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The IQ test ofAttorneys has a standard deviation of $\mathrm{σ}=15$. The level of confidence is 98%, and the sample mean lies within 3 IQ points of the true mean.

The sample size n can be determined by using the following formula.

$n={\left[\frac{z_{\frac{\mathrm{Î\pm }}{2}}×\mathrm{σ}}{E}\right]}^2                          ...\left(1\right)$

Here, E is the margin of error.

The $z_{\frac{\mathrm{Î\pm }}{2}}$is a z-score that separates an area of $\frac{\mathrm{Î\pm }}{2}$in the right tail of the standard normal distribution.

The confidence level of 98% corresponds to $\mathrm{Î\pm }=0.02    \text{and}    \frac{\mathrm{Î\pm }}{2}=0.01$.

The value$z_{\frac{\mathrm{Î\pm }}{2}}$has the cumulative area to its left as $1-\frac{\mathrm{Î\pm }}{2}$.

Mathematically,

$$\begin{gathered} P\left(z<z_{\frac{\mathrm{Î\pm }}{2}}\right)=1-\frac{\mathrm{Î\pm }}{2} \\ =0.99 \end{gathered}$$

From the standard normal table, the area of 0.99 is observed corresponding intersection of the row value 2.3 and column value 0.03, which implies $z_{\frac{\mathrm{Î\pm }}{2}}$is 2.33.

The sample size is calculated by substituting the values of $z_{\frac{\mathrm{Î\pm }}{2}}$,$\mathrm{σ}$, and E in equation (1).

$$\begin{gathered} \text{n}={\left[\frac{z_{\frac{\mathrm{Î\pm }}{2}}×\mathrm{σ}}{\text{E}}\right]}^2 \\ ={\left[\frac{2.33×15}{3}\right]}^2 \\ =135.722 \\ =136   \left(\text{rounded}  \text{up}\right) \end{gathered}$$

Thus, with 136 samples values, you can be 98% confident that your sample mean lies within 3 IQ points of the true mean.

The sample size required to estimate the meanIQ test ofAttorneys is 136. Yes, the sample size does appear to be very practical.