91影视

The mean of the test $渭=100$and the standard deviation is $蟽=15$for the population of adults.

The required confidence level is 99% for the sample mean to lie within 4 IQ points of the true mean.

The sample size n for estimating the true population mean value can be determined by using the following formula.

$n={\left[\frac{z_{\frac{伪}{2}}脳蟽}{E}\right]}^2鈥�鈥�鈥�鈥�鈥�鈥�鈥�鈥�鈥�鈥�鈥�鈥�鈥�鈥�鈥�鈥�鈥�鈥�鈥�鈥�鈥�鈥�鈥�鈥�鈥�鈥�...\left(1\right)$

Here, E is the margin of error.

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

The confidence level of 99% corresponds to $伪=0.01鈥�鈥�鈥�鈥�\text{and}鈥�鈥�鈥�鈥�\frac{伪}{2}=0.005$.

To value$z_{\frac{伪}{2}}$hasthe cumulative area $1-\frac{伪}{2}$to the left.

Mathematically,

$$\begin{gathered} P\left(z<z_{\frac{伪}{2}}\right)=1-\frac{伪}{2} \\ =0.995 \end{gathered}$$

From the standard normal table, the area of 0.995 is observed corresponding to the row value 12.5 and between column values 0.07 and column value 0.08, which $z_{\frac{伪}{2}}$implies is 2.575.

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

$$\begin{gathered} \text{n}={\left[\frac{z_{\frac{伪}{2}}脳蟽}{\text{E}}\right]}^2 \\ ={\left[\frac{2.575脳15}{4}\right]}^2 \\ =93.24 \\ =94鈥�鈥�鈥�\left(\text{rounded}鈥�鈥�\text{up}\right) \end{gathered}$$

Thus, with 94 samples values, we can be 99% confident that the sample mean lies within 4 IQ points of the true mean.

The sample size required to estimate the mean IQ score of college professors is 94. The estimated sample size is neither too large nor too small. So, the sample size does appear to be very practical.