability equal to .90. You do not know ${\mathbf{\sigma }}^{\mathbf{2}}$, but you know that the observations will range in value between 30 and 34.

a. Find the approximate sample size that will produce the desired accuracy of the estimate. You wish to be conservative to ensure that the sample size will be ample to achieve the desired accuracy of the estimate. [Hint: Using your knowledge of data variation from Section 2.6, assume that the range of the observations will equal 4$\mathbf{\sigma }$.]

b. Calculate the approximate sample size, making the less conservative assumption that the range of the observations is equal to $\mathbf{6}\mathbf{\sigma }$.

The standard error is 0.20 with a probability equal to 0.90

Here, ${\mathrm{\sigma }}^2$is not known, but the observations will range in value between 30 and 34.

a.

The range of the observations will equal $4\mathrm{\sigma }$

$\mathrm{Range}\approx 4\mathrm{\sigma }$

$\begin{array}{rcl}\mathrm{Range}& =& \mathrm{Maximum}\mathrm{value}-\mathrm{Minimum}\mathrm{value}\\ 4\mathrm{\sigma }& =& 34-30\\ \mathrm{\sigma }& =& \frac{4}{4}\\ \mathrm{\sigma }& =& 1\end{array}$

The sample size is,

$\begin{array}{rcl}{\mathrm{z}}_{\raisebox{1ex}{$\mathrm{\alpha }$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\left(\frac{\mathrm{\sigma }}{\sqrt{\mathrm{n}}}\right)& =& \mathrm{SE}\\ \mathrm{n}& =& {\left[\frac{\left({\mathrm{z}}_{{\displaystyle \raisebox{1ex}{$\mathrm{\alpha }$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}\right)\mathrm{\sigma }}{\mathrm{SE}}\right]}^2\\ \mathrm{n}& =& {\left[\frac{{\mathrm{z}}_{0.05}\times 1}{0.2}\right]}^2\\ \mathrm{n}& =& {\left[\frac{1.645\times 1}{0.2}\right]}^2\left[\mathrm{From}\mathrm{Standard}\mathrm{Normal}\mathrm{Table}\right]\\ \mathrm{n}& =& \frac{2.706025}{0.04}\\ \mathrm{n}& =& 67.650635\\ n& \approx & 68\end{array}$

Hence, the approximate sample size that will produce the desired accuracy of the estimate is 68.

b.

The range of the observations will equal $6\mathrm{\sigma }$

$\mathrm{Range}\approx 6\mathrm{\sigma }$

$\begin{array}{rcl}\mathrm{Range}& =& \mathrm{Maximum}\mathrm{value}-\mathrm{Minimum}\mathrm{value}\\ 6\mathrm{\sigma }& =& 34-30\\ \mathrm{\sigma }& =& \frac{4}{6}\\ \mathrm{\sigma }& =& 0.6667\end{array}$

The sample size is,

$\begin{array}{rcl}{\mathrm{z}}_{\raisebox{1ex}{$\mathrm{\alpha }$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\left(\frac{\mathrm{\sigma }}{\sqrt{\mathrm{n}}}\right)& =& \mathrm{SE}\\ \mathrm{n}& =& {\left[\frac{\left({\mathrm{z}}_{{\displaystyle \raisebox{1ex}{$\mathrm{\alpha }$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}\right)\mathrm{\sigma }}{\mathrm{SE}}\right]}^2\\ \mathrm{n}& =& {\left[\frac{{\mathrm{z}}_{0.05}\times 0.6667}{0.2}\right]}^2\\ \mathrm{n}& =& {\left[\frac{1.645\times 0.6667}{0.2}\right]}^2\left[\mathrm{From}\mathrm{Standard}\mathrm{Normal}\mathrm{Table}\right]\\ \mathrm{n}& =& \frac{1.2028}{0.04}\\ \mathrm{n}& =& 30.07\\ \mathrm{n}& \approx & 31\end{array}$

Hence, the approximate sample size that will produce the desired accuracy of the estimate is 31.