B. An advertising executive thinks that the proportion of consumer who has?

seen his company’s advertising in news papers is 0.65 .The executive wants to estimate the consumer population proportion to with in ±0.05 and have 98 percent confidence in the estimate. How large a sample should be taken?

B. An advertising executive thinks that the proportion of consumer who has?
n = [z(sub α/2)/E]^2[p(1-p)]


Where z(sub α/2) = 2.33 for α/2


E is your accepted error = ±0.05


Estimate of proportion(binomial) standard deviation = 0.65(1-0.65)


Plugging all this in:


n = [2.33/0.05]^2[0.65(1-0.65)] = 492