Answer to Question #182236 in Statistics and Probability for luna

Question

The dean of a particular college is wanting to find out the proportion of students who are

interested to enroll in online classes. He made a pre-survey and learned that a proportion of 30%

were for the program. How many students will he interview if he is 90% confident and that he

considers a margin of error of 4% in conducting the study?

Answer 1

Suppose $\hat{p}=0.3$ is a sample proportion serving as a point estimate for a population proportion.

The critical value for $\alpha=0.1$ is  $z_c=z_{\alpha/2}=1.6449$

The margin of error in this case can be given by the formula

$$E=z_{\alpha/2}\sqrt\dfrac{\hat{p}(1-\hat{p})}{n}$$

$$n_{min}=\dfrac{z_{\alpha/2}^2\hat{p}(1-\hat{p})}{E_{max}^2}$$

Substitute

$$n_{min}=\dfrac{(1.6445)^2(0.3)(1-0.3))}{(0.04)^2}$$

$$n_{min}=355$$

He will interview 355 students.