Determining Sample Size. In Exercises 31–38, use the given data to find the minimum sample size required to estimate a population proportion or percentage.

Bachelor’s Degree in Four Years

In a study of government financial aid for college students, it becomes necessary to estimate the percentage of full-time college students who earn a bachelor’s degree in four years or less. Find the sample size needed to estimate that percentage. Use a 0.05 margin of error, and use a confidence level of 95%.

a. Assume that nothing is known about the percentage to be estimated.

b. Assume that prior studies have shown that about 40% of full-time students earn bachelor’s degrees in four years or less.

c. Does the added knowledge in part (b) have much of an effect on the sample size?

The percentage of full-time college students who earn a bachelor’s degree in four years or less is to be estimated.

The sample size needs to be determined. The following values are given:

The margin of error is equal to 0.05.

The confidence level is equal to 95%.

a.

Let $\hat{p}$ denote the sample proportion of full-time college students who earn a bachelor’s degree in four years or less.

Let $\hat{q}$ denote the sample proportion of full-time college students who do not earn a bachelor’s degree in four years or less.

Here, nothing is known about the sample proportions.

The formula for finding the sample size is as follows:

$n=\frac{{\left[z_{\frac{\alpha }{2}}\right]}^{2}0.25}{E^2}$

The confidence level is equal to 95%. Thus, the level of significance is equal to 0.05.

The value of $z_{\frac{\alpha }{2}}$ for $\alpha =0.05$ from the standard normal table is equal to 1.96

Substituting the required values, the following value of the sample size is obtained:

$$\begin{gathered} n=\frac{{\left(1.96\right)}^2\times 0.25}{{\left(0.05\right)}^2} \\ =384.16 \\ \approx 384 \end{gathered}$$

Hence, the required sample size is equal to 384.

b.

The value of $\hat{p}$is given to be equal to:

$$\begin{gathered} \hat{p}=40\% \\ =\frac{40}{100} \\ =0.40 \end{gathered}$$

Thus, the value of is computed below:

$$\begin{gathered} \hat{q}=1-\hat{p} \\ =1-0.40 \\ =0.60 \end{gathered}$$

The formula for finding the sample size is as follows:

$n=\frac{{\left[z_{\frac{\alpha }{2}}\right]}^2\hat{p}\hat{q}}{E^2}$

Substituting the required values, the following value of the sample size is obtained:

$$\begin{gathered} n=\frac{{\left(1.96\right)}^2\times 0.40\times 0.60}{{\left(0.05\right)}^2} \\ =368.79 \\ \approx 369 \end{gathered}$$

Hence, the required sample size is equal to 369.

c.

Since the two sample sizes are approximately equal, the knowledge of the sample proportions does not have a significant effect on the sample size.