Question

Sample Size. In Exercises 29–36, find the sample size required to estimate the population mean.

Mean IQ of College Professors the Wechsler IQ test is designed so that the mean is 100 and the standard deviation is 15 for the population of normal adults. Find the sample size necessary to estimate the mean IQ score of college professors. We want to be 99% confident that our sample mean is within 4 IQ points of the true mean. The mean for this population is clearly greater than 100. The standard deviation for this population is less than 15 because it is a group with less variation than a group randomly selected from the general population; therefore, if we use$\sigma =15$ we are being conservative by using a value that will make the sample size at least as large as necessary. Assume then that $\sigma =15$and determine the required sample size. Does the sample size appear to be practical?

Short answer

The required sample size is 94.

The sample size does not appear to be very practical.

Step-by-step solution

Given information

The mean of the test $\mu =100$and the standard deviation is $\sigma =15$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.

Describe the determination of sample size

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

$n={\left[\frac{z_{\frac{\alpha }{2}}\times \sigma }{E}\right]}^2 ...\left(1\right)$

Here, E is the margin of error.

Find the critical value zα2

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

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

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

Mathematically,

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

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 $z_{\frac{\alpha }{2}}$implies is 2.575.

Find the required sample size

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

$$\begin{gathered} \text{n}={\left[\frac{z_{\frac{\alpha }{2}}\times \sigma }{\text{E}}\right]}^2 \\ ={\left[\frac{2.575\times 15}{4}\right]}^2 \\ =93.24 \\ =94 \left(\text{rounded} \text{up}\right) \end{gathered}$$

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

 Step 5: Conclude that the sample size appears to be practical

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.