Question

A school wants to know the proportion of students who choose online classes as their mode of learning with 90% confidence. The first survey showed that 75% choose online classes as their mode of learning. The school wants to be accurate within 3% of the true proportion. What is the total number of sample size?

Answer 1

A 90% confidence interval has a z-score (a critical value) $z_c=1.6449.$

$$ME=z_c\times\sqrt{\dfrac{p(1-p)}{n}}$$

$$=1.6449\times\sqrt{\dfrac{0.75(1-0.75)}{n}}=0.03$$

$$n=\dfrac{1.6449^2(0.75)(0.25)}{0.03^2}$$

$$n=564$$