Question

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

Mean Grade-Point Average Assume that all grade-point averages are to be standardized on a scale between 0 and 4. How many grade-point averages must be obtained so that the sample mean is within 0.01 of the population mean? Assume that a 95% confidence level is desired. If we use the range rule of thumb, we can estimate $\sigma$ to be,

$$\begin{gathered} \sigma =\frac{range}{4} \\ =\frac{4-0}{4} \\ =1 \end{gathered}$$

Does the sample size seem practical?

Short answer

The required sample size is 38416. No, it does not seem practical.

Step-by-step solution

Given information

The results of the testGrade-Point Averageare to be standardized on a scale between 0 and 4. Using the range rule of thumb, the estimated value ofis 1.The confidence level is 95% and it implies that the sample mean is within the 0.01 Grade-Point of the true mean.

Describe the determination of sample size

The sample size n for estimating the true population mean 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 95% corresponds to $\alpha =0.05 \text{and} \frac{\alpha }{2}=0.025$.

The value$z_{\frac{\alpha }{2}}$has the cumulative area of $1-\frac{\alpha }{2}$.

Mathematically,

$$\begin{gathered} P\left(z<z_{\frac{\alpha }{2}}\right)=1-\frac{\alpha }{2} \\ =0.975 \end{gathered}$$

From the standard normal table, the area of 0.99 is observed corresponding to the row value 1.9 and column value 0.06, which implies $z_{\frac{\alpha }{2}}$is 1.96.

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).

$$\begin{gathered} \text{n}={\left[\frac{z_{\frac{\alpha }{2}}\times \sigma }{\text{E}}\right]}^2 \\ ={\left[\frac{1.96\times 1}{0.01}\right]}^2 \\ =38416 \end{gathered}$$

Thus, with 38416 samples values, you can be95% confident that your sample mean lies within 0.01 Grade-points of the true mean.

Conclude that the sample size appears to be practical

The sample size required to estimate the Mean Grade-Point Averageis 38416. The estimated sample size is too large; so the sample size does not seem practical.