Answer to Question #169495 in Statistics and Probability for Sunera

Question

A marketing manager is in the process of deciding whether to introduce a new product. He has concluded that he needs to perform a market survey in which he asks a random sample of people whether they will buy the product. How large a sample should he take if he wants to estimate the proportion of people who will buy the product to within 3%, with 99% confidence?

Answer 1

The critical value for $\alpha=0.01$ is $z_c=z_{1-\alpha/2}=2.576.$

The corresponding confidence interval is computed as shown below:

$$CI(proportion)$$

$$=(\hat{p}-z_c\sqrt{\dfrac{\hat{p}(1-\hat{p})}{n}}, \hat{p}+z_c\sqrt{\dfrac{\hat{p}(1-\hat{p})}{n}})$$

Given

$$z_c\sqrt{\dfrac{\hat{p}(1-\hat{p})}{n}}\leq0.03$$

Then

$$\dfrac{\hat{p}(1-\hat{p})}{n}\leq(\dfrac{0.03}{z_c})^2$$

$$n\geq\dfrac{z_c^2\hat{p}(1-\hat{p})}{(0.03)^2}$$

$\hat{p}=0.5$

$$n\geq\dfrac{(2.576)^2(0.5)(1-0.5)}{(0.03)^2}$$

$$n\geq1844$$