Question #169495

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?


1
Expert's answer
2021-03-18T15:44:42-0400

The critical value for α=0.01\alpha=0.01 is zc=z1α/2=2.576.z_c=z_{1-\alpha/2}=2.576.

The corresponding confidence interval is computed as shown below:


CI(proportion)CI(proportion)

=(p^zcp^(1p^)n,p^+zcp^(1p^)n)=(\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


zcp^(1p^)n0.03z_c\sqrt{\dfrac{\hat{p}(1-\hat{p})}{n}}\leq0.03

Then


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

nzc2p^(1p^)(0.03)2n\geq\dfrac{z_c^2\hat{p}(1-\hat{p})}{(0.03)^2}

p^=0.5\hat{p}=0.5


n(2.576)2(0.5)(10.5)(0.03)2n\geq\dfrac{(2.576)^2(0.5)(1-0.5)}{(0.03)^2}

n1844n\geq1844

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