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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:

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

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

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

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

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

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:

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

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

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

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.